Torque on a Current Carrying Loop in a Magnetic Field

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Torque on a Current Carrying Loop in a Magnetic Field

A complete premium Physics study page by Kumar Sir covering force couple, magnetic dipole moment, angle relation, stable and unstable equilibrium, moving coil galvanometer, graphs, derivations and exam-level questions.torque on current loop in magnetic field

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1. Introduction

When a current carrying rectangular loop is placed in a uniform magnetic field, equal and opposite magnetic forces act on opposite sides of the loop. The net force on the loop is zero, but the two forces act along different lines of action. Therefore they form a couple and produce a turning effect called torque.

Force on straight conductorF = BIL sinφ
Torque of a coupleτ = force × perpendicular distance
Magnetic momentm = NIA

2. Rectangular Current Loop Diagram

Rectangular current loop in a uniform magnetic field
Torque on rectangular current loop uniform magnetic field B II F = BIl into page F = BIl out of page area vector A, magnetic moment m α breadth b length l A = l × b; torque is produced by the force couple

The rectangular loop has length l, breadth b, area A = l × b, current I, uniform magnetic field B, area vector A, magnetic moment m, angle α between area vector and magnetic field, and angle θ between the plane of the loop and magnetic field.

3. Magnetic Force on Sides of Loop

For a current carrying side of length l placed perpendicular to the magnetic field, the magnetic force is F = BIl. The two opposite active sides experience equal and opposite forces. These forces do not cancel rotationally because they act along different lines of action. They form a couple.

Equal magnitude: F = BIl
Opposite direction: net force = 0
Different lines of action: net torque ≠ 0
Couple rotates the loop
DerivationCBSE • NEET • JEE

4. Complete Derivation of Torque

Rectangular current loop in a uniform magnetic field
Torque on rectangular current loop uniform magnetic field B II F = BIl into page F = BIl out of page area vector A, magnetic moment m α breadth b length l A = l × b; torque is produced by the force couple
Given / Symbols:
Length = l, breadth = b, area A = lb, current I, magnetic field B, number of turns N, angle α between area vector and B.
Assumptions:
Uniform magnetic field; rectangular coil; active sides perpendicular to B; wires are rigid and current is steady.
1Force on each active vertical side is F = BIl.
2The perpendicular distance between the two equal and opposite forces is d = b sinα.
3Torque for one turn is τ = F × d.
4Substitute F and d: τ = BIl × b sinα.
5Since A = lb, τ = BIA sinα for one turn.
6For N turns, torque becomes τ = NBIA sinα.
7If θ is angle between plane of loop and B, then α = 90° − θ.
8Therefore sinα = cosθ and τ = NBIA cosθ.
Final Result:τ = NBIA sinα; also τ = NBIA cosθ
Common mistake:
Do not write τ = NBIA sinθ when θ is angle between plane and field. Use τ = NBIA cosθ.

5. Vector Form of Torque

Vector form of torque: tau = m cross B
Magnetic dipole moment and torque direction current I m = NIA B α τ = m × B Right-hand thumb rule: fingers curl with current; thumb gives area vector and m.
Magnetic momentm = NIA
Vector torqueτ = m × B
Magnitudeτ = mB sinα = NBIA sinα

The vector m is along the area vector. Its direction is given by the right-hand thumb rule: curl fingers in the direction of current and the thumb gives the direction of the area vector or magnetic moment.

6. Angle Relation

Relation between alpha and theta
Angle relation: α + θ = 90° B plane of loop area vector A α θ Since A is perpendicular to the plane, α = 90° − θ; hence sinα = cosθ.

α is the angle between area vector and magnetic field. θ is the angle between plane of loop and magnetic field. Since the area vector is perpendicular to the plane of the loop, α + θ = 90°. Hence τ = NBIA sinα = NBIA cosθ.

7. Stable and Unstable Equilibrium

Stable, unstable and maximum torque positions
Equilibrium and maximum torque Bm Stable: α = 0°, U minimum, τ = 0 Bm Unstable: α = 180°, U maximum, τ = 0 90° Maximum: α = 90°, τ = NBIA
Stable equilibriumα = 0°, U = -mB, τ = 0
Unstable equilibriumα = 180°, U = +mB, τ = 0
Maximum torqueα = 90°, τmax = NBIA

8. Potential Energy of Current Loop

The potential energy of a magnetic dipole in a uniform magnetic field is the negative dot product of magnetic moment and magnetic field.

Vector formU = -m · B
Scalar formU = -mB cosα
Current loopU = -NBIA cosα

Minimum potential energy occurs at stable equilibrium, and maximum potential energy occurs at unstable equilibrium.

9. Moving Coil Galvanometer Connection

Moving coil galvanometer torque balance
Moving coil galvanometer: magnetic torque = restoring torque NS radial B m At equilibrium: NBIA = kφ, so I = kφ / NBA
Magnetic torqueτ = NBIA
Restoring torqueτ = kφ
EquilibriumNBIA = kφ
CurrentI = kφ / NBA

A radial magnetic field is used so that the plane of the coil remains parallel to the magnetic field and the area vector remains perpendicular to the field. Therefore sinα = 1 and torque remains maximum for all deflections.

10. Applications

Moving coil galvanometer
Electric motor
Ammeter
Voltmeter
Magnetic dipole behaviour
Torque measurement
Electrical measuring instruments
Current loop in magnetic field

11. Common Student Mistakes

Mistake 1Confusing α and θ.

Correction: Draw B, area vector, current direction and angle definition before substituting formula.

Mistake 2Using sinθ instead of cosθ.

Correction: Draw B, area vector, current direction and angle definition before substituting formula.

Mistake 3Forgetting number of turns N.

Correction: Draw B, area vector, current direction and angle definition before substituting formula.

Mistake 4Forgetting area A = lb.

Correction: Draw B, area vector, current direction and angle definition before substituting formula.

Mistake 5Confusing magnetic moment with torque.

Correction: Draw B, area vector, current direction and angle definition before substituting formula.

Mistake 6Wrong direction of area vector.

Correction: Draw B, area vector, current direction and angle definition before substituting formula.

Mistake 7Forgetting right-hand thumb rule.

Correction: Draw B, area vector, current direction and angle definition before substituting formula.

Mistake 8Confusing stable and unstable equilibrium.

Correction: Draw B, area vector, current direction and angle definition before substituting formula.

Mistake 9Using force instead of couple.

Correction: Draw B, area vector, current direction and angle definition before substituting formula.

Mistake 10Forgetting maximum torque condition.

Correction: Draw B, area vector, current direction and angle definition before substituting formula.

12. Exam Question Bank With Solutions

A. CBSE Board Questions

CBSE Theory Question 1Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 2Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 3Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 4Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 5Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 6Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 7Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 8Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 9State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 10Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 11Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 12Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 13Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 14Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 15Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 16Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 17Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 18Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 19State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 20Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 21Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 22Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 23Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 24Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Theory Question 25Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 1Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 2Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 3Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 4Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 5Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 6Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 7Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 8Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 9State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 10Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 11Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 12Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 13Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 14Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 15Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 16Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 17Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 18Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 19State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Derivation Question 20Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 1Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 2Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 3Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 4Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 5Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 6Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 7Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 8Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 9State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 10Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 11Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 12Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 13Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 14Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 15Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 16Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 17Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 18Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 19State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 20Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 21Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 22Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 23Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 24Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
CBSE Numerical Question 25Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
Case Study 1current loop in uniform magnetic field
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 2electric motor
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 3moving coil galvanometer
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 4magnetic dipole
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 5stable equilibrium
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 6unstable equilibrium
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 7torque-angle graph
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 8potential-energy graph
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 9coil with N turns
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 10rectangular loop numerical
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.

B. NEET Questions

NEET MCQ 1A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 2If θ is the angle between plane of coil and magnetic field, torque is
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 3Maximum torque occurs when area vector is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 4Stable equilibrium of current loop occurs when magnetic moment is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 5Potential energy of current loop is
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 6Magnetic dipole moment of a coil is
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 7In radial magnetic field of galvanometer
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 8If current is doubled, torque becomes
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 9If number of turns and area both double, torque becomes
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 10A loop has zero torque but maximum potential energy when α is
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 11A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 12If θ is the angle between plane of coil and magnetic field, torque is
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 13Maximum torque occurs when area vector is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 14Stable equilibrium of current loop occurs when magnetic moment is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 15Potential energy of current loop is
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 16Magnetic dipole moment of a coil is
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 17In radial magnetic field of galvanometer
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 18If current is doubled, torque becomes
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 19If number of turns and area both double, torque becomes
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 20A loop has zero torque but maximum potential energy when α is
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 21A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 22If θ is the angle between plane of coil and magnetic field, torque is
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 23Maximum torque occurs when area vector is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 24Stable equilibrium of current loop occurs when magnetic moment is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 25Potential energy of current loop is
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 26Magnetic dipole moment of a coil is
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 27In radial magnetic field of galvanometer
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 28If current is doubled, torque becomes
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 29If number of turns and area both double, torque becomes
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 30A loop has zero torque but maximum potential energy when α is
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 31A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 32If θ is the angle between plane of coil and magnetic field, torque is
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 33Maximum torque occurs when area vector is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 34Stable equilibrium of current loop occurs when magnetic moment is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 35Potential energy of current loop is
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 36Magnetic dipole moment of a coil is
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 37In radial magnetic field of galvanometer
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 38If current is doubled, torque becomes
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 39If number of turns and area both double, torque becomes
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 40A loop has zero torque but maximum potential energy when α is
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 41A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 42If θ is the angle between plane of coil and magnetic field, torque is
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 43Maximum torque occurs when area vector is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 44Stable equilibrium of current loop occurs when magnetic moment is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 45Potential energy of current loop is
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 46Magnetic dipole moment of a coil is
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 47In radial magnetic field of galvanometer
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 48If current is doubled, torque becomes
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 49If number of turns and area both double, torque becomes
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 50A loop has zero torque but maximum potential energy when α is
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 51A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 52If θ is the angle between plane of coil and magnetic field, torque is
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 53Maximum torque occurs when area vector is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 54Stable equilibrium of current loop occurs when magnetic moment is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 55Potential energy of current loop is
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 56Magnetic dipole moment of a coil is
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 57In radial magnetic field of galvanometer
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 58If current is doubled, torque becomes
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 59If number of turns and area both double, torque becomes
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 60A loop has zero torque but maximum potential energy when α is
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 61A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 62If θ is the angle between plane of coil and magnetic field, torque is
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 63Maximum torque occurs when area vector is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 64Stable equilibrium of current loop occurs when magnetic moment is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 65Potential energy of current loop is
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 66Magnetic dipole moment of a coil is
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 67In radial magnetic field of galvanometer
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 68If current is doubled, torque becomes
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 69If number of turns and area both double, torque becomes
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 70A loop has zero torque but maximum potential energy when α is
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 71A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 72If θ is the angle between plane of coil and magnetic field, torque is
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 73Maximum torque occurs when area vector is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 74Stable equilibrium of current loop occurs when magnetic moment is
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
NEET MCQ 75Potential energy of current loop is
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Medium
Concept Tested: Torque, magnetic moment and angle relation.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.

C. JEE Main Questions

JEE Main MCQ 1Stable equilibrium of current loop occurs when magnetic moment is A numerical variation may include N=20, I=1 A, A=2×10⁻³ m² and B=0.1 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=1 A, A=2×10⁻³ m² and B=0.1 T.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 2Potential energy of current loop is A numerical variation may include N=30, I=2 A, A=4×10⁻³ m² and B=0.2 T.
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=2 A, A=4×10⁻³ m² and B=0.2 T.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 3Magnetic dipole moment of a coil is A numerical variation may include N=40, I=3 A, A=6×10⁻³ m² and B=0.3 T.
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=3 A, A=6×10⁻³ m² and B=0.3 T.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 4In radial magnetic field of galvanometer A numerical variation may include N=50, I=4 A, A=8×10⁻³ m² and B=0.4 T.
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=4 A, A=8×10⁻³ m² and B=0.4 T.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 5If current is doubled, torque becomes A numerical variation may include N=60, I=1 A, A=10×10⁻³ m² and B=0.5 T.
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=1 A, A=10×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 6If number of turns and area both double, torque becomes A numerical variation may include N=20, I=2 A, A=12×10⁻³ m² and B=0.1 T.
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=2 A, A=12×10⁻³ m² and B=0.1 T.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 7A loop has zero torque but maximum potential energy when α is A numerical variation may include N=30, I=3 A, A=2×10⁻³ m² and B=0.2 T.
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=3 A, A=2×10⁻³ m² and B=0.2 T.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 8A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is A numerical variation may include N=40, I=4 A, A=4×10⁻³ m² and B=0.3 T.
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=4 A, A=4×10⁻³ m² and B=0.3 T.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 9If θ is the angle between plane of coil and magnetic field, torque is A numerical variation may include N=50, I=1 A, A=6×10⁻³ m² and B=0.4 T.
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=1 A, A=6×10⁻³ m² and B=0.4 T.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 10Maximum torque occurs when area vector is A numerical variation may include N=60, I=2 A, A=8×10⁻³ m² and B=0.5 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=2 A, A=8×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 11Stable equilibrium of current loop occurs when magnetic moment is A numerical variation may include N=20, I=3 A, A=10×10⁻³ m² and B=0.1 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=3 A, A=10×10⁻³ m² and B=0.1 T.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 12Potential energy of current loop is A numerical variation may include N=30, I=4 A, A=12×10⁻³ m² and B=0.2 T.
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=4 A, A=12×10⁻³ m² and B=0.2 T.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 13Magnetic dipole moment of a coil is A numerical variation may include N=40, I=1 A, A=2×10⁻³ m² and B=0.3 T.
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=1 A, A=2×10⁻³ m² and B=0.3 T.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 14In radial magnetic field of galvanometer A numerical variation may include N=50, I=2 A, A=4×10⁻³ m² and B=0.4 T.
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=2 A, A=4×10⁻³ m² and B=0.4 T.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 15If current is doubled, torque becomes A numerical variation may include N=60, I=3 A, A=6×10⁻³ m² and B=0.5 T.
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=3 A, A=6×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 16If number of turns and area both double, torque becomes A numerical variation may include N=20, I=4 A, A=8×10⁻³ m² and B=0.1 T.
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=4 A, A=8×10⁻³ m² and B=0.1 T.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 17A loop has zero torque but maximum potential energy when α is A numerical variation may include N=30, I=1 A, A=10×10⁻³ m² and B=0.2 T.
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=1 A, A=10×10⁻³ m² and B=0.2 T.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 18A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is A numerical variation may include N=40, I=2 A, A=12×10⁻³ m² and B=0.3 T.
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=2 A, A=12×10⁻³ m² and B=0.3 T.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 19If θ is the angle between plane of coil and magnetic field, torque is A numerical variation may include N=50, I=3 A, A=2×10⁻³ m² and B=0.4 T.
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=3 A, A=2×10⁻³ m² and B=0.4 T.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 20Maximum torque occurs when area vector is A numerical variation may include N=60, I=4 A, A=4×10⁻³ m² and B=0.5 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=4 A, A=4×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 21Stable equilibrium of current loop occurs when magnetic moment is A numerical variation may include N=20, I=1 A, A=6×10⁻³ m² and B=0.1 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=1 A, A=6×10⁻³ m² and B=0.1 T.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 22Potential energy of current loop is A numerical variation may include N=30, I=2 A, A=8×10⁻³ m² and B=0.2 T.
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=2 A, A=8×10⁻³ m² and B=0.2 T.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 23Magnetic dipole moment of a coil is A numerical variation may include N=40, I=3 A, A=10×10⁻³ m² and B=0.3 T.
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=3 A, A=10×10⁻³ m² and B=0.3 T.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 24In radial magnetic field of galvanometer A numerical variation may include N=50, I=4 A, A=12×10⁻³ m² and B=0.4 T.
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=4 A, A=12×10⁻³ m² and B=0.4 T.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 25If current is doubled, torque becomes A numerical variation may include N=60, I=1 A, A=2×10⁻³ m² and B=0.5 T.
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=1 A, A=2×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 26If number of turns and area both double, torque becomes A numerical variation may include N=20, I=2 A, A=4×10⁻³ m² and B=0.1 T.
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=2 A, A=4×10⁻³ m² and B=0.1 T.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 27A loop has zero torque but maximum potential energy when α is A numerical variation may include N=30, I=3 A, A=6×10⁻³ m² and B=0.2 T.
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=3 A, A=6×10⁻³ m² and B=0.2 T.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 28A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is A numerical variation may include N=40, I=4 A, A=8×10⁻³ m² and B=0.3 T.
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=4 A, A=8×10⁻³ m² and B=0.3 T.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 29If θ is the angle between plane of coil and magnetic field, torque is A numerical variation may include N=50, I=1 A, A=10×10⁻³ m² and B=0.4 T.
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=1 A, A=10×10⁻³ m² and B=0.4 T.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 30Maximum torque occurs when area vector is A numerical variation may include N=60, I=2 A, A=12×10⁻³ m² and B=0.5 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=2 A, A=12×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 31Stable equilibrium of current loop occurs when magnetic moment is A numerical variation may include N=20, I=3 A, A=2×10⁻³ m² and B=0.1 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=3 A, A=2×10⁻³ m² and B=0.1 T.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 32Potential energy of current loop is A numerical variation may include N=30, I=4 A, A=4×10⁻³ m² and B=0.2 T.
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=4 A, A=4×10⁻³ m² and B=0.2 T.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 33Magnetic dipole moment of a coil is A numerical variation may include N=40, I=1 A, A=6×10⁻³ m² and B=0.3 T.
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=1 A, A=6×10⁻³ m² and B=0.3 T.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 34In radial magnetic field of galvanometer A numerical variation may include N=50, I=2 A, A=8×10⁻³ m² and B=0.4 T.
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=2 A, A=8×10⁻³ m² and B=0.4 T.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 35If current is doubled, torque becomes A numerical variation may include N=60, I=3 A, A=10×10⁻³ m² and B=0.5 T.
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=3 A, A=10×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 36If number of turns and area both double, torque becomes A numerical variation may include N=20, I=4 A, A=12×10⁻³ m² and B=0.1 T.
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=4 A, A=12×10⁻³ m² and B=0.1 T.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 37A loop has zero torque but maximum potential energy when α is A numerical variation may include N=30, I=1 A, A=2×10⁻³ m² and B=0.2 T.
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=1 A, A=2×10⁻³ m² and B=0.2 T.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 38A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is A numerical variation may include N=40, I=2 A, A=4×10⁻³ m² and B=0.3 T.
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=2 A, A=4×10⁻³ m² and B=0.3 T.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 39If θ is the angle between plane of coil and magnetic field, torque is A numerical variation may include N=50, I=3 A, A=6×10⁻³ m² and B=0.4 T.
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=3 A, A=6×10⁻³ m² and B=0.4 T.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 40Maximum torque occurs when area vector is A numerical variation may include N=60, I=4 A, A=8×10⁻³ m² and B=0.5 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=4 A, A=8×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 41Stable equilibrium of current loop occurs when magnetic moment is A numerical variation may include N=20, I=1 A, A=10×10⁻³ m² and B=0.1 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=1 A, A=10×10⁻³ m² and B=0.1 T.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 42Potential energy of current loop is A numerical variation may include N=30, I=2 A, A=12×10⁻³ m² and B=0.2 T.
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=2 A, A=12×10⁻³ m² and B=0.2 T.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 43Magnetic dipole moment of a coil is A numerical variation may include N=40, I=3 A, A=2×10⁻³ m² and B=0.3 T.
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=3 A, A=2×10⁻³ m² and B=0.3 T.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 44In radial magnetic field of galvanometer A numerical variation may include N=50, I=4 A, A=4×10⁻³ m² and B=0.4 T.
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=4 A, A=4×10⁻³ m² and B=0.4 T.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 45If current is doubled, torque becomes A numerical variation may include N=60, I=1 A, A=6×10⁻³ m² and B=0.5 T.
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=1 A, A=6×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 46If number of turns and area both double, torque becomes A numerical variation may include N=20, I=2 A, A=8×10⁻³ m² and B=0.1 T.
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=2 A, A=8×10⁻³ m² and B=0.1 T.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 47A loop has zero torque but maximum potential energy when α is A numerical variation may include N=30, I=3 A, A=10×10⁻³ m² and B=0.2 T.
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=3 A, A=10×10⁻³ m² and B=0.2 T.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 48A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is A numerical variation may include N=40, I=4 A, A=12×10⁻³ m² and B=0.3 T.
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=4 A, A=12×10⁻³ m² and B=0.3 T.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 49If θ is the angle between plane of coil and magnetic field, torque is A numerical variation may include N=50, I=1 A, A=2×10⁻³ m² and B=0.4 T.
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=1 A, A=2×10⁻³ m² and B=0.4 T.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 50Maximum torque occurs when area vector is A numerical variation may include N=60, I=2 A, A=4×10⁻³ m² and B=0.5 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=2 A, A=4×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 51Stable equilibrium of current loop occurs when magnetic moment is A numerical variation may include N=20, I=3 A, A=6×10⁻³ m² and B=0.1 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=3 A, A=6×10⁻³ m² and B=0.1 T.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 52Potential energy of current loop is A numerical variation may include N=30, I=4 A, A=8×10⁻³ m² and B=0.2 T.
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=4 A, A=8×10⁻³ m² and B=0.2 T.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 53Magnetic dipole moment of a coil is A numerical variation may include N=40, I=1 A, A=10×10⁻³ m² and B=0.3 T.
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=1 A, A=10×10⁻³ m² and B=0.3 T.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 54In radial magnetic field of galvanometer A numerical variation may include N=50, I=2 A, A=12×10⁻³ m² and B=0.4 T.
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=2 A, A=12×10⁻³ m² and B=0.4 T.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 55If current is doubled, torque becomes A numerical variation may include N=60, I=3 A, A=2×10⁻³ m² and B=0.5 T.
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=3 A, A=2×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 56If number of turns and area both double, torque becomes A numerical variation may include N=20, I=4 A, A=4×10⁻³ m² and B=0.1 T.
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=4 A, A=4×10⁻³ m² and B=0.1 T.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 57A loop has zero torque but maximum potential energy when α is A numerical variation may include N=30, I=1 A, A=6×10⁻³ m² and B=0.2 T.
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=1 A, A=6×10⁻³ m² and B=0.2 T.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 58A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is A numerical variation may include N=40, I=2 A, A=8×10⁻³ m² and B=0.3 T.
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=2 A, A=8×10⁻³ m² and B=0.3 T.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 59If θ is the angle between plane of coil and magnetic field, torque is A numerical variation may include N=50, I=3 A, A=10×10⁻³ m² and B=0.4 T.
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=3 A, A=10×10⁻³ m² and B=0.4 T.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 60Maximum torque occurs when area vector is A numerical variation may include N=60, I=4 A, A=12×10⁻³ m² and B=0.5 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=4 A, A=12×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 61Stable equilibrium of current loop occurs when magnetic moment is A numerical variation may include N=20, I=1 A, A=2×10⁻³ m² and B=0.1 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=1 A, A=2×10⁻³ m² and B=0.1 T.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 62Potential energy of current loop is A numerical variation may include N=30, I=2 A, A=4×10⁻³ m² and B=0.2 T.
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=2 A, A=4×10⁻³ m² and B=0.2 T.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 63Magnetic dipole moment of a coil is A numerical variation may include N=40, I=3 A, A=6×10⁻³ m² and B=0.3 T.
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=3 A, A=6×10⁻³ m² and B=0.3 T.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 64In radial magnetic field of galvanometer A numerical variation may include N=50, I=4 A, A=8×10⁻³ m² and B=0.4 T.
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=4 A, A=8×10⁻³ m² and B=0.4 T.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 65If current is doubled, torque becomes A numerical variation may include N=60, I=1 A, A=10×10⁻³ m² and B=0.5 T.
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=1 A, A=10×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 66If number of turns and area both double, torque becomes A numerical variation may include N=20, I=2 A, A=12×10⁻³ m² and B=0.1 T.
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=2 A, A=12×10⁻³ m² and B=0.1 T.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 67A loop has zero torque but maximum potential energy when α is A numerical variation may include N=30, I=3 A, A=2×10⁻³ m² and B=0.2 T.
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=3 A, A=2×10⁻³ m² and B=0.2 T.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 68A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is A numerical variation may include N=40, I=4 A, A=4×10⁻³ m² and B=0.3 T.
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=4 A, A=4×10⁻³ m² and B=0.3 T.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 69If θ is the angle between plane of coil and magnetic field, torque is A numerical variation may include N=50, I=1 A, A=6×10⁻³ m² and B=0.4 T.
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=1 A, A=6×10⁻³ m² and B=0.4 T.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 70Maximum torque occurs when area vector is A numerical variation may include N=60, I=2 A, A=8×10⁻³ m² and B=0.5 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=2 A, A=8×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 71Stable equilibrium of current loop occurs when magnetic moment is A numerical variation may include N=20, I=3 A, A=10×10⁻³ m² and B=0.1 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=3 A, A=10×10⁻³ m² and B=0.1 T.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 72Potential energy of current loop is A numerical variation may include N=30, I=4 A, A=12×10⁻³ m² and B=0.2 T.
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=4 A, A=12×10⁻³ m² and B=0.2 T.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 73Magnetic dipole moment of a coil is A numerical variation may include N=40, I=1 A, A=2×10⁻³ m² and B=0.3 T.
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=1 A, A=2×10⁻³ m² and B=0.3 T.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 74In radial magnetic field of galvanometer A numerical variation may include N=50, I=2 A, A=4×10⁻³ m² and B=0.4 T.
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=2 A, A=4×10⁻³ m² and B=0.4 T.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Main MCQ 75If current is doubled, torque becomes A numerical variation may include N=60, I=3 A, A=6×10⁻³ m² and B=0.5 T.
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=3 A, A=6×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.

D. JEE Advanced Questions

JEE Advanced Single Correct MCQ 1Stable equilibrium of current loop occurs when magnetic moment is A numerical variation may include N=20, I=1 A, A=2×10⁻³ m² and B=0.1 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=1 A, A=2×10⁻³ m² and B=0.1 T.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 2Potential energy of current loop is A numerical variation may include N=30, I=2 A, A=4×10⁻³ m² and B=0.2 T.
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=2 A, A=4×10⁻³ m² and B=0.2 T.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 3Magnetic dipole moment of a coil is A numerical variation may include N=40, I=3 A, A=6×10⁻³ m² and B=0.3 T.
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=3 A, A=6×10⁻³ m² and B=0.3 T.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 4In radial magnetic field of galvanometer A numerical variation may include N=50, I=4 A, A=8×10⁻³ m² and B=0.4 T.
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=4 A, A=8×10⁻³ m² and B=0.4 T.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 5If current is doubled, torque becomes A numerical variation may include N=60, I=1 A, A=10×10⁻³ m² and B=0.5 T.
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=1 A, A=10×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 6If number of turns and area both double, torque becomes A numerical variation may include N=20, I=2 A, A=12×10⁻³ m² and B=0.1 T.
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=2 A, A=12×10⁻³ m² and B=0.1 T.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 7A loop has zero torque but maximum potential energy when α is A numerical variation may include N=30, I=3 A, A=2×10⁻³ m² and B=0.2 T.
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=3 A, A=2×10⁻³ m² and B=0.2 T.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 8A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is A numerical variation may include N=40, I=4 A, A=4×10⁻³ m² and B=0.3 T.
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=4 A, A=4×10⁻³ m² and B=0.3 T.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 9If θ is the angle between plane of coil and magnetic field, torque is A numerical variation may include N=50, I=1 A, A=6×10⁻³ m² and B=0.4 T.
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=1 A, A=6×10⁻³ m² and B=0.4 T.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 10Maximum torque occurs when area vector is A numerical variation may include N=60, I=2 A, A=8×10⁻³ m² and B=0.5 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=2 A, A=8×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 11Stable equilibrium of current loop occurs when magnetic moment is A numerical variation may include N=20, I=3 A, A=10×10⁻³ m² and B=0.1 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=3 A, A=10×10⁻³ m² and B=0.1 T.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 12Potential energy of current loop is A numerical variation may include N=30, I=4 A, A=12×10⁻³ m² and B=0.2 T.
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=4 A, A=12×10⁻³ m² and B=0.2 T.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 13Magnetic dipole moment of a coil is A numerical variation may include N=40, I=1 A, A=2×10⁻³ m² and B=0.3 T.
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=1 A, A=2×10⁻³ m² and B=0.3 T.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 14In radial magnetic field of galvanometer A numerical variation may include N=50, I=2 A, A=4×10⁻³ m² and B=0.4 T.
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=2 A, A=4×10⁻³ m² and B=0.4 T.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 15If current is doubled, torque becomes A numerical variation may include N=60, I=3 A, A=6×10⁻³ m² and B=0.5 T.
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=3 A, A=6×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 16If number of turns and area both double, torque becomes A numerical variation may include N=20, I=4 A, A=8×10⁻³ m² and B=0.1 T.
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=4 A, A=8×10⁻³ m² and B=0.1 T.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 17A loop has zero torque but maximum potential energy when α is A numerical variation may include N=30, I=1 A, A=10×10⁻³ m² and B=0.2 T.
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=1 A, A=10×10⁻³ m² and B=0.2 T.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 18A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is A numerical variation may include N=40, I=2 A, A=12×10⁻³ m² and B=0.3 T.
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=2 A, A=12×10⁻³ m² and B=0.3 T.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 19If θ is the angle between plane of coil and magnetic field, torque is A numerical variation may include N=50, I=3 A, A=2×10⁻³ m² and B=0.4 T.
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=3 A, A=2×10⁻³ m² and B=0.4 T.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 20Maximum torque occurs when area vector is A numerical variation may include N=60, I=4 A, A=4×10⁻³ m² and B=0.5 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=4 A, A=4×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 21Stable equilibrium of current loop occurs when magnetic moment is A numerical variation may include N=20, I=1 A, A=6×10⁻³ m² and B=0.1 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=1 A, A=6×10⁻³ m² and B=0.1 T.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 22Potential energy of current loop is A numerical variation may include N=30, I=2 A, A=8×10⁻³ m² and B=0.2 T.
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=2 A, A=8×10⁻³ m² and B=0.2 T.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 23Magnetic dipole moment of a coil is A numerical variation may include N=40, I=3 A, A=10×10⁻³ m² and B=0.3 T.
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=3 A, A=10×10⁻³ m² and B=0.3 T.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 24In radial magnetic field of galvanometer A numerical variation may include N=50, I=4 A, A=12×10⁻³ m² and B=0.4 T.
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=4 A, A=12×10⁻³ m² and B=0.4 T.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 25If current is doubled, torque becomes A numerical variation may include N=60, I=1 A, A=2×10⁻³ m² and B=0.5 T.
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=1 A, A=2×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 26If number of turns and area both double, torque becomes A numerical variation may include N=20, I=2 A, A=4×10⁻³ m² and B=0.1 T.
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=2 A, A=4×10⁻³ m² and B=0.1 T.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 27A loop has zero torque but maximum potential energy when α is A numerical variation may include N=30, I=3 A, A=6×10⁻³ m² and B=0.2 T.
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=3 A, A=6×10⁻³ m² and B=0.2 T.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 28A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is A numerical variation may include N=40, I=4 A, A=8×10⁻³ m² and B=0.3 T.
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=4 A, A=8×10⁻³ m² and B=0.3 T.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 29If θ is the angle between plane of coil and magnetic field, torque is A numerical variation may include N=50, I=1 A, A=10×10⁻³ m² and B=0.4 T.
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=1 A, A=10×10⁻³ m² and B=0.4 T.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Single Correct MCQ 30Maximum torque occurs when area vector is A numerical variation may include N=60, I=2 A, A=12×10⁻³ m² and B=0.5 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=2 A, A=12×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 1Stable equilibrium of current loop occurs when magnetic moment is A numerical variation may include N=20, I=1 A, A=2×10⁻³ m² and B=0.1 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=1 A, A=2×10⁻³ m² and B=0.1 T.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 2Potential energy of current loop is A numerical variation may include N=30, I=2 A, A=4×10⁻³ m² and B=0.2 T.
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=2 A, A=4×10⁻³ m² and B=0.2 T.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 3Magnetic dipole moment of a coil is A numerical variation may include N=40, I=3 A, A=6×10⁻³ m² and B=0.3 T.
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=3 A, A=6×10⁻³ m² and B=0.3 T.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 4In radial magnetic field of galvanometer A numerical variation may include N=50, I=4 A, A=8×10⁻³ m² and B=0.4 T.
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=4 A, A=8×10⁻³ m² and B=0.4 T.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 5If current is doubled, torque becomes A numerical variation may include N=60, I=1 A, A=10×10⁻³ m² and B=0.5 T.
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=1 A, A=10×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 6If number of turns and area both double, torque becomes A numerical variation may include N=20, I=2 A, A=12×10⁻³ m² and B=0.1 T.
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=2 A, A=12×10⁻³ m² and B=0.1 T.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 7A loop has zero torque but maximum potential energy when α is A numerical variation may include N=30, I=3 A, A=2×10⁻³ m² and B=0.2 T.
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=3 A, A=2×10⁻³ m² and B=0.2 T.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 8A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is A numerical variation may include N=40, I=4 A, A=4×10⁻³ m² and B=0.3 T.
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=4 A, A=4×10⁻³ m² and B=0.3 T.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 9If θ is the angle between plane of coil and magnetic field, torque is A numerical variation may include N=50, I=1 A, A=6×10⁻³ m² and B=0.4 T.
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=1 A, A=6×10⁻³ m² and B=0.4 T.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 10Maximum torque occurs when area vector is A numerical variation may include N=60, I=2 A, A=8×10⁻³ m² and B=0.5 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=2 A, A=8×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 11Stable equilibrium of current loop occurs when magnetic moment is A numerical variation may include N=20, I=3 A, A=10×10⁻³ m² and B=0.1 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. opposite current
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=3 A, A=10×10⁻³ m² and B=0.1 T.
Detailed Explanation: U = -mB cosα is minimum at α = 0°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 12Potential energy of current loop is A numerical variation may include N=30, I=4 A, A=12×10⁻³ m² and B=0.2 T.
  1. mB cosα
  2. -mB cosα
  3. mB sinα
  4. -mB sinα
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=4 A, A=12×10⁻³ m² and B=0.2 T.
Detailed Explanation: For magnetic dipole, U = -m · B.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 13Magnetic dipole moment of a coil is A numerical variation may include N=40, I=1 A, A=2×10⁻³ m² and B=0.3 T.
  1. NIA
  2. NBA
  3. NBI
  4. IA/N
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=1 A, A=2×10⁻³ m² and B=0.3 T.
Detailed Explanation: m = NIA along area vector.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 14In radial magnetic field of galvanometer A numerical variation may include N=50, I=2 A, A=4×10⁻³ m² and B=0.4 T.
  1. torque is zero
  2. torque remains maximum
  3. B is zero
  4. current is not needed
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=2 A, A=4×10⁻³ m² and B=0.4 T.
Detailed Explanation: The area vector remains perpendicular to B, so sinα = 1.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 15If current is doubled, torque becomes A numerical variation may include N=60, I=3 A, A=6×10⁻³ m² and B=0.5 T.
  1. half
  2. same
  3. double
  4. four times
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=3 A, A=6×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ ∝ I.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 16If number of turns and area both double, torque becomes A numerical variation may include N=20, I=4 A, A=8×10⁻³ m² and B=0.1 T.
  1. 2 times
  2. 4 times
  3. 8 times
  4. same
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=20, I=4 A, A=8×10⁻³ m² and B=0.1 T.
Detailed Explanation: τ ∝ NA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 17A loop has zero torque but maximum potential energy when α is A numerical variation may include N=30, I=1 A, A=10×10⁻³ m² and B=0.2 T.
  1. 0°
  2. 45°
  3. 90°
  4. 180°
Correct Answer: D
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=30, I=1 A, A=10×10⁻³ m² and B=0.2 T.
Detailed Explanation: At α = 180°, τ = 0 and U = +mB.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 18A rectangular coil of N turns and area A carries current I in uniform B. If its area vector makes angle α with B, torque is A numerical variation may include N=40, I=2 A, A=12×10⁻³ m² and B=0.3 T.
  1. NBIA sinα
  2. NBIA cosα
  3. NBI/A
  4. NBA/I
Correct Answer: A
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=40, I=2 A, A=12×10⁻³ m² and B=0.3 T.
Detailed Explanation: Use τ = mB sinα and m = NIA.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 19If θ is the angle between plane of coil and magnetic field, torque is A numerical variation may include N=50, I=3 A, A=2×10⁻³ m² and B=0.4 T.
  1. NBIA sinθ
  2. NBIA cosθ
  3. NBIA tanθ
  4. zero always
Correct Answer: B
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=50, I=3 A, A=2×10⁻³ m² and B=0.4 T.
Detailed Explanation: Since α + θ = 90°, sinα = cosθ.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Multiple Correct MCQ 20Maximum torque occurs when area vector is A numerical variation may include N=60, I=4 A, A=4×10⁻³ m² and B=0.5 T.
  1. parallel to B
  2. antiparallel to B
  3. perpendicular to B
  4. zero
Correct Answer: C
Difficulty: Difficult
Concept Tested: Torque, magnetic moment and angle relation. A numerical variation may include N=60, I=4 A, A=4×10⁻³ m² and B=0.5 T.
Detailed Explanation: τ = NBIA sinα is maximum at α = 90°.
Common Student Mistake: Students often confuse α, the angle with area vector, with θ, the angle with the plane.
JEE Advanced Integer Question 1Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 2Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 3Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 4Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 5Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 6Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 7Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 8Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 9State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 10Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 11Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 12Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 13Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 14Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Integer Question 15Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Matrix Match Question 1Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Matrix Match Question 2Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Matrix Match Question 3Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Matrix Match Question 4Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Matrix Match Question 5Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Matrix Match Question 6Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Matrix Match Question 7Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Matrix Match Question 8Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Matrix Match Question 9State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Matrix Match Question 10Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Paragraph Question 1Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Paragraph Question 2Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Paragraph Question 3Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Paragraph Question 4Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Paragraph Question 5Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Paragraph Question 6Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Paragraph Question 7Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Paragraph Question 8Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Paragraph Question 9State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
JEE Advanced Paragraph Question 10Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.

E. IB Physics Questions

IB Question 1Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 2Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 3Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 4Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 5Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 6Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 7Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 8Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 9State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 10Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 11Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 12Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 13Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 14Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 15Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 16Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 17Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 18Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 19State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 20Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 21Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 22Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 23Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 24Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IB Question 25Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.

F. ICSE Physics Questions

ICSE Question 1Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 2Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 3Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 4Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 5Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 6Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 7Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 8Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 9State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 10Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 11Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 12Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 13Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 14Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 15Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 16Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 17Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 18Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 19State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 20Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 21Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 22Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 23Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 24Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
ICSE Question 25Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.

G. IGCSE Physics Questions

IGCSE Question 1Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 2Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 3Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 4Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 5Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 6Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 7Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 8Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 9State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 10Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 11Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 12Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 13Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 14Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 15Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 16Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 17Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 18Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 19State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 20Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 21Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 22Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 23Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 24Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
IGCSE Question 25Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.

H. British Curriculum / A-Level Physics

A-Level Question 1Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 2Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 3Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 4Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 5Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 6Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 7Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 8Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 9State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 10Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 11Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 12Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 13Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 14Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 15Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 16Explain the relation between angle α and angle θ.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 17Draw torque versus angle graph and explain important points.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 18Draw potential energy versus angle graph and explain minima and maxima.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 19State applications of torque on current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 20Explain why net force on a closed loop in uniform magnetic field is zero but net torque may be non-zero.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 21Explain why equal and opposite forces on a current loop can produce torque.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 22Define magnetic dipole moment of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 23Derive torque on a current carrying rectangular coil.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 24Explain stable and unstable equilibrium of a current loop.
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.
A-Level Question 25Why is radial magnetic field used in moving coil galvanometer?
Answer: Draw the current loop, mark the area vector by right-hand thumb rule, use F = BIl on the active sides, and then apply τ = force × perpendicular distance. The final formula is τ = NBIA sinα = NBIA cosθ. For energy use U = -NBIA cosα. In board answers, define every symbol and mention the physical meaning.

13. Case Study Section

Case Study 1current loop in uniform magnetic field
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 2electric motor
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 3moving coil galvanometer
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 4magnetic dipole
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 5stable equilibrium
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 6unstable equilibrium
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 7torque-angle graph
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 8potential-energy graph
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 9coil with N turns
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 10rectangular loop numerical
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 11current loop in uniform magnetic field
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 12electric motor
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 13moving coil galvanometer
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 14magnetic dipole
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 15stable equilibrium
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 16unstable equilibrium
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 17torque-angle graph
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 18potential-energy graph
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 19coil with N turns
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.
Case Study 20rectangular loop numerical
Scenario: A rectangular coil is placed in a uniform magnetic field. Its current, area, number of turns and angle may be varied.
Questions: calculate torque, identify equilibrium, compare energy, and predict rotation direction.
Solution: Use m = NIA, τ = mB sinα, U = -mB cosα. Stable equilibrium is α = 0°, unstable equilibrium is α = 180°, and maximum torque occurs at α = 90°.

14. Graphs

Torque and potential energy graphs
Graphs: τ = NBIA sinα and U = −NBIA cosα τα 0°90°180° τmax Uα 0°90°180° −mB+mB

The torque graph follows a sine curve. The potential energy graph follows negative cosine behaviour. These graphs are very useful for NEET and JEE conceptual questions.

15. Final Revision Sheet

AreaA = l × b
Force on sideF = BIl
Torqueτ = NBIA sinα
Plane angle formulaτ = NBIA cosθ
Magnetic momentm = NIA
Vector torqueτ = m × B
Potential energyU = -mB cosα
GalvanometerI = kφ/NBA

CBSE derivation: force couple method. NEET trap: angle relation. JEE trap: energy and equilibrium analysis. Advanced type: combine torque, potential energy and restoring torque.

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