Force Between Two Parallel Current Carrying Conductors - Complete Physics Guide

CBSE • NEET • IIT JEE • JEE Advanced • IB • AP • IGCSE • ICSE • A-Level

Force Between Two Parallel Current Carrying Conductors

A complete premium Physics guide by Kumar Sir covering attraction, repulsion, direction rules, derivation, ampere definition, rectangular loop near wire, suspended conductor equilibrium, inclined-plane conductor problems and exam-level questions.

Talk to Kumar SirKumarPhysicsClasses.com
If force between two parallel current carrying conductors is not clear, students can contact Kumar Sir for one-to-one Physics guidance.
Phone / WhatsApp: +91-9958461445 | Email: kumarsirphysics@gmail.com | Website: kumarphysicsclasses.com

1. Introduction

When two long parallel conductors carry currents, each conductor produces a magnetic field at the position of the other conductor. Since a current-carrying conductor placed in a magnetic field experiences force, the two wires exert equal and opposite forces on each other.

Magnetic field of long wireB = μ₀I / 2πr
Force on current conductorF = ILB

2. Same Direction Currents

Same direction currents: attraction
I₁I₂ separation r F₁F₂ B₁ at wire 2 = ⊙ B₂ at wire 1 = ⊗ Same downward current direction ⇒ attraction

When currents in two parallel conductors are in the same direction, the conductors attract each other.

3. Opposite Direction Currents

Opposite direction currents: repulsion
I₁ downwardI₂ upward separation r F₁F₂ B₁ at wire 2 = ⊙ B₂ at wire 1 = ⊙ Opposite current direction ⇒ repulsion

When currents in two parallel conductors are in opposite directions, the conductors repel each other.

DerivationCBSE • NEET • JEE

4. Derivation of Force Between Two Parallel Conductors

Derivation geometry: field of wire 1 acts on wire 2
wire 1, I₁wire 2, I₂ r length L B₁ = μ₀I₁/2πr F₂ = I₂LB₁ Force per unit length: F/L = μ₀I₁I₂/2πr
Given / Symbols:
Two long parallel conductors carry currents I₁ and I₂, separated by distance r. Consider length L.
1Magnetic field due to conductor 1 at conductor 2 is B₁ = μ₀I₁/2πr.
2Force on length L of conductor 2 is F₂ = I₂LB₁.
3Substitute B₁ into the force expression.
4F₂ = I₂L × μ₀I₁/2πr.
5Therefore F = μ₀I₁I₂L/2πr.
6Force per unit length is obtained by dividing by L.
7Equivalent form: F/L = (μ₀/4π) × 2I₁I₂/r.
Final Result:F = μ₀I₁I₂L/2πr; F/L = μ₀I₁I₂/2πr
Common mistake:
B due to conductor 1 acts on conductor 2. A conductor does not exert magnetic force on itself.

5. Direction of Force

Use right-hand thumb rule to find magnetic field due to one conductor.
Use F = ILB or Fleming's left hand rule to find force on the other conductor.
Same direction currents attract.
Opposite direction currents repel.
⊙ means field out of page.
⊗ means field into page.

6. Definition of One Ampere

One ampere is that constant current which, if maintained in each of two straight parallel conductors of infinite length, negligible circular cross-section, and placed one metre apart in vacuum, produces a force of 2 × 10⁻⁷ newton per metre length between them.

1Start with F/L = μ₀I₁I₂/2πr.
2Put I₁ = I₂ = 1 A and r = 1 m.
3Use μ₀ = 4π × 10⁻⁷ T m A⁻¹.
4F/L = (4π×10⁻⁷)/(2π) = 2×10⁻⁷ N/m.
CBSE Definition ResultF/L = 2 × 10⁻⁷ N/m
DerivationCBSE • NEET • JEE

7. Rectangular Current Loop Near a Long Straight Wire

Rectangular current loop near a long straight wire
I₁ loop current I₂ a h L F_near: attraction F_far: repulsion F_net = μ₀I₁I₂L/2π [1/a − 1/(a+h)]
Given / Symbols:
Long wire current I₁, rectangular loop current I₂, near distance a, loop height h, parallel side length L.
1Magnetic field at nearer side is B_near = μ₀I₁/2πa.
2Force on nearer side is F_near = I₂LB_near = μ₀I₁I₂L/2πa.
3Magnetic field at farther side is B_far = μ₀I₁/2π(a+h).
4Force on farther side is F_far = μ₀I₁I₂L/2π(a+h).
5Forces on horizontal sides cancel.
6Net force equals difference between near and far side forces.
7Direction depends on whether nearer side current is same as or opposite to long wire current.
Final Result:F_net = μ₀I₁I₂L/2π [1/a - 1/(a+h)]
Common mistake:
Near side distance is a; far side distance is a+h.
DerivationCBSE • NEET • JEE

8. Two Horizontal Parallel Conductors in Equilibrium

Two horizontal parallel conductors in equilibrium
lower fixed conductor I₁ upper conductor I₂ r F_magnetic mg Equilibrium: μ₀I₁I₂L/2πr = mg, or μ₀I₁I₂/2πr = λg
Given / Symbols:
Upper conductor has weight mg or mass per unit length λ. Magnetic force balances weight.
1Force on length L is F = μ₀I₁I₂L/2πr.
2For equilibrium of a length L, F_magnetic = mg.
3Therefore μ₀I₁I₂L/2πr = mg.
4If mass per unit length is λ, mass of length L is λL.
5Then μ₀I₁I₂L/2πr = λLg.
6Cancel L to get μ₀I₁I₂/2πr = λg.
Final Result:μ₀I₁I₂L/2πr = mg; per unit length: μ₀I₁I₂/2πr = λg
Common mistake:
Do not mix total mass m with mass per unit length λ.
DerivationCBSE • NEET • JEE

9. Current Carrying Conductor on an Inclined Plane

Current carrying conductor on an inclined plane in magnetic field
wire, I ILB mg N B = ⊙ θ No friction case: ILB = mg sinθ
Given / Symbols:
Inclined plane angle θ, conductor length L, current I, magnetic field B, mass m.
1Magnetic force on conductor is F = ILB when conductor is perpendicular to B.
2Component of weight along incline is mg sinθ.
3Without friction, equilibrium along incline requires ILB = mg sinθ.
4Therefore B = mg sinθ/IL.
5Normal reaction may change as N = mg cosθ ± ILB depending on field direction.
6With friction, equilibrium equation becomes ILB + friction = mg sinθ or ILB = mg sinθ + friction depending on direction.
Final Result:No friction: ILB = mg sinθ, so B = mg sinθ/IL
Common mistake:
Choose magnetic field direction carefully so force acts along required direction.

10. Applications

Definition of ampere
Current balance
Electrodynamic force measurement
Parallel busbars
Short-circuit forces in power lines
Magnetic levitation concept
Moving coil instruments

11. Common Student Mistakes

Mistake 1Confusing attraction and repulsion.

Correction: Draw currents, field direction and force direction before using formulas.

Mistake 2Forgetting force per unit length.

Correction: Draw currents, field direction and force direction before using formulas.

Mistake 3Using wrong distance r.

Correction: Draw currents, field direction and force direction before using formulas.

Mistake 4Forgetting length L.

Correction: Draw currents, field direction and force direction before using formulas.

Mistake 5Wrong direction of magnetic field.

Correction: Draw currents, field direction and force direction before using formulas.

Mistake 6Wrong use of right-hand thumb rule.

Correction: Draw currents, field direction and force direction before using formulas.

Mistake 7Confusing B due to one wire with force on same wire.

Correction: Draw currents, field direction and force direction before using formulas.

Mistake 8Forgetting forces on horizontal sides cancel.

Correction: Draw currents, field direction and force direction before using formulas.

Mistake 9Using a+h incorrectly in rectangular loop problem.

Correction: Draw currents, field direction and force direction before using formulas.

Mistake 10Mixing mg and λg.

Correction: Draw currents, field direction and force direction before using formulas.

12. Exam Question Bank With Solutions

A. CBSE Board Questions

CBSE Theory Question 1Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 2Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 3State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 4Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 5Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 6Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 7Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 8Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 9Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 10Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 11State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 12Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 13Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 14Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 15Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 16Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 17Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 18Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 19State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 20Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 21Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 22Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 23Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 24Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Theory Question 25Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 1Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 2Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 3State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 4Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 5Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 6Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 7Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 8Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 9Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 10Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 11State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 12Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 13Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 14Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 15Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 16Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 17Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 18Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 19State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Derivation Derivation 20Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 1Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 2Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 3State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 4Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 5Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 6Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 7Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 8Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 9Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 10Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 11State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 12Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 13Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 14Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 15Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 16Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 17Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 18Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 19State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 20Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 21Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 22Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 23Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 24Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Numerical Numerical 25Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Case Case 1Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Case Case 2Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Case Case 3State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Case Case 4Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Case Case 5Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Case Case 6Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Case Case 7Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Case Case 8Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Case Case 9Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
CBSE Case Case 10Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.

B. NEET Questions

NEET Q1Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q2Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q3Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q4If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q5The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q6In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q7Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q8If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q9Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q10Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q11Variant 2: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q12Variant 2: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q13Variant 2: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q14Variant 2: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q15Variant 2: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q16Variant 2: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q17Variant 2: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q18Variant 2: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q19Variant 2: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q20Variant 2: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q21Variant 3: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q22Variant 3: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q23Variant 3: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q24Variant 3: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q25Variant 3: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q26Variant 3: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q27Variant 3: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q28Variant 3: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q29Variant 3: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q30Variant 3: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q31Variant 4: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q32Variant 4: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q33Variant 4: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q34Variant 4: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q35Variant 4: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q36Variant 4: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q37Variant 4: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q38Variant 4: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q39Variant 4: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q40Variant 4: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q41Variant 5: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q42Variant 5: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q43Variant 5: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q44Variant 5: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q45Variant 5: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q46Variant 5: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q47Variant 5: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q48Variant 5: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q49Variant 5: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q50Variant 5: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q51Variant 6: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q52Variant 6: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q53Variant 6: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q54Variant 6: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q55Variant 6: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q56Variant 6: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q57Variant 6: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q58Variant 6: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q59Variant 6: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q60Variant 6: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q61Variant 7: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q62Variant 7: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q63Variant 7: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q64Variant 7: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q65Variant 7: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q66Variant 7: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q67Variant 7: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q68Variant 7: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q69Variant 7: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q70Variant 7: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q71Variant 8: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q72Variant 8: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q73Variant 8: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q74Variant 8: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
NEET Q75Variant 8: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.

C. JEE Main Questions

JEE Main Q1Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q2Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q3Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q4If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q5The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q6In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q7Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q8If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q9Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q10Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q11Variant 2: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q12Variant 2: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q13Variant 2: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q14Variant 2: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q15Variant 2: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q16Variant 2: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q17Variant 2: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q18Variant 2: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q19Variant 2: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q20Variant 2: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q21Variant 3: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q22Variant 3: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q23Variant 3: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q24Variant 3: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q25Variant 3: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q26Variant 3: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q27Variant 3: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q28Variant 3: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q29Variant 3: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q30Variant 3: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q31Variant 4: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q32Variant 4: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q33Variant 4: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q34Variant 4: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q35Variant 4: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q36Variant 4: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q37Variant 4: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q38Variant 4: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q39Variant 4: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q40Variant 4: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q41Variant 5: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q42Variant 5: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q43Variant 5: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q44Variant 5: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q45Variant 5: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q46Variant 5: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q47Variant 5: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q48Variant 5: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q49Variant 5: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q50Variant 5: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q51Variant 6: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q52Variant 6: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q53Variant 6: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q54Variant 6: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q55Variant 6: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q56Variant 6: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q57Variant 6: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q58Variant 6: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q59Variant 6: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q60Variant 6: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q61Variant 7: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q62Variant 7: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q63Variant 7: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q64Variant 7: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q65Variant 7: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q66Variant 7: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q67Variant 7: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q68Variant 7: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q69Variant 7: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q70Variant 7: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q71Variant 8: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q72Variant 8: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q73Variant 8: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q74Variant 8: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Main Q75Variant 8: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.

D. JEE Advanced Questions

JEE Advanced Single Correct Q1Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q2Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q3Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q4If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q5The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q6In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q7Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q8If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q9Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q10Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q11Variant 2: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q12Variant 2: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q13Variant 2: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q14Variant 2: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q15Variant 2: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q16Variant 2: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q17Variant 2: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q18Variant 2: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q19Variant 2: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q20Variant 2: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q21Variant 3: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q22Variant 3: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q23Variant 3: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q24Variant 3: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q25Variant 3: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q26Variant 3: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q27Variant 3: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q28Variant 3: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q29Variant 3: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Single Correct Q30Variant 3: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q1Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q2Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q3Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q4If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q5The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q6In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q7Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q8If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q9Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q10Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q11Variant 2: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q12Variant 2: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q13Variant 2: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q14Variant 2: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q15Variant 2: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q16Variant 2: In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q17Variant 2: Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q18Variant 2: If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q19Variant 2: Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Multiple Correct Q20Variant 2: Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q1Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q2Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q3Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q4If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q5The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q6In a rectangular loop near a straight wire, forces on horizontal sides
  1. add
  2. cancel
  3. are maximum only
  4. are zero always
Correct Answer: B
Detailed Explanation: They are equal and opposite for the standard rectangular loop setup.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q7Net force on rectangular loop depends mainly on
  1. near and far parallel sides
  2. only horizontal sides
  3. only length h
  4. no current
Correct Answer: A
Detailed Explanation: Near and far vertical sides have different distances and hence different forces.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q8If magnetic force balances weight for upper wire, condition per unit length is
  1. μ₀I₁I₂/2πr = λg
  2. I₁I₂=mg
  3. B=0
  4. r=0
Correct Answer: A
Detailed Explanation: Use F/L = λg.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q9Wrong distance in rectangular loop formula is often
  1. using a+h for near side
  2. using a for near side
  3. using a+h for far side
  4. using L for length
Correct Answer: A
Detailed Explanation: Near side is at a and far side is at a+h.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q10Force on a current carrying conductor in magnetic field is
  1. ILB sinφ
  2. IR
  3. qE
  4. μ₀I/2πr
Correct Answer: A
Detailed Explanation: F = ILB sinφ; here perpendicular cases use sinφ=1.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q11Variant 2: Two long parallel wires carrying currents in the same direction
  1. attract
  2. repel
  3. do not interact
  4. become neutral
Correct Answer: A
Detailed Explanation: Same direction currents attract.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q12Variant 2: Two long parallel wires carrying currents in opposite directions
  1. attract
  2. repel
  3. do not interact
  4. short circuit
Correct Answer: B
Detailed Explanation: Opposite direction currents repel.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q13Variant 2: Force per unit length between two wires is
  1. μ₀I₁I₂/2πr
  2. μ₀I₁I₂r/2π
  3. μ₀I₁/2πr
  4. I₂LB
Correct Answer: A
Detailed Explanation: F/L = μ₀I₁I₂/2πr.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q14Variant 2: If separation is doubled, force per unit length becomes
  1. double
  2. half
  3. four times
  4. zero
Correct Answer: B
Detailed Explanation: F/L is inversely proportional to r.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Integer Type Q15Variant 2: The SI definition of ampere is based on force per metre equal to
  1. 2×10⁻⁷ N/m
  2. 4π×10⁻⁷ N/m
  3. 1 N/m
  4. 9.8 N/m
Correct Answer: A
Detailed Explanation: For I₁=I₂=1 A and r=1 m, F/L=2×10⁻⁷ N/m.
Common Student Mistake: First decide direction of currents, then field direction, then force direction.
JEE Advanced Matrix Match Matrix 1Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Matrix Match Matrix 2Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Matrix Match Matrix 3State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Matrix Match Matrix 4Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Matrix Match Matrix 5Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Matrix Match Matrix 6Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Matrix Match Matrix 7Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Matrix Match Matrix 8Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Matrix Match Matrix 9Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Matrix Match Matrix 10Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Paragraph Paragraph 1Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Paragraph Paragraph 2Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Paragraph Paragraph 3State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Paragraph Paragraph 4Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Paragraph Paragraph 5Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Paragraph Paragraph 6Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Paragraph Paragraph 7Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Paragraph Paragraph 8Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Paragraph Paragraph 9Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
JEE Advanced Paragraph Paragraph 10Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.

E. IB Physics Questions

IB Structured 1Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 2Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 3State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 4Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 5Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 6Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 7Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 8Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 9Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 10Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 11State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 12Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 13Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 14Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 15Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 16Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 17Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 18Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 19State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 20Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 21Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 22Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 23Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 24Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IB Structured 25Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.

F. ICSE Physics Questions

ICSE Question 1Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 2Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 3State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 4Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 5Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 6Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 7Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 8Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 9Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 10Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 11State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 12Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 13Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 14Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 15Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 16Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 17Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 18Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 19State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 20Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 21Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 22Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 23Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 24Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
ICSE Question 25Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.

G. IGCSE Physics Questions

IGCSE Question 1Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 2Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 3State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 4Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 5Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 6Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 7Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 8Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 9Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 10Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 11State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 12Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 13Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 14Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 15Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 16Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 17Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 18Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 19State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 20Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 21Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 22Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 23Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 24Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
IGCSE Question 25Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.

H. British Curriculum / A-Level Physics

A-Level Advanced 1Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 2Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 3State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 4Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 5Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 6Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 7Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 8Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 9Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 10Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 11State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 12Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 13Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 14Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 15Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 16Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 17Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 18Derive force per unit length between two long parallel conductors.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 19State and derive the SI definition of one ampere.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 20Derive net force on a rectangular loop near a long straight wire.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 21Explain suspended wire equilibrium using magnetic force and weight.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 22Explain the inclined plane conductor equilibrium condition.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 23Compare attraction and repulsion cases with diagrams.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 24Explain why B due to one conductor acts on the other conductor, not on itself.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.
A-Level Advanced 25Explain why two parallel current-carrying conductors exert force on each other.
Solution: Use B = μ₀I/2πr for the magnetic field due to one long straight wire and F = ILB for the force on the other conductor. For unit length, divide by L. Direction is decided by right-hand thumb rule and Fleming's left hand rule.

13. Case Study Section

Case Study 1two parallel wires
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 2current balance
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 3ampere definition
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 4power transmission lines
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 5rectangular loop near wire
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 6suspended wire equilibrium
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 7inclined plane wire
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 8two parallel wires
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 9current balance
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 10ampere definition
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 11power transmission lines
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 12rectangular loop near wire
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 13suspended wire equilibrium
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 14inclined plane wire
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 15two parallel wires
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 16current balance
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 17ampere definition
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 18power transmission lines
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 19rectangular loop near wire
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.
Case Study 20suspended wire equilibrium
Scenario: Two current-carrying conductors or a current loop are placed in a magnetic interaction setup. Data: currents I₁, I₂, separation r, length L, distance a, height h, mass m or mass per unit length λ may be given. Questions: find force, force per unit length, direction, equilibrium current or separation. Solution: Use F/L = μ₀I₁I₂/2πr. For rectangular loop use F_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]. For equilibrium use magnetic force = weight or component of weight.

14. Final Revision Sheet

Field due to long wireB = μ₀I/2πr
Force on conductorF = ILB
Parallel wiresF = μ₀I₁I₂L/2πr
Force per unit lengthF/L = μ₀I₁I₂/2πr
Ampere definitionF/L = 2×10⁻⁷ N/m
Rectangular loopF_net = μ₀I₁I₂L/2π[1/a - 1/(a+h)]
Suspended wireμ₀I₁I₂/2πr = λg
Inclined planeILB = mg sinθ

Direction summary: same direction currents attract; opposite direction currents repel. NEET trap: use force per unit length. JEE trap: in rectangular loop, near and far sides have different fields.

Need Personal Help?

If force between two parallel current carrying conductors, ampere definition, rectangular loop near a long wire, suspended conductor equilibrium, inclined plane magnetic force problem, derivation, numerical problem, MCQ, case study, or advanced question is not clear, students can directly contact Kumar Sir for one-to-one personalized Physics guidance.

Kumar Sir provides personal doubt-solving, conceptual clarity, derivation practice, numerical problem-solving techniques, and advanced Physics mentoring for CBSE, NEET, IIT JEE, JEE Advanced, IB, AP, IGCSE, ICSE and British Curriculum Physics.

Phone / WhatsApp: +91-9958461445kumarsirphysics@gmail.comkumarphysicsclasses.com
Scroll to Top