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Biot Savart Law – Complete Guide for CBSE, NEET, IIT JEE and Advanced Physics

Biot-Savart Law is the fundamental method used to calculate magnetic field due to a current element, circular loop, semicircular loop, arc-shaped conductor, finite straight conductor, square loop, rectangular loop, triangular loop and mixed wire shapes.

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1. Introduction to Biot-Savart Law

Biot-Savart Law tells us how a small element of current produces a small magnetic field at an observation point. It is especially powerful when the conductor is made of arcs, straight pieces, loops or mixed shapes where each part can be added vectorially.

Current element Id𝐥
Circular loop
Semicircular loop
Arc conductor
Finite straight conductor
Square loop
Triangular loop
Mixed wire shapes

2. Statement and Symbol Meaning

Scalar formdB = μ₀/4π × I dℓ sinθ / r²
Vector formd𝐁 = μ₀/4π × I d𝐥 × r̂ / r²
Alternative vector formd𝐁 = μ₀/4π × I d𝐥 × 𝐫 / r³

Here I is current, d𝐥 is the directed length element, Id𝐥 is the current element, r is distance from current element to observation point, r̂ is unit vector along 𝐫, θ is the angle between d𝐥 and 𝐫, and d𝐁 is the small magnetic field.

Vector diagram of Biot-Savart Law
I dℓ P r, r̂ θ dB = ⊙ out Current-carrying wire element d𝐁 = μ₀/4π · I d𝐥 × r̂ / r²

3. Direction of Magnetic Field

The direction of d𝐁 is found by the right-hand rule for the cross product d𝐥 × r̂. Dot ⊙ means field coming out of the page; cross ⊗ means field going into the page.

Dot and cross convention
⊙ field out of paper ⊗ field into paper
Magnetic field around a straight current-carrying wire
I Right-hand thumb rule: thumb along current, curled fingers show B
Core LawCBSE • NEET • JEE

Statement of Biot-Savart Law

Vector diagram of Biot-Savart Law
I dℓ P r, r̂ θ dB = ⊙ out Current-carrying wire element d𝐁 = μ₀/4π · I d𝐥 × r̂ / r²
Given / Symbols:
I is current, d𝐥 is current element, r is distance to observation point, θ is angle between d𝐥 and 𝐫.
Assumptions:
Current is steady and geometry is fixed.
1Magnetic field due to a small current element is proportional to I, dℓ, sinθ and inversely proportional to r².
2Scalar form: dB = μ₀/4π × I dℓ sinθ / r².
3Vector form: d𝐁 = μ₀/4π × I d𝐥 × r̂ / r² = μ₀/4π × I d𝐥 × 𝐫 / r³.
4Direction is perpendicular to the plane containing d𝐥 and 𝐫.
Final Result:d𝐁 = μ₀/4π · I d𝐥 × r̂ / r²
Common mistake:
Do not drop sinθ; it becomes zero when d𝐥 is parallel to r.
Derivation 1CBSE • NEET • JEE

Magnetic Field at Centre of Circular Loop

Circular loop: magnetic field at centre
R O Anticlockwise current ⇒ B = ⊙ at centre
Given / Symbols:
Loop radius R, current I, centre O, N turns if required.
Assumptions:
Every current element is at distance R and angle θ = 90° from the radius vector.
1For each element, dB = μ₀/4π × I dℓ/R².
2All dB vectors have the same direction at centre, so B = ∫dB.
3B = μ₀I/4πR² ∫dℓ.
4For a full circle, ∫dℓ = 2πR.
5Therefore B = μ₀I/2R. For N turns multiply by N.
Final Result:B = μ₀NI / 2R
Common mistake:
Clockwise and anticlockwise currents give opposite dot/cross directions.
Derivation 2CBSE • NEET • JEE

Magnetic Field at Centre of Semicircular Loop

Semicircular loop at centre
R B = μ₀I / 4R
Given / Symbols:
Semicircular arc radius R carrying current I.
Assumptions:
Straight diameter part, if present, gives zero at centre because d𝐥 ∥ r.
1Arc length of semicircle is πR.
2dB = μ₀/4π × I dℓ/R².
3B = μ₀I/4πR² × πR.
4Hence B = μ₀I/4R. For N turns multiply by N.
Final Result:B = μ₀NI / 4R
Common mistake:
Do not use full-circle length 2πR for a semicircle.
Derivation 3CBSE • NEET • JEE

Magnetic Field at Centre of Circular Arc

Circular arc subtending angle α
α B = μ₀Iα / 4πR
Given / Symbols:
Arc radius R, current I, angle α in radians.
Assumptions:
Point is at the centre of curvature and all elements have θ = 90°.
1Arc length l = Rα.
2B = μ₀I/4πR² ∫dℓ.
3Substitute ∫dℓ = Rα.
4B = μ₀Iα/4πR.
5Full circle α = 2π, semicircle α = π, quarter circle α = π/2.
Final Result:B = μ₀Iα / 4πR
Common mistake:
α must be in radians, not degrees.
Derivation 4CBSE • NEET • JEE

Finite Straight Current-Carrying Conductor

Finite straight conductor
AB P a θ₁ θ₂ B = μ₀I/4πa (sinθ₁ + sinθ₂)
Given / Symbols:
Finite wire AB, current I, point P at perpendicular distance a, end angles θ₁ and θ₂.
Assumptions:
Wire is straight and current is steady.
1Using Biot-Savart Law and integrating along the wire gives the standard finite-wire result.
2The field direction is perpendicular to the page by right-hand thumb rule.
3For infinite wire, θ₁ = θ₂ = 90°.
4For semi-infinite wire, one angle is 90° and the other is 0°.
Final Result:B = μ₀I/4πa (sinθ₁ + sinθ₂)
Common mistake:
Use perpendicular distance a, not the distance from P to an end.
Derivation 5CBSE • NEET • JEE

Magnetic Field at Centre of Square Loop

Square loop at centre
O a/2 B = 2√2 μ₀I / πa
Given / Symbols:
Square loop side a, centre O, current I.
Assumptions:
Each side is a finite straight conductor; all four contributions have the same direction at O.
1Distance from centre to each side is a/2.
2For one side, θ₁ = θ₂ = 45°.
3B₁ = μ₀I/[4π(a/2)](sin45° + sin45°).
4B₁ = μ₀I√2/2πa.
5Total field B = 4B₁.
Final Result:B = 2√2 μ₀I / πa
Common mistake:
Do not calculate only one side and forget to multiply by four.
Derivation 6CBSE • NEET • JEE

Wire of Length L Formed into N Turns

Single loop and N-turn loop comparison
One turn: R = L/2π N turns: R = L/2πN Same wire length L, same current I
Given / Symbols:
Wire length L, current I, number of turns N.
Assumptions:
Same wire length is used and current remains the same.
1For one loop, L = 2πR, so R = L/2π.
2B₁ = μ₀I/2R = μ₀πI/L.
3For N turns, L = N × 2πR, so R = L/2πN.
4B_N = μ₀NI/2R = μ₀πN²I/L.
5Therefore B_N/B₁ = N².
Final Result:B_N = μ₀πN²I/L, so B_N/B₁ = N²
Common mistake:
This result assumes current remains same after rewinding.

10. Important Shapes Using Biot-Savart Law

A. Equilateral triangular loop

Each side contributes by the finite-wire formula. At the centre the perpendicular distance is a√3/6 and both end angles are 60°. Add three identical contributions.

B = 3(μ₀I/4πd)(sin60° + sin60°), d = a√3/6

B. Rectangular loop

Treat the rectangle as four finite conductors. The two long sides and two short sides are added with the same field direction at the centre.

B = 2B_long + 2B_short

C. Quarter circular loop

Only the arc contributes at the centre; radial joining wires have dℓ parallel to r, so their contribution is zero.

B = μ₀I/8R

D. Three-quarter circular loop

Use α = 3π/2 in the arc formula. Direction is decided by the sense of current.

B = 3μ₀I/8R

E. Straight wire joined with a semicircular arc

Add the semicircular arc field and the finite straight-wire contribution, if the straight part is not radial with respect to the point.

B_total = B_arc + B_straight

F. Two concentric semicircular arcs

The two arcs usually carry current in opposite circular senses at the common centre; subtract magnitudes according to directions.

B = μ₀I/4(1/R₁ - 1/R₂) with sign by direction

G. Loop made from wire of length L

For one circular loop R = L/2π. Substitute in B = μ₀I/2R.

B = μ₀πI/L
Equilateral triangular loop
O At centre: add field of three finite sides
Rectangular loop carrying current
length l breadth b Use finite-wire formula for each side
Two concentric semicircular arcs
O Radial parts give zero at O because dℓ ∥ r

12. Limitations of Biot-Savart Law

  • It is mainly useful for steady currents.
  • Direct integration becomes difficult for complex current distributions.
  • For high symmetry, Ampere's Law is often faster.
  • It does not directly handle time-varying fields in school-level form.
  • The geometry of d𝐥 and observation point must be defined carefully.

13. Common Student Mistakes

Mistake 1Forgetting sinθ

Correction: Always draw geometry first, mark current direction, mark r vector, then apply d𝐥 × r̂.

Mistake 2Taking wrong d𝐥 direction

Correction: Always draw geometry first, mark current direction, mark r vector, then apply d𝐥 × r̂.

Mistake 3Confusing r and R

Correction: Always draw geometry first, mark current direction, mark r vector, then apply d𝐥 × r̂.

Mistake 4Using α in degrees instead of radians

Correction: Always draw geometry first, mark current direction, mark r vector, then apply d𝐥 × r̂.

Mistake 5Adding radial wire contribution

Correction: Always draw geometry first, mark current direction, mark r vector, then apply d𝐥 × r̂.

Mistake 6Wrong dot/cross direction

Correction: Always draw geometry first, mark current direction, mark r vector, then apply d𝐥 × r̂.

Mistake 7Wrong finite-wire angles

Correction: Always draw geometry first, mark current direction, mark r vector, then apply d𝐥 × r̂.

Mistake 8Missing N turns factor

Correction: Always draw geometry first, mark current direction, mark r vector, then apply d𝐥 × r̂.

Mistake 9Confusing L with circumference

Correction: Always draw geometry first, mark current direction, mark r vector, then apply d𝐥 × r̂.

Mistake 10Reversing clockwise/anticlockwise field

Correction: Always draw geometry first, mark current direction, mark r vector, then apply d𝐥 × r̂.

14. Complete Question Bank With Solutions

A. CBSE Questions

CBSE Theory Question 1Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 2Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 3Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 4Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 5Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 6Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 7Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 8Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 9Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 10Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 11Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 12Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 13Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 14Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 15Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 16Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 17Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 18Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 19Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 20Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 21Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 22Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 23Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 24Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Theory Question 25Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 1Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 2Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 3Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 4Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 5Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 6Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 7Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 8Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 9Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 10Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 11Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 12Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 13Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 14Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 15Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 16Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 17Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 18Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 19Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 20Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 21Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 22Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 23Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 24Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Derivation Derivation 25Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 1Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 2Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 3Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 4Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 5Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 6Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 7Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 8Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 9Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 10Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 11Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 12Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 13Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 14Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 15Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 16Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 17Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 18Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 19Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 20Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 21Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 22Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 23Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 24Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Numerical Numerical 25Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Case Study Case 1Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Case Study Case 2Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Case Study Case 3Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Case Study Case 4Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Case Study Case 5Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Case Study Case 6Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Case Study Case 7Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Case Study Case 8Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Case Study Case 9Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
CBSE Case Study Case 10Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.

B. NEET Tough MCQs

NEET Q1For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q2At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q3A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q4The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q5For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q6Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q7If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q8For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q9The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q10If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q11 Variant 2: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q12 Variant 2: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q13 Variant 2: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q14 Variant 2: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q15 Variant 2: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q16 Variant 2: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q17 Variant 2: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q18 Variant 2: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q19 Variant 2: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q20 Variant 2: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q21 Variant 3: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q22 Variant 3: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q23 Variant 3: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q24 Variant 3: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q25 Variant 3: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q26 Variant 3: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q27 Variant 3: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q28 Variant 3: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q29 Variant 3: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q30 Variant 3: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q31 Variant 4: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q32 Variant 4: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q33 Variant 4: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q34 Variant 4: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q35 Variant 4: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q36 Variant 4: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q37 Variant 4: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q38 Variant 4: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q39 Variant 4: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q40 Variant 4: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q41 Variant 5: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q42 Variant 5: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q43 Variant 5: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q44 Variant 5: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q45 Variant 5: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q46 Variant 5: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q47 Variant 5: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q48 Variant 5: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q49 Variant 5: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q50 Variant 5: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q51 Variant 6: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q52 Variant 6: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q53 Variant 6: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q54 Variant 6: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q55 Variant 6: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q56 Variant 6: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q57 Variant 6: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q58 Variant 6: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q59 Variant 6: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q60 Variant 6: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q61 Variant 7: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q62 Variant 7: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q63 Variant 7: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q64 Variant 7: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q65 Variant 7: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q66 Variant 7: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q67 Variant 7: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q68 Variant 7: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q69 Variant 7: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q70 Variant 7: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q71 Variant 8: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q72 Variant 8: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q73 Variant 8: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q74 Variant 8: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
NEET Q75 Variant 8: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.

C. JEE Main MCQs

JEE Main Q1For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q2At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q3A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q4The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q5For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q6Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q7If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q8For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q9The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q10If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q11 Variant 2: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q12 Variant 2: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q13 Variant 2: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q14 Variant 2: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q15 Variant 2: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q16 Variant 2: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q17 Variant 2: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q18 Variant 2: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q19 Variant 2: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q20 Variant 2: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q21 Variant 3: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q22 Variant 3: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q23 Variant 3: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q24 Variant 3: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q25 Variant 3: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q26 Variant 3: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q27 Variant 3: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q28 Variant 3: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q29 Variant 3: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q30 Variant 3: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q31 Variant 4: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q32 Variant 4: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q33 Variant 4: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q34 Variant 4: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q35 Variant 4: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q36 Variant 4: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q37 Variant 4: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q38 Variant 4: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q39 Variant 4: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q40 Variant 4: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q41 Variant 5: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q42 Variant 5: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q43 Variant 5: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q44 Variant 5: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q45 Variant 5: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q46 Variant 5: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q47 Variant 5: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q48 Variant 5: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q49 Variant 5: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q50 Variant 5: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q51 Variant 6: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q52 Variant 6: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q53 Variant 6: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q54 Variant 6: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q55 Variant 6: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q56 Variant 6: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q57 Variant 6: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q58 Variant 6: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q59 Variant 6: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q60 Variant 6: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q61 Variant 7: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q62 Variant 7: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q63 Variant 7: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q64 Variant 7: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q65 Variant 7: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q66 Variant 7: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q67 Variant 7: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q68 Variant 7: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q69 Variant 7: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q70 Variant 7: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q71 Variant 8: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q72 Variant 8: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q73 Variant 8: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q74 Variant 8: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Main Q75 Variant 8: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.

D. JEE Advanced Questions

JEE Advanced Single Correct Q1For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q2At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q3A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q4The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q5For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q6Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q7If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q8For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q9The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q10If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q11 Variant 2: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q12 Variant 2: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q13 Variant 2: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q14 Variant 2: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q15 Variant 2: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q16 Variant 2: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q17 Variant 2: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q18 Variant 2: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q19 Variant 2: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q20 Variant 2: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q21 Variant 3: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q22 Variant 3: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q23 Variant 3: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q24 Variant 3: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q25 Variant 3: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q26 Variant 3: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q27 Variant 3: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q28 Variant 3: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q29 Variant 3: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Single Correct Q30 Variant 3: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q1For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q2At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q3A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q4The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q5For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q6Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q7If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q8For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q9The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q10If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q11 Variant 2: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q12 Variant 2: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q13 Variant 2: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q14 Variant 2: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q15 Variant 2: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q16 Variant 2: Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q17 Variant 2: If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q18 Variant 2: For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q19 Variant 2: The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Multiple Correct Q20 Variant 2: If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q1For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q2At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q3A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q4The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q5For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q6Radial wires connected to an arc contribute at the centre
  1. maximum
  2. minimum
  3. zero
  4. same as arc
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For radial parts d𝐥 is parallel to r, so d𝐥 × r̂ = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q7If same wire is made into N turns and current is unchanged, centre field becomes
  1. N times
  2. N² times
  3. 1/N times
  4. unchanged
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Radius reduces by N and turn factor also gives N.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q8For an infinitely long straight wire, B at distance a is
  1. μ₀I/4πa
  2. μ₀I/2πa
  3. μ₀I/πa
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Both end angles are 90° in finite wire formula.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q9The SI unit of μ₀ is
  1. T m A⁻¹
  2. T A m⁻¹
  3. N C⁻¹
  4. Wb m
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: From B = μ₀I/2πa, μ₀ has unit T m A⁻¹.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q10If current direction in a loop is reversed, magnetic field at centre
  1. doubles
  2. becomes zero
  3. reverses direction
  4. is unchanged
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Magnitude remains same but vector direction reverses.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q11 Variant 2: For a current element, the magnetic field is zero when θ is
  1. 0°
  2. 30°
  3. 60°
  4. 90°
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Because dB ∝ sinθ; sin0° = 0.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q12 Variant 2: At the centre of a circular loop of radius R carrying current I, B is
  1. μ₀I/4R
  2. μ₀I/2R
  3. μ₀I/R
  4. 2μ₀I/R
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Integrating all current elements gives B = μ₀I/2R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q13 Variant 2: A semicircular arc of radius R gives field at centre
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. zero
Correct Answer: B
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: For α = π, B = μ₀Iα/4πR = μ₀I/4R.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q14 Variant 2: The field due to a quarter circular arc at centre is
  1. μ₀I/2R
  2. μ₀I/4R
  3. μ₀I/8R
  4. μ₀I/16R
Correct Answer: C
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Use α = π/2 in B = μ₀Iα/4πR.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Integer Type Q15 Variant 2: For a square loop of side a, field at centre is
  1. 2√2 μ₀I/πa
  2. √2 μ₀I/πa
  3. μ₀I/2a
  4. μ₀I/πa
Correct Answer: A
Concept Tested: Biot-Savart Law and standard magnetic field results.
Detailed Explanation: Add equal fields of four finite sides.
Common Student Mistake: Students often use a memorised formula without checking angle, direction and geometry.
JEE Advanced Matrix Match Matrix 1Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Matrix Match Matrix 2Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Matrix Match Matrix 3Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Matrix Match Matrix 4Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Matrix Match Matrix 5Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Matrix Match Matrix 6Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Matrix Match Matrix 7Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Matrix Match Matrix 8Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Matrix Match Matrix 9Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Matrix Match Matrix 10Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Paragraph Paragraph 1Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Paragraph Paragraph 2Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Paragraph Paragraph 3Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Paragraph Paragraph 4Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Paragraph Paragraph 5Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Paragraph Paragraph 6Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Paragraph Paragraph 7Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Paragraph Paragraph 8Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Paragraph Paragraph 9Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
JEE Advanced Paragraph Paragraph 10Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.

E. ICSE Physics Questions

ICSE Question 1Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 2Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 3Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 4Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 5Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 6Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 7Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 8Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 9Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 10Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 11Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 12Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 13Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 14Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 15Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 16Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 17Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 18Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 19Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 20Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 21Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 22Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 23Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 24Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
ICSE Question 25Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.

F. IGCSE Physics Questions

IGCSE Question 1Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 2Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 3Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 4Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 5Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 6Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 7Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 8Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 9Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 10Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 11Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 12Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 13Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 14Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 15Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 16Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 17Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 18Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 19Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 20Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 21Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 22Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 23Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 24Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IGCSE Question 25Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.

G. IB Physics Questions

IB Question 1Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 2Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 3Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 4Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 5Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 6Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 7Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 8Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 9Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 10Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 11Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 12Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 13Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 14Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 15Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 16Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 17Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 18Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 19Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 20Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 21Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 22Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 23Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 24Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
IB Question 25Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.

H. British Curriculum / A-Level Physics

A-Level Question 1Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 2Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 3Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 4Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 5Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 6Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 7Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 8Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 9Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 10Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 11Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 12Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 13Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 14Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 15Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 16Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 17Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 18Write a complete answer to explain right-hand rule.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 19Write a complete answer to derive circular loop field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 20Write a complete answer to derive finite wire formula.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 21Write a complete answer to find field for an arc.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 22Write a complete answer to discuss limitations.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 23Write a complete answer to compare Biot-Savart and Ampere's Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 24Write a complete answer to explain why radial wires give zero field.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.
A-Level Question 25Write a complete answer to state Biot-Savart Law.
Answer: Use the law d𝐁 = μ₀/4π · I d𝐥 × r̂/r², explain geometry, direction and integration where required. For derivations, define symbols first, state assumptions and then integrate the contribution of all current elements. Exam note: Draw a labelled diagram before the derivation.

15. Final Revision Sheet

Biot-Savart LawdB = μ₀I dℓ sinθ / 4πr²
Circular LoopB = μ₀NI / 2R
SemicircleB = μ₀NI / 4R
ArcB = μ₀Iα / 4πR
Finite WireB = μ₀I/4πa(sinθ₁+sinθ₂)
Square LoopB = 2√2 μ₀I / πa

Direction rule: Use right-hand thumb rule for wire and right-hand curl rule for loops. Dot means out; cross means into the page.

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