Magnetic Field on Axis of Circular Current Loop - Complete Physics Guide

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Magnetic Field on the Axis of a Circular Current Carrying Loop

A complete premium Physics guide by Kumar Sir covering geometry, Biot-Savart Law, vector resolution, symmetry cancellation, derivation, special cases, graph analysis, applications and exam-oriented questions.

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1. Introduction

An axial point is a point lying on the straight line passing through the centre of a circular loop and perpendicular to its plane. This line is called the axis of the circular current loop. The derivation is very important because it combines Biot-Savart Law, vector resolution, symmetry and integration. It is repeatedly asked in CBSE derivations, NEET formula-based MCQs and JEE conceptual problems.

CBSE derivation
NEET formula MCQs
JEE ratio problems
JEE Advanced maximum field
IB data interpretation
IGCSE diagrams
ICSE board answers
A-Level calculus

2. Geometry of the Problem

Geometry of magnetic field on the axis of a circular current loop
X'X-axis I O P OP = x R r = √(R²+x²) I dℓ tangent B_net dB dB cosθ axial dB sinθ transverse θ Correct geometry: dℓ is tangential, r goes from current element to P, net B is axial

The circular loop has radius R and current I. Point P is on the axis at distance x from centre O. For every current element I dℓ, the distance to P is r = √(R²+x²).

3. Biot-Savart Law for the Loop

Biot-Savart LawdB = μ₀/4π × I dℓ sin90° / r²
Since dℓ ⟂ rdB = μ₀I dℓ / 4πr²
Geometryr² = R² + x²

Here μ₀ is permeability of free space, I is current, dℓ is the small length element, r is the distance between current element and observation point, and dB is the small magnetic field produced by that element.

4. Vector Resolution and Cancellation

Vector resolution of dB into axial and perpendicular components
axis P dB axial component = dB cosθ transverse = dB sinθ θ dB is resolved at P: axial part adds, transverse part cancels by symmetry
Cancellation of transverse components by symmetry
Idℓ₁Idℓ₂ P B_net dB₁dB₂ axial components add transversetransverse ΣB_transverse = 0 ΣB_axial ≠ 0, so net magnetic field is along the axis

The magnetic field due to one element is resolved into axial and perpendicular components. Due to symmetry, every element has a diametrically opposite element. Their perpendicular components are equal and opposite, so Σ(dBᵧ) = 0 and Σ(dBᶻ) = 0. Only axial components survive, so B = Σ(dBₓ).

5. Complete Step-by-Step Derivation

Geometry of magnetic field on the axis of a circular current loop
X'X-axis I O P OP = x R r = √(R²+x²) I dℓ tangent B_net dB dB cosθ axial dB sinθ transverse θ Correct geometry: dℓ is tangential, r goes from current element to P, net B is axial
Cancellation of transverse components by symmetry
Idℓ₁Idℓ₂ P B_net dB₁dB₂ axial components add transversetransverse ΣB_transverse = 0 ΣB_axial ≠ 0, so net magnetic field is along the axis
Given: Radius = R, current = I, axial distance OP = x, observation point = P.
Key idea: The loop is symmetric. Perpendicular components cancel and axial components add.
1Consider a circular loop of radius R carrying current I. Point P lies on the axis at distance OP = x.
2For a small current element I dℓ, the position vector from the element to P has magnitude r = √(R² + x²).
3The current element dℓ is tangential to the loop and is perpendicular to r, so sin90° = 1.
4By Biot-Savart Law: dB = μ₀/4π × I dℓ/r².
5The magnetic field dB is resolved into axial and perpendicular components.
6For diametrically opposite elements, perpendicular components cancel by symmetry.
7Only axial components add: dBₓ = dB cosθ.
8From geometry, cosθ = R/r.
9Therefore dBₓ = μ₀/4π × I dℓ/r² × R/r.
10So dBₓ = μ₀IR dℓ / 4πr³.
11Since r is constant for the entire loop, B = ∫dBₓ = μ₀IR/4πr³ ∫dℓ.
12For a complete circular loop, ∫dℓ = 2πR.
13Therefore B = μ₀IR/4πr³ × 2πR.
14Hence B = μ₀IR² / 2r³.
15Substitute r² = R² + x², so r³ = (R² + x²)³ᐟ².
Final ResultB = μ₀IR² / 2(R² + x²)³ᐟ²

6. Special Cases

Special case: magnetic field at the centre of the loop
O = P R B at centre ⊙ out of page x = 0 ⇒ B = μ₀I / 2R for one turn
At centre, x = 0B = μ₀I / 2R

This is the most important CBSE special case.

Far away, x >> RB ≈ μ₀IR² / 2x³

The loop behaves like a magnetic dipole.

Near centreB decreases slowly and symmetrically

The field is maximum at O and falls on both sides.

7. Maximum Magnetic Field Position

1B = μ₀IR² / 2(R²+x²)³ᐟ².
2Let C = μ₀IR²/2, so B = C(R²+x²)⁻³ᐟ².
3dB/dx = C × (-3/2)(R²+x²)⁻⁵ᐟ² × 2x.
4dB/dx = -3Cx(R²+x²)⁻⁵ᐟ².
5Set dB/dx = 0, so x = 0.
ConclusionMaximum magnetic field occurs at the centre of the loop.

8. Graphical Analysis

Graph of B versus x on the axis
xB Maximum at x = 0 O B = μ₀IR² / 2(R²+x²)³ᐟ²

The graph is symmetric about x = 0. The maximum value occurs at the centre. As |x| increases, the denominator (R²+x²)³ᐟ² increases, so B decreases.

9. Physical Interpretation

Perpendicular components cancel because opposite elements produce opposite sideways fields.
Axial components add because they point in the same direction.
Field is maximum at centre because distance from every element is minimum in the symmetric axial sense.
Field decreases away from centre because r increases and dB follows inverse-square geometry plus component reduction.

10. Most Important Applications

Application: Helmholtz coil idea
common axis nearlyuniformB Same current direction in both coils gives a nearly uniform central field
Helmholtz coils
Electromagnets
MRI machines
Magnetic field generation
Circular current loops
Particle beam systems
Experimental Physics

11. Types of Questions Asked

CBSE: direct derivation, special case x=0, diagram and graph questions.

NEET: formula MCQs, radius-distance ratio, centre field and N-turn questions.

JEE Main: numerical calculations, comparison and multi-concept questions.

JEE Advanced: differentiation, maximum field, multiple-loop systems and reasoning.

IB: structured answers, real-life applications and data interpretation.

IGCSE: simplified applications and diagram labels.

ICSE: board derivation and formula substitution.

A-Level: calculus-based derivation and advanced applications.

12. Question Bank With Solutions

CBSE Theory Questions

CBSE Theory Question 1Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 2Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 3Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 4Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 5Draw and label the geometry used in the derivation.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 6Explain why this derivation is important for board and competitive exams.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 7Find the field ratio at x = R and x = 0.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 8Explain the shape of the B-x graph.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 9Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 10Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 11Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 12Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 13Draw and label the geometry used in the derivation.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 14Explain why this derivation is important for board and competitive exams.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 15Find the field ratio at x = R and x = 0.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 16Explain the shape of the B-x graph.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 17Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 18Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 19Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Theory Question 20Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.

CBSE Numericals

CBSE Numerical Numerical 1Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 2Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 3Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 4Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 5Draw and label the geometry used in the derivation.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 6Explain why this derivation is important for board and competitive exams.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 7Find the field ratio at x = R and x = 0.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 8Explain the shape of the B-x graph.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 9Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 10Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 11Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 12Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 13Draw and label the geometry used in the derivation.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 14Explain why this derivation is important for board and competitive exams.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 15Find the field ratio at x = R and x = 0.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 16Explain the shape of the B-x graph.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 17Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 18Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 19Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Numerical Numerical 20Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.

CBSE Derivation Questions

CBSE Derivation Derivation 1Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Derivation Derivation 2Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Derivation Derivation 3Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Derivation Derivation 4Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Derivation Derivation 5Draw and label the geometry used in the derivation.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Derivation Derivation 6Explain why this derivation is important for board and competitive exams.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Derivation Derivation 7Find the field ratio at x = R and x = 0.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Derivation Derivation 8Explain the shape of the B-x graph.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Derivation Derivation 9Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
CBSE Derivation Derivation 10Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.

NEET MCQs

NEET Q1At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q2The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
NEET Q3For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q4The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q5In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q6For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
NEET Q7At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
NEET Q8The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
NEET Q9Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q10The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q11Variant 2: At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q12Variant 2: The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
NEET Q13Variant 2: For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q14Variant 2: The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q15Variant 2: In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q16Variant 2: For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
NEET Q17Variant 2: At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
NEET Q18Variant 2: The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
NEET Q19Variant 2: Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q20Variant 2: The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q21Variant 3: At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q22Variant 3: The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
NEET Q23Variant 3: For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q24Variant 3: The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q25Variant 3: In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q26Variant 3: For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
NEET Q27Variant 3: At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
NEET Q28Variant 3: The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
NEET Q29Variant 3: Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q30Variant 3: The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q31Variant 4: At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q32Variant 4: The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
NEET Q33Variant 4: For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q34Variant 4: The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q35Variant 4: In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q36Variant 4: For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
NEET Q37Variant 4: At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
NEET Q38Variant 4: The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
NEET Q39Variant 4: Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q40Variant 4: The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q41Variant 5: At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q42Variant 5: The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
NEET Q43Variant 5: For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q44Variant 5: The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q45Variant 5: In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q46Variant 5: For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
NEET Q47Variant 5: At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
NEET Q48Variant 5: The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
NEET Q49Variant 5: Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
NEET Q50Variant 5: The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.

JEE Main MCQs

JEE Main Q1At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q2The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q3For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q4The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q5In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q6For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q7At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q8The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q9Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q10The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q11Variant 2: At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q12Variant 2: The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q13Variant 2: For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q14Variant 2: The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q15Variant 2: In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q16Variant 2: For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q17Variant 2: At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q18Variant 2: The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q19Variant 2: Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q20Variant 2: The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q21Variant 3: At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q22Variant 3: The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q23Variant 3: For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q24Variant 3: The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q25Variant 3: In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q26Variant 3: For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q27Variant 3: At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q28Variant 3: The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q29Variant 3: Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q30Variant 3: The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q31Variant 4: At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q32Variant 4: The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q33Variant 4: For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q34Variant 4: The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q35Variant 4: In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q36Variant 4: For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q37Variant 4: At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q38Variant 4: The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q39Variant 4: Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q40Variant 4: The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q41Variant 5: At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q42Variant 5: The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q43Variant 5: For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q44Variant 5: The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q45Variant 5: In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q46Variant 5: For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q47Variant 5: At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q48Variant 5: The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q49Variant 5: Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
JEE Main Q50Variant 5: The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.

JEE Advanced Single Correct

JEE Advanced Single Correct Q1At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q2The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q3For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q4The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q5In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q6For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q7At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q8The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q9Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q10The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q11Variant 2: At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q12Variant 2: The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q13Variant 2: For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q14Variant 2: The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q15Variant 2: In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q16Variant 2: For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q17Variant 2: At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q18Variant 2: The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q19Variant 2: Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Single Correct Q20Variant 2: The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.

JEE Advanced Multiple Correct

JEE Advanced Multiple Correct Q1At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Multiple Correct Q2The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Multiple Correct Q3For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Multiple Correct Q4The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Multiple Correct Q5In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Multiple Correct Q6For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Multiple Correct Q7At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Multiple Correct Q8The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Multiple Correct Q9Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Multiple Correct Q10The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.

JEE Advanced Integer Type

JEE Advanced Integer Q1At the centre of a loop, x is
  1. R
  2. 0
  3. ∞
  4. R/2
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: At centre, the axial distance OP is zero, so B = μ₀I/2R.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Integer Q2The axial field of a circular loop is proportional to
  1. R
  2. R²/(R²+x²)³ᐟ²
  3. x²/R²
  4. 1/x
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: The derived formula is B = μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Integer Q3For x >> R, the magnetic field varies approximately as
  1. 1/x
  2. 1/x²
  3. 1/x³
  4. x²
Correct Answer: C
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: For x much larger than R, (R²+x²)³ᐟ² ≈ x³, so B ∝ 1/x³.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Integer Q4The magnetic field on the axis is maximum at
  1. x = R
  2. x = R/2
  3. x = 0
  4. x = ∞
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Differentiating B with respect to x gives dB/dx = 0 at x = 0, and field decreases away from centre.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Integer Q5In the derivation, perpendicular components cancel because of
  1. Ohm's law
  2. symmetry
  3. resistance
  4. capacitance
Correct Answer: B
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Diametrically opposite current elements produce equal and opposite transverse components.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Integer Q6For N turns, axial field becomes
  1. N times
  2. N² times always
  3. 1/N times
  4. unchanged
Correct Answer: A
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Each turn contributes the same field, so B_N = N μ₀IR²/2(R²+x²)³ᐟ².
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Integer Q7At x = R, B/B_center equals
  1. 1/2
  2. 1/(2√2)
  3. 1/√2
  4. 2
Correct Answer: B
Difficulty: Difficult
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: B = μ₀IR²/2(2R²)³ᐟ² = μ₀I/(2·2³ᐟ²R). Divide by μ₀I/2R to get 1/(2√2).
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Integer Q8The correct value of r in the geometry is
  1. R+x
  2. R-x
  3. √(R²+x²)
  4. R²+x²
Correct Answer: C
Difficulty: Easy
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: R and x are perpendicular, so Pythagoras gives r = √(R²+x²).
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Integer Q9Which components survive after summing over the whole loop?
  1. Only transverse
  2. Only axial
  3. Both always cancel
  4. None
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Transverse components cancel; axial components add.
Common Mistake: Do not use centre formula unless x = 0.
JEE Advanced Integer Q10The field at P is directed along the axis for a complete loop because
  1. charge is zero
  2. loop symmetry removes non-axial components
  3. current is zero
  4. radius changes
Correct Answer: B
Difficulty: Medium
Concept Tested: Magnetic field on axis of circular loop.
Detailed Solution: Every element has an opposite partner canceling sideways components.
Common Mistake: Do not use centre formula unless x = 0.

IB Structured Questions

IB Structured 1Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 2Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 3Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 4Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 5Draw and label the geometry used in the derivation.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 6Explain why this derivation is important for board and competitive exams.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 7Find the field ratio at x = R and x = 0.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 8Explain the shape of the B-x graph.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 9Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 10Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 11Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 12Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 13Draw and label the geometry used in the derivation.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 14Explain why this derivation is important for board and competitive exams.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 15Find the field ratio at x = R and x = 0.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 16Explain the shape of the B-x graph.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 17Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 18Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 19Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IB Structured 20Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.

IGCSE Questions

IGCSE Question 1Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 2Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 3Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 4Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 5Draw and label the geometry used in the derivation.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 6Explain why this derivation is important for board and competitive exams.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 7Find the field ratio at x = R and x = 0.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 8Explain the shape of the B-x graph.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 9Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 10Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 11Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 12Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 13Draw and label the geometry used in the derivation.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 14Explain why this derivation is important for board and competitive exams.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 15Find the field ratio at x = R and x = 0.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 16Explain the shape of the B-x graph.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 17Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 18Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 19Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
IGCSE Question 20Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.

ICSE Questions

ICSE Question 1Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 2Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 3Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 4Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 5Draw and label the geometry used in the derivation.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 6Explain why this derivation is important for board and competitive exams.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 7Find the field ratio at x = R and x = 0.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 8Explain the shape of the B-x graph.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 9Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 10Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 11Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 12Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 13Draw and label the geometry used in the derivation.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 14Explain why this derivation is important for board and competitive exams.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 15Find the field ratio at x = R and x = 0.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 16Explain the shape of the B-x graph.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 17Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 18Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 19Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
ICSE Question 20Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.

A-Level Questions

A-Level Question 1Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 2Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 3Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 4Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 5Draw and label the geometry used in the derivation.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 6Explain why this derivation is important for board and competitive exams.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 7Find the field ratio at x = R and x = 0.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 8Explain the shape of the B-x graph.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 9Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 10Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 11Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 12Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 13Draw and label the geometry used in the derivation.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 14Explain why this derivation is important for board and competitive exams.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 15Find the field ratio at x = R and x = 0.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 16Explain the shape of the B-x graph.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 17Explain why transverse components cancel for a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 18Derive the magnetic field at an axial point of a circular current loop.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 19Show that at the centre the formula reduces to B = μ₀I/2R.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.
A-Level Question 20Discuss the far-away approximation of the axial magnetic field.
Solution: Start from B = μ₀IR²/2(R²+x²)³ᐟ². Use geometry, symmetry, component resolution and the required special condition. For numerical questions, substitute values carefully and keep SI units. For derivations, draw the loop, axis, P, R, x, r, dB and its axial component.

13. Case Study Section

Case Study 1Helmholtz coil based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 2MRI magnetic field based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 3magnetic sensor based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 4school laboratory loop based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 5particle beam steering based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 6Helmholtz coil based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 7MRI magnetic field based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 8magnetic sensor based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 9school laboratory loop based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 10particle beam steering based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 11Helmholtz coil based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 12MRI magnetic field based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 13magnetic sensor based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 14school laboratory loop based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.
Case Study 15particle beam steering based axial field problem
Scenario: A circular coil is used to generate a controlled magnetic field on its axis. Data: radius R, current I, number of turns N and axial distance x are given. Questions: find B at centre, B at distance x, compare with centre field and explain symmetry. Solution: Use B = Nμ₀IR²/2(R²+x²)³ᐟ². At centre set x = 0. For far points use x >> R, giving B ≈ Nμ₀IR²/2x³.

14. Common Student Mistakes

Mistake 1Using sinθ incorrectly in Biot-Savart Law.

Correction: Draw the full geometry first and then apply B = μ₀IR²/2(R²+x²)³ᐟ².

Mistake 2Forgetting that dℓ is perpendicular to r.

Correction: Draw the full geometry first and then apply B = μ₀IR²/2(R²+x²)³ᐟ².

Mistake 3Resolving dB in the wrong direction.

Correction: Draw the full geometry first and then apply B = μ₀IR²/2(R²+x²)³ᐟ².

Mistake 4Forgetting symmetry cancellation.

Correction: Draw the full geometry first and then apply B = μ₀IR²/2(R²+x²)³ᐟ².

Mistake 5Skipping integration over the full loop.

Correction: Draw the full geometry first and then apply B = μ₀IR²/2(R²+x²)³ᐟ².

Mistake 6Using centre formula for axial point.

Correction: Draw the full geometry first and then apply B = μ₀IR²/2(R²+x²)³ᐟ².

Mistake 7Confusing R, r and x.

Correction: Draw the full geometry first and then apply B = μ₀IR²/2(R²+x²)³ᐟ².

Mistake 8Thinking maximum occurs away from centre.

Correction: Draw the full geometry first and then apply B = μ₀IR²/2(R²+x²)³ᐟ².

Mistake 9Drawing graph as one-sided only.

Correction: Draw the full geometry first and then apply B = μ₀IR²/2(R²+x²)³ᐟ².

Mistake 10Forgetting N turns factor.

Correction: Draw the full geometry first and then apply B = μ₀IR²/2(R²+x²)³ᐟ².

15. Final Revision Sheet

Axial fieldB = μ₀IR² / 2(R²+x²)³ᐟ²
N turnsB = μ₀NIR² / 2(R²+x²)³ᐟ²
At centreB = μ₀NI / 2R
Far pointB ≈ μ₀NIR² / 2x³
Maximumx = 0
SymmetryΣ transverse components = 0

Quick revision: Draw loop, axis, P, R, x, r, dB. Use Biot-Savart Law. Resolve dB. Cancel perpendicular components. Integrate axial components over 2πR.

Need Personal Help?

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