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Magnetic Field on the Axis of a Circular Current Loop

The magnetic field at a point on the axis of a circular loop of radius R carrying current I is B = mu0 I R^2/[2(R^2 + x^2)^(3/2)], directed along the axis of the loop.

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Student-friendly explanation

Each current element of the loop produces a magnetic field at the axial point. By symmetry, components perpendicular to the axis cancel, while components along the axis add. At the centre of the loop, x = 0, so the expression becomes B = mu0 I/(2R). For N turns, the field is multiplied by N.

How to write this in exams

  1. 1

    Start with the exact idea

    The magnetic field at a point on the axis of a circular loop of radius R carrying current I is B = mu0 I R^2/[2(R^2 + x^2)^(3/2)], directed along the axis of the loop.

  2. 2

    Then show how to use it

    Choose an element dl. Write dB using Biot-Savart law. Resolve dB into axial and transverse components. Cancel transverse components by symmetry. Integrate the axial component around the loop. Put x = 0 for centre if required.

  3. 3

    Add one concrete example

    A circular coil produces a strong field near its centre. Increasing the number of turns or current increases the field, while increasing radius decreases the centre field.

  4. 4

    Avoid this incomplete answer

    Using B = mu0 I/(2pi R) at the centre confuses the circular loop result with the long straight wire formula.

Definition

The magnetic field at a point on the axis of a circular loop of radius R carrying current I is B = mu0 I R^2/[2(R^2 + x^2)^(3/2)], directed along the axis of the loop.

Example

A circular coil produces a strong field near its centre. Increasing the number of turns or current increases the field, while increasing radius decreases the centre field.

Rule to remember

On-axis field: B = mu0 I R^2/[2(R^2 + x^2)^(3/2)]. At centre: B = mu0 I/(2R). For N turns: B = mu0 N I R^2/[2(R^2 + x^2)^(3/2)]. R and x are in metre, I in ampere, B in tesla. Use only for points lying on the axis of a circular loop.

Memory hook

For a loop, sideways parts cancel; axis parts collect.

Examples and method

Worked example

A circular coil of radius 0.10 m has 50 turns and carries 2 A. At the centre, B = mu0 N I/(2R) = (4pi x 10^-7 x 50 x 2)/(0.20) = 6.28 x 10^-4 T. The field is along the axis as decided by right-hand thumb rule.

Method to apply

Choose an element dl. Write dB using Biot-Savart law. Resolve dB into axial and transverse components. Cancel transverse components by symmetry. Integrate the axial component around the loop. Put x = 0 for centre if required.

Diagram support

Draw a circular loop in a plane, axis through centre O, point P at distance x from O, radius R, current direction, dB components, and resultant B along the axis.

How CBSE asks it

Commonly asked as a derivation, a centre-field numerical, or a conceptual question on why transverse components cancel.

Avoid common mistakes

Common confusion

Students often substitute diameter in place of radius or forget the power 3/2 in (R^2 + x^2)^(3/2).

Common wrong answer

Using B = mu0 I/(2pi R) at the centre confuses the circular loop result with the long straight wire formula.

Exam tip

In derivation questions, state the symmetry cancellation clearly. This is often where marks are awarded before the final formula.

Quick check

Why do only axial components of dB add for a circular current loop?

Only axial components add because for every current element, an opposite element produces an equal transverse component in the opposite direction. These transverse components cancel, while the components along the axis are in the same direction.

Answer writing and exam use

1-mark answer

The magnetic field at a point on the axis of a circular loop of radius R carrying current I is B = mu0 I R^2/[2(R^2 + x^2)^(3/2)], directed along the axis of the loop.

2-mark answer

The magnetic field at a point on the axis of a circular loop of radius R carrying current I is B = mu0 I R^2/[2(R^2 + x^2)^(3/2)], directed along the axis of the loop. On-axis field: B = mu0 I R^2/[2(R^2 + x^2)^(3/2)]. At centre: B = mu0 I/(2R). For N turns: B = mu0 N I R^2/[2(R^2 + x^2)^(3/2)]. R and x are in metre, I in ampere, B in tesla. Use only for points lying on the axis of a circular loop. A circular coil produces a strong field near its centre. Increasing the number of turns or current increases the field, while increasing radius decreases the centre field.

3-mark answer

Each current element of the loop produces a magnetic field at the axial point. By symmetry, components perpendicular to the axis cancel, while components along the axis add. At the centre of the loop, x = 0, so the expression becomes B = mu0 I/(2R). For N turns, the field is multiplied by N. On-axis field: B = mu0 I R^2/[2(R^2 + x^2)^(3/2)]. At centre: B = mu0 I/(2R). For N turns: B = mu0 N I R^2/[2(R^2 + x^2)^(3/2)]. R and x are in metre, I in ampere, B in tesla. Use only for points lying on the axis of a circular loop. A circular coil of radius 0.10 m has 50 turns and carries 2 A. At the centre, B = mu0 N I/(2R) = (4pi x 10^-7 x 50 x 2)/(0.20) = 6.28 x 10^-4 T. The field is along the axis as decided by right-hand thumb rule. Commonly asked as a derivation, a centre-field numerical, or a conceptual question on why transverse components cancel. Using B = mu0 I/(2pi R) at the centre confuses the circular loop result with the long straight wire formula.
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