JEE Mains Physics · Moving Charges and Magnetism

Circular Loops and Coils

A coil of N turns and radius R gives μ₀NI/2R at its centre; along its axis the field falls as R³/(R² + x²)^(3/2), so most questions are a ratio to the centre value.

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Why this matters

Twenty PYQs, sixteen of them multiple choice, and three from 2026. Ten work at the centre of a coil: three rewind the same wire into a different number of turns, three put two coils in perpendicular planes with a common centre, two use the formula directly, one asks for the energy stored there and one for a flat spiral. Ten go along the axis, and eight of those are a ratio of the field at two points on the axis, usually with the centre as one of them.

Concept 1 of 2: Field at the centre of a circular coil

Every piece of a circular loop is at the same distance R from the centre and at right angles to the line joining it, so all the Biot–Savart pieces add in the same direction. N turns give N times one turn. If the same length of wire is rewound into more turns, each turn is smaller as well, so the field grows faster than the number of turns.

Definition

Centre of a coil of N turns: $B = \dfrac{\mu_0 N I}{2R}$ , along the axis (right-hand rule: fingers along the current, thumb gives B).
Rewinding a fixed length L of wire: $L = N(2\pi R)$ , so $R \propto 1/N$ and $B \propto N/R \propto N^{2}$ .
Two coils with a common centre and perpendicular planes: their fields are perpendicular, so $B = \sqrt{B_1^{2} + B_2^{2}}$ .
Flat spiral wound evenly from radius a to radius b: treat it as rings, with $\dfrac{N}{b - a}$ turns per unit radius, and add $\dfrac{\mu_0 I\,dN}{2r}$ from a to b; the result carries a logarithm, $\ln(b/a)$ .
Energy stored per unit volume where the field is B: $u = \dfrac{B^{2}}{2\mu_0}$ ; multiply by a small volume for the energy in it.

Centre of a coil and energy density

B = \frac{\mu_0 N I}{2R} \qquad u = \frac{B^{2}}{2\mu_0}

Worked example

A coil of 50 turns and radius 10 cm carries 2 A. (a) Find the field at its centre. (b) The same wire is rewound into a coil of 25 turns carrying the same current. Find the new field at the centre. (

\mu_0 = 4\pi \times 10^{-7}

T m/A)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q100Moderate

Example 1 · Moving Charges and Magnetism · Circular Loops and Coils

A long conducting wire having a current I flowing through it, is bent into a circular coil of

N

turns. Then it is bent into a circular coil of

n

turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is:

Rewinding changes the radius too

The same wire made into more turns makes each turn smaller. B depends on N/R and R goes as 1/N, so B goes as N². Scaling only by N misses half the change.

Perpendicular coils add as vectors

Two coils in perpendicular planes give fields along two perpendicular axes. The net field is √(B₁² + B₂²), not B₁ + B₂.

Do not forget the number of turns

μ₀I/2R is one turn. A coil of N closely wound turns gives N times that; leaving out N is a common slip when the turns are given in a separate sentence.

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Concept 2 of 2: Field on the axis of a circular loop

Away from the centre, along the axis, the pieces of the loop are farther away and their fields tilt, so the sideways parts cancel and only the parts along the axis add. The field is still along the axis but smaller. Written as a fraction of the centre field, it depends only on the ratio x/R, and falls off as the cube of distance far away.

Definition

On the axis at distance x from the centre: $B = \dfrac{\mu_0 N I R^{2}}{2(R^{2} + x^{2})^{3/2}}$ .
As a fraction of the centre field: $\dfrac{B}{B_c} = \dfrac{R^{3}}{(R^{2} + x^{2})^{3/2}} = \sin^{3}\theta$ , where $\theta$ is the angle at the point between the axis and the line to the rim.
Far away ( $x \gg R$ ): $B \approx \dfrac{\mu_0 N I R^{2}}{2x^{3}}$ .
Two coaxial loops on either side of a point O: a current that looks clockwise from O gives a field at O pointing away from O, towards that loop. Add the two with signs.
Just off the axis there is also a small sideways part of the field. It is zero in the plane of the loop and points opposite ways above and below it.

Axis of a loop

B = \frac{\mu_0 N I R^{2}}{2\left(R^{2} + x^{2}\right)^{3/2}} \qquad \frac{B}{B_c} = \frac{R^{3}}{\left(R^{2} + x^{2}\right)^{3/2}}

Worked example

The field at the centre of a loop of radius 3 cm is 125 μT. Find the field on its axis 4 cm from the centre.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q18Moderate

Example 2 · Moving Charges and Magnetism · Circular Loops and Coils

The magnetic field at the center of a current carrying circular loop of radius

R

16\mu\text{ }T

. The magnetic field at a distance

x = \sqrt{3}R

on its axis from the centre is

\_\_\_\_

\mu T

The power is 3/2, not 1/2

The axis field has (R² + x²)^(3/2) in the denominator and R² on top. Using the square root, or leaving out R², gives an expression with the wrong units.

An approximation needs x much less than R

A binomial approximation of the axis field is valid only very close to the centre. When x is comparable with R, work out (R/√(R² + x²))³ exactly.

The direction of each coaxial loop's field

Seen from the point between two loops, a clockwise current gives a field pointing towards that loop. Two loops whose currents both look clockwise from the point give opposite fields there, which subtract.

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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Field at the centre of a circular coil
Centre of a coil and energy density
$B = \frac{\mu_0 N I}{2R} \qquad u = \frac{B^{2}}{2\mu_0}$
Field on the axis of a circular loop
Axis of a loop
$B = \frac{\mu_0 N I R^{2}}{2\left(R^{2} + x^{2}\right)^{3/2}} \qquad \frac{B}{B_c} = \frac{R^{3}}{\left(R^{2} + x^{2}\right)^{3/2}}$

Watch out for (6)

Rewinding changes the radius too
→ Field at the centre of a circular coil
Perpendicular coils add as vectors
→ Field at the centre of a circular coil
Do not forget the number of turns
→ Field at the centre of a circular coil
The power is 3/2, not 1/2
→ Field on the axis of a circular loop
An approximation needs x much less than R
→ Field on the axis of a circular loop
The direction of each coaxial loop's field
→ Field on the axis of a circular loop

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