Toroid — find magnetic field inside using Ampere's law

Question

A toroid has $N$ turns, inner radius $r_1$ , outer radius $r_2$ , and carries current $I$ . Using Ampere’s circuital law, find the magnetic field (a) inside the toroid, (b) in the open space inside the toroid (at the centre), and (c) outside the toroid.

(NCERT Class 12 — Moving Charges and Magnetism)

Solution — Step by Step

Consider a circular Amperian loop of radius $r$ where $r_1 < r < r_2$ (inside the toroid winding). By symmetry, $\vec{B}$ is tangential and constant along this loop.

Ampere’s law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$

Left side: $B \times 2\pi r$ (since $\vec{B}$ is parallel to $d\vec{l}$ everywhere on the loop).

Right side: The loop threads through all $N$ turns, each carrying current $I$ . So $I_{enclosed} = NI$ .

B \times 2\pi r = \mu_0 NI

\mathbf{B = \frac{\mu_0 NI}{2\pi r}}

If we define $n = N/(2\pi r)$ as the number of turns per unit length, then $B = \mu_0 n I$ — same as a solenoid.

Take an Amperian loop of radius $r < r_1$ (inside the hole of the toroid). No current passes through this loop — wires go up on one side and come back down on the other, but none actually crosses this inner loop.

$I_{enclosed} = 0$ , so $B = 0$ .

The magnetic field in the central open space of a toroid is zero.

Take an Amperian loop with $r > r_2$ . Each turn carries current $I$ going in one direction (say inward) and comes back carrying $I$ outward. The net current through the loop: $NI - NI = 0$ .

$I_{enclosed} = 0$ , so $B = 0$ .

The magnetic field outside the toroid is also zero.

Why This Works

A toroid is essentially a solenoid bent into a circle. Inside the winding, the field is confined and follows the circular path. Outside and in the central hole, the symmetry of the winding ensures all currents cancel out.

This field confinement makes toroids ideal for transformers and inductors — the magnetic field does not leak out, reducing electromagnetic interference.

The key difference from a solenoid: in a toroid, $B$ varies with $r$ (it is stronger near the inner edge). In an ideal solenoid, $B$ is uniform throughout.

Alternative Method

For a thin toroid ( $r_2 - r_1 \ll r_1$ ), the field is approximately uniform and equal to $B = \mu_0 nI$ where $n = N/(\text{mean circumference})$ . This approximation treats the toroid as a straight solenoid, which is valid when the cross-section is small compared to the radius.

CBSE board exams often ask you to compare a solenoid and a toroid. Key differences: solenoid has field leaking from ends, toroid confines field entirely inside. Solenoid has uniform $B$ , toroid has $B \propto 1/r$ . Both give $B = 0$ outside (ideal cases).

Common Mistake

Students often confuse the total turns $N$ with turns per unit length $n$ . For a toroid: $n = N/(2\pi r)$ . The formula $B = \mu_0 NI/(2\pi r)$ uses total turns $N$ . The formula $B = \mu_0 nI$ uses turns per unit length $n$ . Both are correct but mixing $N$ and $n$ in the wrong formula gives the wrong answer. Stick to one version consistently.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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