A toroidal coil has a total of $2000$ turns wound uniformly on a ring whose mean radius is $0.25\text{m}$. The coil carries a current of $3.0\text{A}$. Using Ampère's circuital law, calculate the magnitude of the magnetic field at a point on the central circle inside the toroid (i.e. at a radial distance of $0.25\text{m}$ from the symmetry axis).

Step-by-step solution

1. Write Ampère's circuital law: $\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_{\text{enc}}$. For a toroid, by symmetry $B$ is constant in magnitude and tangential along a circular Amperian loop of radius $r$ concentric with the toroid.
2. Evaluate the left-hand side on the circular path of radius $r=0.25\text{m}$: $\oint \vec{B}\cdot d\vec{l}=B\times 2\pi r$ because $\vec{B}$ is everywhere parallel to $d\vec{l}$ and has the same magnitude $B$ on the circle.
3. Identify the enclosed current: the Amperian circle threads the toroid core $N=2000$ times, so $I_{\text{enc}}=NI=2000\times 3.0=6000\text{A}$.
4. Set the two sides equal: $B\times 2\pi r={\mu }_{0}NI$.
5. Solve for $B$: $B=\frac{{\mu }_{0}NI}{2\pi r}=\frac{(4\pi \times 10^{-7})(2000)(3.0)}{2\pi \times 0.25}$.
6. Simplify: $B=\frac{4\pi \times 10^{-7}\times 6000}{2\pi \times 0.25}=\frac{24\pi \times 10^{-7}}{0.5\pi }=48\times 10^{-7}/0.5=0.048\times \frac{1}{0.5}=0.050\text{T}$.

Concept tested: Ampere's law applied to a toroid

Published 22 September 2026 · How this solution was prepared and checked