# Magnetic Field of a Toroid

A toroid is a solenoid bent in a circular shape. The magnetic field outside the solenoid is zero. The magnetic field inside the solenoid is given by \begin{align} B=\mu_0 n I, \end{align} where $I$ is the current and $n$ is the number of turns per unit length. Note that magnetic field inside a toroid is independent of its radius. A toroid of radius r have $N=(2\pi r)n$ turns.

The magnetic field inside a toroid is obtained by using Ampere's circuital law. Let us find the field at a point P inside the toroid. The point P is at a distance r from the centre of the toroid. Take a closed path of radius r passing through P and concentric with the solenoid. By symmetry, the field is tangential to the circle and its magnitude is the same at all points of the path. Thus, \begin{align} \oint \vec{B}\cdot\mathrm{d}\vec{l}=\int B\mathrm{d}l=B\int\mathrm{d}l=2\pi r B. \end{align} If the total number of turns is $N$, the current crossing the area bounded by the circle is $NI$ where $I$ is the current in the toroid. Using Ampere's law on the circular path, \begin{align} \oint\vec{B}\cdot\mathrm{d}l=\mu_0 NI. \end{align} Above equations gives the magnetic field inside the toroid as \begin{align} B=\frac{\mu_0 N I}{2\pi r}. \end{align}