# Explain the working of a toroid

A toroid or a toroidal coil or an anchor ring consists of a wire uniformly wound on a torus or a doughnut-shaped structure. A toroid can be considered to be a close solenoid of small radius.

Let us consider a toroid with N turns of wire, all equally spaced and let I be the electric current along them.

(a) Magnetic field at a point inside the core of the toroid

To calculate the field at any point inside the toroid, we evaluate the line integral of $\overrightarrow{B}\cdot \overrightarrow{dl}$ over an Amperean loop L of radius r passing through that point. By symmetry, we see that the magnetic field is constant in magnitude on this loop and is tangent to it so that $\overrightarrow{B}\cdot \overrightarrow{dl}=Bdl$

Thus, $\oint{\overrightarrow{B}\cdot \overrightarrow{dl}=\oint{Bdl}=B\oint{dl}}=B\times 2\pi r$

Current linked with the Amperean loop L = total number of turns × current linking each turn = NI

From Ampere law,

$B\times 2\pi r=\mu _{0}NI$ or $B=\frac{{{\mu}_{0}}NI}{2\pi r}$ ………………(1)

Thus, the field inside the toroid varies as 1/r and hence is non-uniform.

On the other hand, if the cross-sectional area of the toroid is very-very small compared to r, we can neglect any variation in r. Considering 2πr to be the circumference of the toroid, N/2πr will be a constant and equal to the number of turns per unit length (n). In this case, equation (1) takes the form,

$B={{\mu }_{0}}nI$

Which is the same as for the long solenoid

(b) Magnetic field at a point in the open space inside the toroid

Consider the Amperean loop L’ passing through any point in the open space inside the toroid.

Since total current linking the loop L’ is zero

$\oint{\overrightarrow{B}\cdot \overrightarrow{dl}=0\,\,\,\,\,\,or\,\,\,\,\,B=0\,\,\,\,\,\,\,}$

(c) Magnetic field at a point outside the toroid

For any point outside the toroid, the net current threading the loop L” through that point is zero. This is due to the reason that each turn of the winding passes twice through the area bounded by the path, carrying equal currents in opposite directions as shown in fig 7.34 (b). Clearly,

$\oint{\overrightarrow{B}\cdot \overrightarrow{dl}=0\,\,\,\,\,\,or\,\,\,\,\,B=0\,\,\,\,\,\,\,}$

Thus, in an ideal toroid (in which the turns of the wire are very closely spaced) the magnetic field is confined entirely within its core and is almost uniform. For regions exterior to the coil, B = 0. In other words, magnetic field does not come out anywhere in case of a toroid as it does at the ends of an open solenoid. The magnetic field (B) within the toroid is tangent to the circular field lines in the torus.