A toroid is a long coil of wire ( $N$ turns) wound over a circular core. The coefficient of self-induction…
A toroid is a long coil of wire ( $N$ turns) wound over a circular core. The coefficient of self-induction of the toroid is [The magnetic field in it is uniform and $R>>r$, where $r=$ radius of wire, $R=$ radius of coil] ( $\mu_0=$ permeability of free space $)$
$\frac{\mu_0 N^2 R^2}{2 r}$
$\frac{\mu_0 N r}{2 R}$
$\frac{\mu_0 N^2 r^2}{R}$
$\frac{\mu_0 N^2 r^2}{2 R}$
Solution
We know relation between the flux and the induction is: $L=\frac{\phi}{I}$
Flux through the given area $A$ with magnetic field $B$ is given by,
$\phi=N A B$
The field at the center of the toroid is given by $B=\mu_0 n I$, where
$\begin{aligned} & n=\frac{N}{2 \pi R} \\ & \therefore \phi=N\left(\pi r^2\right)\left[\mu_0\left(\frac{N}{2 \pi R}\right) I\right] \\ & \Rightarrow \phi=\frac{\mu_0 N^2 r^2 I}{2 R}\end{aligned}$
Using the definition of self-induction: $L=\frac{\phi}{I}=\frac{\mu_0 N^2 r^2}{2 R}$