Consider a toroid with rectangular cross section, of inner radius $a$, outer radius $b$ and height $h$, carrying $n$ number of turns. Then the self-inductance of the toroidal coil when current $I$ passing through the toroid is

• Correct answer
  A

  $\frac{\mu_0 n^2 h}{2 \pi} \ln \left(\frac{b}{a}\right)$
• B

  $\frac{\mu_0 n h}{2 \pi} \ln \left(\frac{b}{a}\right)$
• C

  $\frac{\mu_0 n^2 h}{2 \pi} \ln \left(\frac{a}{b}\right)$
• D

  $\frac{\mu_0 n h}{2 \pi} \ln \left(\frac{a}{b}\right)$

Solution

Given, a toroid with a rectangular cross-section of inner radius $a$ and outer radius $b$. Height of the solenoid $=h$ Magnetic field inside a rectangular toroid is given by $B=\frac{\mu_0 n I}{2 \pi r}$
using the infinitesimal cross-sectional area element, $d x=h d r$ $\therefore$ Flux passing through the cross-section of toroid. $\begin{aligned} \phi & =\int B \cdot d x=\int_a^b \frac{\mu_0 n I}{2 \pi r} \cdot(h d r) \\ & \phi=\frac{\mu_0 n I h}{2 \pi} \int_a^b \frac{1}{r} d r \\ \Rightarrow \quad & \phi=\frac{\mu_0 n h I}{2 \pi}\left[\log r l_a^b\right. \\ \phi & =\frac{\mu_0 n h I}{2 \pi}(\log b-\log a) \\ & \phi=\frac{\mu_0 n h I}{2 \pi} \ln \left(\frac{b}{a}\right)\end{aligned}$ Now, self inductance of rectangular toroid, $L=\frac{n \phi}{I}$ Putting the value of $\phi$, we get $L=\frac{\mu n^2 h}{2 \pi} \ln \left(\frac{b}{a}\right)$

Hence, the correct answer is option: 'A'.