Question

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

• 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. $\frac{\mu_{0} n h}{2 \pi} \ln \left(\frac{a}{b}\right)$

Answer 1

Answer: (A)

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.
$\phi=\int B \cdot d x=\int\limits_{a}^{b} \frac{\mu_{0} n I}{2 \pi r} \cdot(h d r)$
$ \phi=\frac{\mu_{0} \,n I h}{2 \pi} \int\limits_{a}^{b} \frac{1}{r} d r$
$\Rightarrow \phi=\frac{\mu_{0} \,n h I}{2 \pi}[\log r]_{a}^{b}$
$\Rightarrow \phi=\frac{\mu_{0} \,n h I}{2 \pi}(\log b-\log a)$
$\phi=\frac{\mu_{0}\, n h I}{2 \pi} \ln \left(\frac{b}{a}\right)$
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)$