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. $\frac{{\mu }_{0}nh}{2\pi }\ln \left(\frac{b}{a}\right)$
• C. $\frac{{\mu }_0{n}^{2}h}{2\pi }\ln \left(\frac{a}{b}\right)$
• D. $\frac{{\mu }_{0}nh}{2\pi }\ln \left(\frac{a}{b}\right)$

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
$B=\frac{{\mu }_{0}nI}{2\pi r}$

using the infinitesimal cross-sectional area element,
$dx=hdr$
$\therefore$ Flux passing through the cross-section of toroid.
$\varphi =\int B\cdot dx=\underset{a}{\overset{b}{\int }}\frac{{\mu }_{0}nI}{2\pi r}\cdot (hdr)$
$\varphi =\frac{{\mu }_{0}nIh}{2\pi }\underset{a}{\overset{b}{\int }}\frac{1}{r}dr$
$\Rightarrow \varphi =\frac{{\mu }_{0}nhI}{2\pi }[\log r{]}_a^b$
$\Rightarrow \varphi =\frac{{\mu }_{0}nhI}{2\pi }(\log b-\log a)$
$\varphi =\frac{{\mu }_{0}nhI}{2\pi }\ln \left(\frac{b}{a}\right)$
Now, self inductance of rectangular toroid,
$L=\frac{n\varphi }{I}$
Putting the value of $\varphi$, we get
$L=\frac{\mu n^{2}h}{2\pi }\ln \left(\frac{b}{a}\right)$