Question

A toroidal coil of rectangular crossection has an inner radius a, outer radius b, height h and total number of turns N. Current in the toroid is $i$, and say $r$ is distance of a point from the center of toroid.

• A. Magnetic field at the point decreases with r for 0 < r < a
• B. Magnetic field at the point increases with r for a < r < b
• C. Magnetic field is zero for r > b
• D. Self inductance of toroid is $\frac{{\mu }_0{N}^{2}h}{2\pi }ln\left(\frac{b}{a}\right)$

Answer 1

Answer: (C), (D)

Self inductance of toroid is $\frac{{\mu }_0{N}^{2}h}{2\pi }ln\left(\frac{b}{a}\right)$

The magnetic field is confined inside the toroid, so the field outside is zero.

By Amperes law $\oint B.dl={\mu }_{0}I$
$B\times 2\pi r={\mu }_{0}Ni$
$B=\frac{{\mu }_{0}Ni}{2\pi r}$
$\{\begin{array}{c}\text{So}B\alpha \frac{1}{r}\text{for}a<r<b\\ \&B=0\text{for}r>b\&r<a\end{array}$
Flux in the toroid is,
$d\varphi =B.dA=\frac{{\mu }_{0}Ni}{2\pi r}hdr$
$\varphi =\int d\varphi =\frac{{\mu }_{0}Ni}{2\pi }h{\int }_a^b\frac{dr}{r}$
= $\frac{{\mu }_{0}Nih}{2\pi }ln\left(\frac{b}{a}\right)$
${\varphi }_{net}=N\varphi \&{\varphi }_{net}=Li$
$\therefore L=\frac{{\mu }_0{N}^{2}h}{2\pi }ln\left(\frac{b}{a}\right)$