For a toroid $N=500$, radius $=40 \mathrm{~cm}$, and area of cross section $=10 \mathrm{~cm}^2$. Find inductance.

• Correct answer
  A

  $125 \mu \mathrm{H}$
• B

  $250 \mu \mathrm{H}$
• C

  $0.00248 \mathrm{H}$
• D

  zero

Solution

If we flow a current $i$, then $\begin{aligned} & \phi_{\text {self }}=(B \times A) \times N=\left(\frac{\mu_0 N i}{2 \pi R}\right) A \times N \\ & \Rightarrow \quad \phi_{\text {self }}=\left(\frac{\mu_0 N^2 A}{2 \pi R}\right) i \text { and } \phi_{\text {self }}=L i\end{aligned}$ $\begin{aligned} & L=\frac{\mu_0 N^2 A}{2 \pi R}=\frac{\mu_0}{4 \pi} \frac{2 N^2 A}{R} \\ & \Rightarrow L=\left(10^{-7}\right) \times \frac{2 \times(500)^2 \times 10 \times\left(10^{-2}\right)^2}{0.4} \\ &=125 \mu \mathrm{H}\end{aligned}$

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