Question

The current in a solenoid of 240 turns, having a length of 12 cm and a radius of 2 cm, changes at a rate of 0.8 A s⁻¹. Find the emf induced in it.

Answer 1

Given:
Number of turns, N = 240
Radius of the solenoid, r = 2 cm
Length of the solenoid, l = 12 cm

The emf induced in the solenoid is given by
$e=L\frac{\mathrm{d}i}{\mathrm{d}t}$

The self-inductance of the solenoid is given by
$$\begin{gathered} L=\frac{{\mu }_0{N}^{2}A}{l} \\ L=\frac{4\mathrm{\pi }\times {10}^{-7}\times {240}^2\times \mathrm{\pi }\times (2\times {10}^{-2}{)}^2}{12\times {10}^{-2}} \end{gathered}$$

Thus, the emf induced in the solenoid is given by

$$\begin{gathered} e=\frac{4\mathrm{\pi }\times {10}^{-7}\times {240}^2\times \mathrm{\pi }\times (2\times {10}^{-2}{)}^2}{12\times {10}^{-2}}\times 0.8 \\ =60577.3824\times {10}^{-8}=6\times {10}^{-4}\mathrm{V} \end{gathered}$$