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

$$e=L\frac{di}{dt}$$

The self-inductance of the solenoid is given by

$$L=\frac{{\mu }_0{N}^{2}A}{l}$$

$$L=\frac{4\pi \times {10}^{-7}\times {240}^2\times \pi \times (2\times {10}^{-2}{)}^2}{12\times {10}^{-2}}$$

Thus, the emf induced in the solenoid is given by

$$e=\frac{4\pi \times {10}^{-7}\times {240}^2\times \pi \times (2\times {10}^{-2}{)}^2}{12\times {10}^{-2}}\times 0.8$$

$$=60577.3824\times {10}^{-8}=6\times {10}^{-4}V$$