Question

A coil of wire of a certain radius has 600 turns and a self inductance of 108 mH. The self inductance of a 2nd similar coil of 500 turns will be(a) 74 mH(b) 75 mH(c) 76 mH(d) 77 mH

Answer 1

answer : option (b) 75mH

explanation : we know, self inductance of a solenoid/coil is given by, $L=\frac{\mu_0N^2A}{l}$

where L is self inductance of solenoid, $\mu_0$ is permeability of medium, A is cross sectional area , N is number of turns and l is length of solenoid/coil

it is clear that, self inductance of coil is directly proportional to square of number of turns.

e.g., $\frac{L_1}{L_2}=\left(\frac{n_1}{n_2}\right)^2$

so, 108mH/$L_2$ = 600²/500²

or, $L_2$ = 108 × 500²/600²

or, $L_2$ = 108 × 25/36

hence, $L_2$ = 75mH