Question

How many turns should a solenoid of cross-sectional area 3.0×10−2 m2 and length 0.24 m have if its inductance is to be 36 mH ?

Answer 1

Answer:

The number of turns of the solenoid is 478.7 turns.

Explanation:

Given;

area of the solenoid, A = 3.0 x 10⁻² m²

length of the solenoid, l = 0.24 m

inductance of the solenoid, L = 36 mH = 0.036 H

The number of turns of the solenoid is calculated as follows;

$L = \frac{\mu N^2 A}{l} \\\\N^2 = \frac{Ll}{\mu A} \\\\N = \sqrt{\frac{Ll}{\mu A}} \\\\N = \sqrt{\frac{0.036 \ \times \ 0.24}{4\pi \times 10^{-7} \ \times \ 3.0 \times 10^{-2} }}\\\\N = 478.7 \ turns$

Therefore, the number of turns of the solenoid is 478.7 turns.