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

The number of turns of the solenoid is 478.7 turns.

Step-by-step 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 = (\mu N^2 A)/(l) \\\\N^2 = (Ll)/(\mu A) \\\\N = \sqrt{(Ll)/(\mu A)} \\\\N = \sqrt{(0.036 \ * \ 0.24)/(4\pi * 10^(-7) \ * \ 3.0 * 10^(-2) )}\\\\N = 478.7 \ turns`

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