Question

Suppose you wish to make a solenoid whose self-inductance is 1.8 mH. The inductor is to have a cross-sectional area of 1.6 x 10-3 m2 and a length of 0.066 m. How many turns of wire are needed

Answer 1

The number of turns of the wire needed is 243 turns

Step-by-step explanation:

Given;

self inductance of the solenoid, L = 1.8 mH

cross sectional area of the inductor, A = 1.6 x 10⁻³ m²

length of the inductor, l = 0.066 m

The self inductance of long solenoid is given by;

L = μ₀n²Al

where;

μ₀ is permeability of free space = 4π x 10⁻⁷ H/m

n is number of turns per length

A is the area of the solenoid

l is length of the solenoid

`n = \sqrt{(L)/(\mu_o Al) } \\\\n = \sqrt{(1.8*10^(-3))/((4\pi*10^(-7)) (1.6*10^(-3))(0.066)) } \\\\n = √(13562583.184) \\\\n = 3682.74 \ turns/m`

The number of turns is given by;

N = nL

N = (3682.74)(0.066)

N = 243 turns

Therefore, the number of turns of the wire needed is 243 turns