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

The number of turns of the wire needed is 243 turns

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

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

magnitude and direction

d is the answer i came up with