Question

A solenoid is wound with a single layer of insulated copper wire of diameter 3.000 mm and has a diameter of 3.000 cm and is 1.800 m long. Assume that adjacent wires touch and that insulation thickness is negligible.

R

1. How many turns are on the solenoid?

2. What is the inductance per meter (H/m) for the solenoid near its center?

Answer 1

1] Number of turns x diameter of the wire = length of the solenoid

=> N (3 x 10^-3) = 1.8

=> N = 600 turns

2]

Inductance is given by:

$L = \frac{\mu_o N^2 A}{l}$

=> $L = \frac{4\pi \times 10^{-7} (600)^2 (\pi (0.03)^2)}{1.8} = 7.106 \times 10^{-4}H$ = 0.7106 mH.

so, Inductance per meter will be: L/1.8 = 0.395 mH.