Question

A coiled telephone cord forms a spiral with 52.0 turns, a diameter of 1.30 cm, and an unstretched length of 57.5 cm. Determine the inductance of one conductor in the unstretched cord.

Answer 1

The inductance of one conductor in the unstretched cord is 0.7849 μH

Step-by-step explanation:

Given;

number of turns of the coil, N = 52 turns

diameter of the coil, d = 1.30 cm

radius of the coil, r = d/2 = 1.30 cm / 2 = 0.65 cm = 0.0065 m

length of the unstretched cord, l = 57.5 cm = 0.575 m

The inductance of one conductor in the unstretched cord is given by;

`L = (N^2 \mu_o A)/(l)`

where;

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

A is area of coil, = πr² = π x (0.0065)² = 1.328 x 10⁻⁴ m²

`L = ((52)^2(4\pi *10^(-7))(1.328*10^(-4)))/(0.575) \\\\L = 7.849*10^(-7) \ H`

L = 0.7849 μH

Therefore, the inductance of one conductor in the unstretched cord is 0.7849 μH