Question

A coiled telephone cord forms a spiral with 90.0 turns, a diameter of 1.30 cm, and an unstretched length of 57.0 cm. Determine the inductance of one conductor in the unstretched cord.

Answer 1

Answer:

2.36 μ H

Explanation:

Given,

Number of turns= 90

diameter = 1.3 cm = 0.013 m

unscratched length = 57 cm  = 0.57 m

Area, A = π r²

            = π x 0.0065² = 1.32 x 10⁻⁴ m²

 we know,

   $L = \dfrac{\mu_0N^2A}{l}$

   $L = \dfrac{4\pi \times 10^{-7}\times 90^2\times 1.32\times 10^{-4}}{0.57}$

    L = 2.36 μ H

Hence, the inductance of the unstretched cord is equal to 2.36 μ H