Question

The self-inductance of a closely wound coil of 200 turns is 10 mH. Determine the value of magnetic flux through the cross-section of the coil when the current passing through the coil is 4 mA.

Answer 1

Given: Self inductance of coil

= L = 10 mH = 10 x 10^-3 H

Number of turns, N = 200 and Current through the coil, i = 4 mA = 4 x 10^-3 A The total value of magnetic flux `(phi)` associated with the coil is,

`phi = Li`

= (10 x 10^-3) H x (4 x 10^-3) A

= 4 x 10^-5 Wb

The flux per turn (or flux through the crosssection of the coil)

`= phi/N`

`= (4xx10^-5 Wb)/ 200`

= 2 x 10^-7 Wb