Answer to Question #122630 in Electrical Engineering for Varun Bhutada

Question

A steel ring has a mean diameter of 15 cm, a cross-section of 20 cm2
and a radial airgap of 0.5
mm cut in it. The ring is uniformly wound with 1500 turns of insulated wire and a magnetizing
current of 1 A produces a flux of 1 mWb in the airgap. Neglecting the effect of magnetic leakage
and fringing, calculate: (a) the reluctance of the magnetic circuit; (b) the relative permeability
of the steel.

Answer 1

Given

L = 0.4712 m = 3.14* d

A = 20 *10^-4 m²

Lg = 0.5 mm

N = 1500

I = 1 A

Solution

a) the reluctance of the magnetic circuit

R = (0.4712/u * 20 * 10^-4) + (0.0005/u_o *20 *10^-4) ........ expression 1

Flux = MMF/R = 1500/R

R = 1500/(1*10^-3)

= 1.5*10^-6 A/wb expression 2

b) the relative permeability

Equating expression 1 and 2

1.5 *10^-6 = (0.4712/u * 20 * 10^-4) + (0.0005/u_o *20 *10^-4)

By simplifying above

ur = 143.9