Answer to Question #122630 in Electrical Engineering for Varun Bhutada

Question #122630
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.
1
Expert's answer
2020-06-18T14:51:40-0400

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/uo *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/uo *20 *10-4)

By simplifying above

ur = 143.9


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