Answer to Question #161612 in Electrical Engineering for Tony

Question

An iron ring has a mean diameter of 15 cm, a cross-section of 20 cm ^ 2 and a radial air gap of 0.5 mm cut on it. The ring is uniformly wound with 1500 turns of insulated wire and a magnetizing current of 1A produces a flax of 1 mWb in the air gap. Calculate: (i) the reluctance of the magnetic circuit and fii the relative normoabilit of t [5]

Answer 1

Given quantities:

$d = 0.15m \space \space L = \pi d = 0.4712m$ 

$A = 20*10^{-4}m^2$

$Lg = 0.5mm$

$N = 1500$

$I = 1A$

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