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.

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.

Expert's answer

Given

L = 0.4712 m = 3.14* d

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

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

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