Answer to Question #103572 in Electrical Engineering for Apoel

Question

A mild steel ring has a mean circumference of 500 mm
and a uniform cross-sectional area of 300 mm2
.
Calculate the m.m.f. required to produce a flux of
500 µWb. An airgap, 1.0 mm in length, is now cut in
the ring. Determine the flux produced if the m.m.f.
remains constant. Assume the relative permeability
of the mild steel to remain constant at 220.

Answer 1

The magnetomotive force can be expressed in terms of flux and magnetic reluctance as

$$F=\Phi R,$$

where the magnetic reluctance is

$$R=\frac{l}{\mu\mu_0A}.$$

Therefore, for the ring without a gap the MMF is

$$F=\Phi\frac{l}{\mu\mu_0A}=\\=500\cdot10^{-6}\cdot\frac{500\cdot10^{-3}}{220\cdot4\pi\cdot10^{-7}\cdot300\cdot10^{-6}}=3014\text{ A}.$$

When we have the airgap, the reluctance changes:

$$R=\frac{l_{steel}}{\mu\mu_0A}+\frac{l_{air}}{\mu_0A}.$$

$$F=\frac{\Phi}{\mu_0A}\bigg(\frac{l_{steel}}{\mu}+l_{air}\bigg),$$

where the length of the steel ring will be, of course, 1 mm less than in the previous case.

Calculations give us the answer of

$$F=4335\text{ A}.$$