Answer to Question #103572 in Electrical Engineering for Apoel

Question #103572
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.
Expert's answer

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

"F=\\Phi R,"

where the magnetic reluctance is


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:



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}."

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Assignment Expert
26.06.20, 13:26

Dear Varun Bhutada, Questions in this section are answered for free. We can't fulfill them all and there is no guarantee of answering certain question but we are doing our best. And if answer is published it means it was attentively checked by experts. You can try it yourself by publishing your question. Although if you have serious assignment that requires large amount of work and hence cannot be done for free you can submit it as assignment and our experts will surely assist you.

Varun Bhutada
16.06.20, 08:58

An air-gap between two pole pieces is 20 mm in length and the area of the flux path across the gap is 5 cm2. If the flux required in the air-gap is 0.75 mWb. Find the mmf necessary.

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