# Magnetic Permeability & Reluctance on old exam question

This is a past exam question from one of our lectures, and we have an issue with (i), I believe I need to use the equation $\rho=\frac{RA}{l}$, but I am not sure - could someone enlighten me on the issue?

A mild steel ring of magnetic permeability 380, having a cross sectional area of $500mm^2$ and a mean circumference of $400mm$, has a coil of $200$ turns wound uniformly around it.

Given that the magnetic permeability of free space is $400 nH/m$ determine:

• (i) The reluctance of the ring.

• (ii) The current required to produce a flux of $800\mu$ Wb in the ring.

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The equation above has nothing to do with the problem below. In case the problem would be in SI units, I might be able to solve it... :/ –  Pygmalion May 7 '12 at 17:03
That figure for magnetic permeability of free space doesn't seem to be right. –  Mark Beadles May 7 '12 at 19:19
I was puzzled by N=200 in the numerator and I found it was NOT necessary. Please check codecogs.com/library/physics/magnetism/magnetic-reluctance.php for the explanation. –  user34961 Nov 29 '13 at 13:54

## 1 Answer

As Pygmalion says, that equation is not related to the question. $\rho=R\frac{A}{l}$ is electrical resistivity, not magnetic.

As for part (i), this is simply answered from all the values given just plugged in to the appropriate formula for magnetic reluctance:

$$\mathcal{R}=\frac{\mathcal{F}}{\Phi}=\frac{l}{\mu_0\mu_rA}=\frac{(200\cdot400mm)}{(4\pi\times10^{-7}Hm^{-1})(380)(500mm^2)}=\text{answer}$$

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