Question

A thin iron ring with mean diameter $d=50cm$ supports a winding consisting of $N=800$ turns carrying current $I=3.0A$. The ring has a cross-cut of width $b=2.0mm$. Neglecting the scattering of the magnetic flux at the gap edges, and using the plot shown in Fig., find the permeability of iron under these conditions.

Answer 1

From the theorem on circulation of vector $\overrightarrow{H}$,
$H\pi d+\frac{Bb}{{\mu }_0}=NI$ or, $B=\frac{{\mu }_{0}NI}{b}-\frac{{\mu }_0\pi d}{b}H=(1.51-0.987)H$,
Where $B$ is in Tesla and $H$ in $kA/m$. Besides, $B$ and $H$ are interrelated as in the figure shown below of the text. Thus we have to solve for $B$, $H$ graphically by simultaneously drawing the two curves (the hysterisis curve and the straight line, given above) and find the point of intersection. It is at
$H\approx 0.26kA/m$, $B=1.25T$
Then $\mu =\frac{B}{{\mu }_{0}H}\approx 4000$.