A thin iron ring with mean diameter $d = 50 c m$ supports a winding consisting of $N = 800$ turns carrying current $I = 3.0 A$ . The ring has a cross-cut of width $b = 2.0 m m$ . 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.

Question

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Answer 1

From the theorem on circulation of vector $H$ ,

$H π d + \frac{B b}{μ _{0}} = N I$ or, $B = \frac{μ _{0} N I}{b} - \frac{μ _{0} π d}{b} H = (1.51 - 0.987) H$ ,

Where $B$ is in Tesla and $H$ in $k A / m$ . 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
$H \approx 0.26 k A / m$ , $B = 1.25 T$
Then $μ = \frac{B}{μ _{0} H} \approx 4000$ .

Answer 2

From the theorem on circulation of vector

H

,

H π d + \frac{B b}{μ _{0}} = N I

or,

B = \frac{μ _{0} N I}{b} - \frac{μ _{0} π d}{b} H = (1.51 - 0.987) H

,

Where

B

is in Tesla and

H

in

k A / 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 0.26 k A / m

,

B = 1.25 T

Then

μ = \frac{B}{μ _{0} H} \approx 4000

.

The B-H curve (i) and (ii) shown in fig., associated with

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