A thin iron ring with mean diameter d = 50 cm, supports a winding consisting of 800 turns, with current 3.A. Thering has a cross-cut of width 2.0 mm. The pemeability of iron is (graph between B and H is given)

For small gap, total line integral of $\overrightarrow{\mathrm{H}}$ is

$\int {\mathrm{\mu }}_0 \overrightarrow{\mathrm{H}} \overrightarrow{\mathrm{dl}}={\mathrm{\mu }}_0\mathrm{NI}-\mathrm{Bb}$

or $\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{NI}}{\mathrm{b}}-\frac{\mathrm{\pi d} {\mathrm{\mu }}_0\mathrm{H}}{\mathrm{b}}$

After solving, we get 

B = 1.51 - 0.987 H

B and H are also related to given graph. The required values of B and H must simultaneously satisfy both the relations. Solving this system of equation by means of plotting we get

B = 1.257 T

H = 0.26 kA/m

$\mathrm{\mu }=\frac{\mathrm{B}}{{\mathrm{\mu }}_0\mathrm{H}}=4 \mathrm{x} {10}^3$