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 H→ is∫μ0 H→ dl→=μ0NI-Bbor B=μ0NIb-πd μ0HbAfter solving, we get B = 1.51 - 0.987 HB 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 getB = 1.257 TH = 0.26 kA/mμ=Bμ0H=4 x 103