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in Physics by (63.5k points)

A soft iron ring having a mean circumference of 40 cm and cross-sectional area of 3 cm2 has two radial saw cuts made at diametrically opposite points. A brass plate 0.5 mm thick is inserted in each gap. The ring is wound with 800 turns. Calculate the magnetic leakage and fringing. Assume the following data for soft iron :

B (Wb/m2 ) : 0.76 1.13 1.31  1.41 1.5
H (AT/m) : 50  100 150  200 250 

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It should be noted that brass is a non-magnetic material. Force at one separation = B2A/2μ0 newton. 

Force at both separations = B2A/μ0 newton. 

Now F = 12 kg wt = 12 × 9.81 = 117.7 N 

∴ 117.7 = B2 × 3 × 10− 4/4π × 10− 7 ; B = 0.7 Wb/m2 

If B/H curve is drawn, it will be found that for B = 0.7 Wb/m2 , value of H = 45 AT/m. 

Now, length of iron path = 40 cm = 0.4 m. 

AT required for iron path = 45 × 0.4 = 18 

Value of H in the non-magnetic brass plates 

= B/μ0 = 0.7/4π × 10− 7 = 557,042 AT/m 

Total thickness of brass plates = 0.5 × 2 = 1 mm 

AT required = 557,042 × 1 × 10− 3 = 557, 

Total AT needed = 18 + 557 = 575 

∴ magnetising current required = 557/800 = 0.72 A

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