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Problem 1: Two identical coils A and B of 500 turns each has parallel planes such that 70% of flux produced by one coil links with the other. A current of 6 Amp flowing in coil A produces a flux of 0.06 Wb in it. If the current in coil A changes from 10 A to -10 A in .03 s, calculate (a) the mutual inductance and (b) the emf induced in coil B.

Solution

Given data:

No of turns in the coils = N = 500 Current flowing in A = 6 Amp

Magnetic flux produced = 0.06 m Wb = 0.06 ×  10-3 Wb

Flux due to current in coil A which links coil B = 0.06 ×  10-3 Wb ×  70% = 0.06 ×  10-3 ×  0.70 = 0.042 ×  10-3 Wb

 Change in current= ΔI = 10 – (-10) = 20 Amp

 Time interval = 0.03 s

Required: (a) Mutual Inductance, M =?         Induced emf, ε =?

Formulae

So apply first formula to get the mutual inductance.

Apply second formula to find induced emf.

4 Comments

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