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
![](https://mashalscienceacademy.com/wp-content/uploads/2020/10/e69c14p12.png"?w=319")
So apply first formula to get the mutual inductance.
Apply second formula to find induced emf.
Thanks so much sir ❣️ for ur great efforts may allah give u best rewards of it
Thanks Umair.
Great work
Thank you Tariq.
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