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An armature coil consists of 30 turns of wire, each of area A = 0.05 m^2 and total resistance of 10 Ω. It rotates in a magnetic field of 0.15T at a constant frequency of \(\frac {140}{\pi }\) Hz. Determine the value of maximum induced emf produced in the coil.(a) 1 V(b) 500 V(c) 63 V(d) 43 VThis question was posed to me at a job interview.Enquiry is from AC Generator topic in division Alternating Current of Physics – Class 12

Answer»

Correct option is (c) 63 V

The best I can explain: e0 = nBAω.

e0 = 30 × 0.15 × 0.05 × 2π × (\(\FRAC {140}{\PI }\))

e0 = 63 V.



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