Question

A circular coil having an area of 3.14 × 10^-2 m^2  and 30 turns is rotated about its vertical diameter with an angular speed of 70 rad^-1 in a uniform horizontal magnetic field of magnitude 5 × 10^-2 T. If the coil forms a closed loop of resistance 15 Ω, what is the average power loss due to Joule heating?

(a) 0.360

(b) 0.362

(c) 0.724

(d) 0.726

I had been asked this question during an interview.

Query is from Energy Consideration: A Quantitative Study in section Electromagnetic Induction of Physics – Class 12

Answer 1

Correct answer is (b) 0.362

The best explanation: Given: N = 30; A = 3.14 × 10^-2 m^2 ; Angular velocity (ω) = 70 rad^-1; B = 5 × 10^-2 T; R = 15 Ω

Emf (E) = NABω

E = 30 × 3.14 × 10^-2 × 5 × 10^-2  × 70

E = 3.297 V

Current (I) = \(\frac {E}{R}\)

I = \(\frac {3.297}{15}\)

I = 0.2198 A

Average power loss (P) = \(\frac {I^2 R}{2}\)

P = 0.2198^2 × \(\frac {15}{2}\)

P = 0.362 W

Therefore, the average power loss due to Joule heating is 0.362 W.