Question

A solenoid of 10 turns, cross section 36 cm² and of resistance 10 Ω is placed in magnetic field which is varying at constant rate of 0.5 T/sec .Find power of heat dissipation.

(1) 1.8 W

(2) 3.8 W

(3) 2.34 W

(4) 7.6 W

Answer 1

Correct Option is (3) 2.34 W 

\(V=\varepsilon=\frac{d\phi}{dt}=\frac{dB}{dt}\) NA 

= \(0.5\times10\times36\times10^{-4}\) 

\(P=\frac{V^2}{R}= \frac{(0.5)^2\times10^2\times(36)^2\times10^{-8}}{10\times10\times10^{-6}}\) 

= \(\frac{1}{4}\times36\times10^{-2}\) 

= 2.34 W