Question

A rectangular loop of sides 10 cm and 5 cm with a cut is stationary between the pole pieces of an electromagnet.
The magnetic field of the magnet is normal to the loop. The current feeding the electromagnet is reduced, so that the field decreased from its initial value of 0.2 T at the rate of `0.02Omega`. If the cut is joined and the loop has a resistance of `2.0Omega`, then the power dissipated by the loop as heat is
A. 5 nW
B. 4 nW
C. 3 nW
D. 2 nW

Answer 1

Correct Answer - A
Here, area, `A=10xx5=50cm^(2)=50xx10^(-4)m^(2)`
`(dB)/(dt)=0.2Ts^(-1)rArrR=2Omega`
`therefore" emf, e"=(dphi)/(dt)=A.(dB)/(dt)=50xx10^(-4)xx0.02=10^(-4)V`
Power dissipated in the form of heat
`=(e^(2))/(R)=(10^(-4)xx10^(-4))/(2)=0.5xx10^(-8)W=5xx10^(-9)W=5nW`