A rectangular loop of sides `10cm` and `5cm` 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 decreases from its initial value of `3.0 T` at the rate of `0.02 T s^(-1)`. If the cut is joined and the loop has a resistance of `2.0 Omega`, the power dissipated by the loop as heat is

Options

A. 5 nW
B. 4 nW
C. 3 nW
D. 2 nW

Answer

A

Explanation

Here, `A=10xx5=50cm^(2)`
`=50xx10^(-4)m^(2)`
`(dB)/(dt)=0.2Ts^(-1)`
`R=2 Omega`
`E=(d phi)/(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`