Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction `B=B_(0)e^(-t)` is established inside the coil. If the key (K) is closed, the electric power developed right after closing the switch is equal to:-

Options

A. `(B_(0)^(2)pi r^(2))/R`
B. `(B_(0)10 r^(3))/R`
C. `(B_(0)^(2) pi^(2) r^(4)R)/5`
D. `(B_(0)^(2) pi^(2)r^(4))/R`

Answer

D

Explanation

Induced emf,
`E= (d phi)/(dt)`
`=d/(dt)(BA) = A(dB)/(dt)`
`=pi r^(2) B_(0) d/(dt)(e^(-t))`
`=-pi r^(2) B_(0)e^(-t)`
At t=0, `E_(0) = B_(0)pir^(2)`
The electric power developed in the resistor R just at the instant of closing the key is:
`P=E_(0)^(2)/R = (B_(0)^(2) pi^(2)r^(4))/R`