Shown in the figure is a circular loop of radius `r` and resistance `R`. A varible magnetic field of induction `B=B_(0)e^(-1)` is established inside the coil. If the key `(K)` is closed, the electrical power devloped right after closing the switch is equal to

Options

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

Answer

D

Explanation

`P=(e^(2))/(R ),e=-(d)/(dt)(BA)=A(d)/(dt)(B_(0)e^(-t))=AB_(0)e^(-t)`
`P=(1)/(R )(AB_(0)e^(-t)^(2))/(R )=(B_(0)^(2)pi^(2)r^(4))/(R )`