A conducting metal circular-wire-loop of radius `r` is placed perpendicular to a magnetic field which varies with time as `B = B_0e^(-t//tau)`, where `B_0` and `tau` are constants, at time = 0. If the resistance of the loop is `R`,then the heat generated in the loop after a long time `(t to oo)` is :

Options

A. `(pi^(2)r^(4)B_(0)^(4))/(2tauR)`
B. `(pi^(2)r^(4)B_(0)^(2))/(2tauR)`
C. `(pi^(2)r^(4)B_(0)^(2)R)/(tau)`
D. `(pi^(2)r^(4)B_(0)^(2))/(tauR)`

Answer

B

Explanation

Here, `B=B_(0)e^((t)/(tau))`
Area of the circular loop, `A=pir^(2)`
Flux linked with the loop at any time, t,
`phi=BA=pir^(2)B_(0)e^((t)/(tau))`
Emf is induced in the loop , `epsi=-(dphi)/(dt)=pir^(2)B_(0)(1)/(tau)e^((t)/(tau))`
Net heat generated in the loop
`underset(0)overset(oo)int(epsi^(2))/(R)=(pi^(2)r^(4)B_(0)^(4))/(tau^(2)R)underset(0)overset(oo)int e^((2t)/(tau))dt`
`=(pi^(2)r^(2)B_(0)^(2))/(tau^(2)R)xx(1)/((-(2)/(tau)))xx[e^((2t)/(tau))]_(0)^(oo)`
`(-pi^(2)r^(4)B_(0)^(2))/(2tau^(2)R)xxtau(0-1)=(pi^(2)r^(4)B_(0)^(2))/(2tauR)`