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

Flux passing through
`phi= Bpi r^(2)`
`phi= B_(0)e^(-t//tau)pi r^(2)`
`(dphi)/(dt)= pi r^(2)B_(0)e^(-t//T)((-1)/(tau))`
`V= -(dphi)/(dt)= (pi r^(2) B_(0)e^(-t//tau))/(tau)`
Heat generated in the loop from t = 0 lo `t = oo`
`H= int (V^(2))/(R)dt`
`= (pi^(2)B_(0)^(2) r^(4) )/(Rtau^(2))int_(0)^(oo) e^(-2t//tau)dt= (pi^(2)B_(0)^(2) r^(4) )/(2Rtau)`