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

Answer

b

Explanation

Magnetic flux linked with the circular wire loop can be written as follows :
`phi = B pi r^(2) rArr phi = pi B_(0) r^(2) e^(-t//tau)`
Emf induced can be written as follows :
`epsi = |(d phi)/(dt)| rArr epsi = (pi B_(0)r^(2))/tau e^(-t//tau) `
Heat developed across the resistor can be written as follows :
`H = underset(0) overset(infty) int(epsi^(2))/R dt = (pi^(2) B_(0)^(2) r^(4) )/(R tau^(2)) underset(0)overset( infty) int e^(-2t//tau) dt`
`rArr" " H = (pi^(2) B_(0)^(2)r^(4))/(R tau^(2)) [ - tau/2 e^( 2t//tau)]_(0)^(infty)`
`rArr" " H = (pi^(2)B_(0)^(2) r^(4))/(2R tau)`
Hence option (b) is correct.