A stationary circular loop of radius a is located in a magnetic field which varies with time from t = 0 to t = T according to law `B = B_(0) t(T - t)`. If plane of loop is normal to the direction of field and resistance of the loop is R, calculate
magnitude of charge flown through the loop from instant t = 0 to the instant when current reverses its direction.

Options

A. `(pi^(2)a^(2)B_(0)T^(2))/(R)`
B. `(pi^(2)a^(2)B_(0)T^(2))/(4R)`
C. `(4pi^(2)a^(2)B_(0)T^(2))/(R)`
D. None of these

Answer

B

Explanation

Since magnetic field strength B varies with time, therefore an emf is incuced in the loop ans a current starts flowing through the loop. Due to flow of current heat is generated.
At an instant t, flux linked with the loop.
`phi=pi s^(2)B_(0)(tT-t^(2))`
Induced emf, `e=-(d phi)/(dt) =-pi ^(2)B_(0)(T-2t)`
Induced current , `i=e/R = (pi a^(2)B_(0))/(R) (2t-T)`
Thermal power generated at this instant,
`P=(pi^(2)a^(4)B_(0)^(2))/(R ) (2t-t)^(2)`
During an elementary time interval dt, heat generated,
`pdt = (pi^(2)a^(4)B_(0)^(2))/(R) (2t-T)^(2)*dt`
Total heat generated from `t=0 to t=T`.
`Q=int Pdt = (pi^(2)a^(4)B_(0)^(2))/(R) int_(0)^(T)(2t-T)^(2)dt`
`=(pi^(2)a^(4)B_(0)^(2)T^(3))/(3R)`
The current reverses its sign when its magnitude reduces to zero. Let this happen at instant `t=t_(0)`. Substituting t by `t_(0)` in equation (i).
`(pia^(2)B_(0))/(R) (2t-t_(0)) =0`
`t_(0)=T/2`
`:.` Charge that flows from `t=0` to `t=T//2`
`q=(pi a^(2)B_(0))/(R) int_(0)^(T//2) (2t-T)dt = (pi a^(2)B_(0)T^(2))/(4R)`.