A uniform conducting ring of mass `pi` kg and radius 1 m is kept on smooth horizontal table. A uniform but time varying magnetic field `B = (hat (i) + t^(2) hat (j))T` is present in the region, where t is time in seconds. Resistance of ring is `2 (Omega)`. Then

Induced electric field (in volt/meter) at the circumference of ring at the instant ring start toppling is

Options

A. `(10)/(pi)`
B. `(20)/(pi)`
C. `(5)/(pi)`
D. `(25)/(pi)`

Answer

A

Explanation

Given, `vec(A)= pi a^(2) hat(j) and B = (hat(i) +t^(2) hat(j))`
Associated flux is `phi = vec(B).vec(A)= pi a^(2)t^(2)`
Differentiating , `e= -2 pi a^(2)t`
Thus, induced electric field is
`E=(e)/(2 pi a) = at`
If induced current is I, Torque due to its weight is `pi g a`
Torque due to applied field is `vec(tau) = I vec(A) xx vec(B)=-pi a^(2)Ivec(k)`
where `I=e//R`
Ring just begins to topple when torque due to field become numerically equal to torque due to weight of ring.
`pi a^(2)* (2 pi a^(2)t)/(R)=pi ga`. Putting `a=1m` and `R=2 Omega`, we get `t= (10)/(pi)s`
Further, heat produced in time t is
`H=int (e^2)/(R) dt = (4Pi^(2)a^(4))/(R) * (T^3)/(3)`
Substituting value of time,
Subsituting value of time,
`H=(2000)/(3 pi)J`
Induced electric field at the circumference of ring at the instant ring start toppling is `E=at = (10)/(pi)`.