A metal rod OA of mann 'm' and length 'r' is kept rotating with a constantangular speed `omega` in a vertical plane about a horizontal axis at the end O. The free end A is arraged to slide without friction along fixed conduction circular ring in the same plane as that of rotation. A uniform and constant magnetic induction `vec(B)` is applied perpendicular and into the plane of rotation as shown in the figure below. An inductor L and an external resistance R are connected through a swithch S between the point O and a point C on the ring to form an electrical circuit. Neglect the resistance of the ring and the rod. Initially, the switch is open.

(a) What is the induced emf across the teminal of the switch?
(b) The switch S is closed at time t=0.
(i) Obtain an expression for the current as a function of time.
(ii) In the steady state, obtin the time dependence of the torque required to maintain the constant angular speed, given that the rod OA was along th positive X-axis at t=0.

Explanation

a. The emf induced across the whole rod is given by:
`E = (1)/(2) B omega r^(2)`
`= B omega [(x^(2))/(2)]_(0)^(r ) = (1)/(2) B omega r^(2)` (i)
b. i. When the rod roates anticlockwise (as shows), the end `O` becomes positive and `A` negative. As resistance of ring and rod is negligible, therefore, the equivenent circuit is shows in Fig. When switch `S` is closed, let `i` be the current and `di//dt` the rate of change of current in the circuit, then from kirchhoff's second law, the equation of emf is

`E = Ri + L(di)/(dt)`
which upon integation gives `i = (E)/(R) (1 - e^(-Rt//L))`
Using Eq.(i), `i = (Br^(2) omega)/(2R) (1 - e^(-Rt//L))`
ii. In steady state, current is `I_(0) = E//R`

Magnetic force, `F_(B) = I_(0) rB = (ErB)/(R)`, `theta = omega t`
Required torque
`tau = tau_(due "to" F_(B)) + tau_(due "to" mg)`
`rArr tau = F_(B)(r)/(2) + mg(r)/(2) cos theta = (EBr^(2))/(2R) + (mgr)/(2) cos omega t`
`= (B^(2)omega r^(4))/(4R) + (mgr cos omega t)/(2)`