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.

Answer

A, B, C, D

Explanation

(a) Let us consider a small length of metal rod dx at a distance x from the origin. Small amount of emf (de) induced in this small length (due to metallic rod cutting megnetic lines of force) is
`de=B(dx)v` ...(i)
where v is the velocity of small length dx
`v=x(Omega)` ...(ii)
`:. Total emf across the whole metallic rod OA is

`e=int_(0)^(r) Bx(omega)dx=B(omega)[(x^20/(2)]_(0)^(r)=(Br^(2)omega)/(2)`
(b) The above diagram can be reconstruced as the adjacent figure, e is a constant. O will accumulate postive charge and A negative. When the switch S is closed, transient current at any time t, when current I is flowing in the circuit,
`I=(I_0)(1-e^(t//tau))`
Here,
`(I_0)=e/R =( B omegar^2)/(2R)`
and `(tau)=L/R`
Therefore, I=(B omegar^(2))/(2R)[1-e^(-(R)/(L))t)]`
(ii) In steady state,
`I=(B omegar^(2))/(2R) [:. t has a large value and e^(-(R/T)t)rarr 0]`
When current flows in the circuit in steady state, there is a power loss through the resistor,
Also since the rid ritating in a vertical planem work needs Power loss due to current I will be
`P=(I^2)R=((Br^2omega)/(2R))^(2)R`
If torque required for this Power
is `(tau_(1))` then
`P=(tau_1)(omega)`
`implies (tau_1)=(B^(2)r^(4)omega)/(4R)`
Torque required to move the rod in circular motion against gravitational field
`(tau_2)=mgxxr/2 cos(theta)`
The totla torque
`(tau)=(tau_1)+(tau_2)` (clockwise).

`(tau)=(b^(2)r^(2)omega)/(4R)+(mgr)/(2) cos omega t`
The required torque will be of same magnitude and in anticlockwise direction. The second term will change signs as the value of `cos(theta)` can be positive as well as negative.