Question

A metal rod OA and mass m and length r kept rotating with a constant angular speed to in
a vertical plane about horizontal axis at the end O. The free end A is arranged to slide
without friction along a fixed conducting circular ring in the same plane as that of rotation
. A uniform and constant magnetic induction B is applied perpendicular and into the plane
of rotation as shown in figure. An inductor L and an external resistance R are connected through a switch 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 terminals 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, obtain the time dependence of the torque required to maintain the
constant angular speed. Given that the rod OA was along the positive x-axis at
t = 0.

Answer 1

Consider a small element of length dx of the rod OA
situated at a distance x from O.
Speed of this element, v = x $\omega$
Therefore, induced emf developed across this element in
uniform magnetic field B
de = (B) ( $\omega )$ dx ( e = Bvl)
Hence, total induced emf across OA,
e = $\int \limits_{ x = 0 }^{ x = y } de = \int \limits_0^r \, B \omega x dx = \frac{ B \omega r^2 }{ 2 } \Rightarrow \, e = \frac{ B \omega r^2 }{ 2 }$
(b) (i) A constant emf or PD, e = $\frac{ B \omega r^2 }{ 2 }$ is induced across O and A
The equivalent circuit can be drawn as shown in the figure.
Switch S is closed at time t = 0. Therefore, it is case of
growth of current in an L-R circuit. Current at any time t
is given by
i = $i_0 (1 - e^{ - t / \pi L} ) , i_0 = \frac{ e}{R} = \frac{ B \omega r^2 }{ 2 R}$
$T_L = L/R$
i = $\frac{ B \omega r^2 }{ 2 R} \Bigg [ 1 - e^{ - \bigg( \frac{ R}{ L} \bigg)t } \Bigg ]$
The -t graph will be as follows
(ii) At constant angular speed, net torque =0
The steady state current will be i = $i_0 = \frac{ B \omega r^2 }{ 2 R }$
From right hand rule we can see that this current would
be inwards (from circumference to centre) and
corresponding magnetic force will be in the direction
shown in figure and its magnitude is given by
$F_m = ( i) \, ( r) \, (B) = \frac{ B^2 \omega r^3 }{ 2 R}$ $( F_m = ilB )$
Torque of this force about centre O is
$T_{ F^m} = F_m . \frac{ F}{ 2 } = \frac{ B^2 \omega r^4 }{ 4 R }$ ( clockwise)
Similarly, torque of weight (mg) about centre O is
$T_{ mg } = ( mg) \frac{ r }{ 2 } cos \, \theta = \frac{ mgr }{ 2 } cos \omega t$ ( clockwise)
Therefore, net torque at any time t ( after steady state
condition is achieved) about centre O will be
$T_{ net } = T_{ F^m} + T_{ mg}$
= $\frac{ B^2 \omega r^4 }{ 4 R } + \frac{ mgr }{ 2 } cos \omega t$ ( clockwise)
Hence, the external torque applied to maintain a constant
angular speed is $T_{ ext} = \frac{ B^2 \, \omega r^4 }{ 4 R} + \frac{ mgr }{ 2} cos \omega t$
(but in 2 anticlockwise direction).
Note that for $\frac{ \pi }{ 2 } < \theta < \frac{ 3 \pi}{ 2 },$ torque of weight will be
anti-clockwise the sign of which is automatically
adjusted because cos $\theta$ = negative for $\frac{ \pi }{ 2 } < \theta < \frac{ 3 \pi}{ 2 },$