Question

A metal rod $OA$ of mass $m$ and length $r$ is kept rotating with a constant angular speed $w$ in a vertical plane about a 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 $\overrightarrow{B}$ 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. What is the induced emf across the terminals of the switch?
(i) Obtain an expression for the current as a function of time after switch $S$ is closed.
(ii) 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

When switch is not closed current through circuit is zero and em induced in the rod is equal to the emf induced across switch.

$e=\left(\frac{wr}{2}\right)B\left(r\right)=\frac{B\omega +2}{2}$ towards center 0.

i) When switch S closed current through the circuit $i$

$I=I_0(1-e\frac{-t}{T})$ where $I_0=\frac{e}{R}$

$I=\frac{Bw{r}^2}{2R}(1-e\frac{-Rt}{L})T=\frac{L}{R}$ (time constant) current flow anty clockwise.

ii)Electromagnetic force $F_m$ act and equal to $F_m=Idl\times \overrightarrow{B}=IrB$ downward direction between the OA.

Torque $=\frac{r}{2}{F}_m$ directly outward

Torque $=\frac{B^{2}w{r}^4}{4}(1-e\frac{-Rt}{L})$