Question

The figure shows a conducting circular loop of radius $50\text{cm}$ is placed in a uniform, perpendicular magnetic field $4\text{T}.$ A metal rod $\text{OA}$ is pivoted at the centre $\text{O}$ of the loop. The other end $\text{A}$ of the rod touches the loop. The rod $\text{OA}$ and loop are resistanceless but a resistor having a resistance $5\Omega$ is connected between $\text{O}$ and a fixed point $\text{C}$ on the loop. The rod $\text{OA}$ is made to rotate anticlockwise at a uniform angular speed $20\text{rad/s}$ by an external force. Find the external torque $\left(\text{in Nm}\right)$ needed to keep the rod rotating with the constant angular velocity of $20\text{rad/s}$.
(Give only integer value)

Answer 1

Given:
$a=50\text{cm};B=4\text{T};R=5\Omega$
$\omega =20\text{rad/s}$

Emf induced between the ends of the rod,

$E=\frac{B\omega l^2}{2}=\frac{B\omega a^2}{2}[\because l=a]$

The equivalent circuit can be drawn as,


So the current induced in the circuit

$i=\frac{E}{R}=\frac{B\omega a^2}{2R}$



The force on the rod is,

$F=ilB$

As this force is uniformly distributed over $\text{OA}$, it may be assumed to act at the middle point of $\text{OA}$.

Therefore, the torque on it about $\text{O}$ is,

$\tau =\left(iaB\right)\times \frac{a}{2}=\frac{B^2\omega a^4}{4R}$

To keep the rod rotating with constant angular velocity, an external torque of the same magnitude should be applied on the rod in anticlockwise direction.

$\therefore \tau =\frac{(4{)}^2\times 20\times (0.5{)}^4}{4\times 5}=1\text{Nm}$

Accepted answer: $1$