Question

Consider a variation of the previous problem (figure). Suppose the circular loop lies in a vertical plane. The rod has a mass m. The rod and the loop have negligible resistances but the wire connecting O and C has a resistance R. The rod is made to rotate with a uniform angular velocity ω in the clockwise direction by applying a force at the midpoint of OA in a direction perpendicular to it. Find the magnitude of this force when the rod makes an angle θ with the vertical.

Answer 1

When the circular loop is in the vertical plane, it tends to rotate in the clockwise direction because of its weight.

Let the force applied be F and its direction be perpendicular to the rod.
The component of mg along F is mg sin θ.
The magnetic force is in perpendicular and opposite direction to mg sin θ.
Now,
Current in the rod:
$i=\frac{B{a}^2\omega }{2R}$
The force on the rod is given by
$F_{\mathrm{B}}=iBl=\frac{B^2{a}^2\omega }{2R}$

Net force = F− $\frac{B^2{a}^2\omega }{2R}$ + mg sin θ
The net force passes through the centre of mass of the rod.
Net torque on the rod about the centre O:
$\tau =\left(F-\frac{B^2{a}^3\omega }{2R}+mg\sin \theta \right)\frac{\mathrm{OA}}{2}$
Because the rod rotates with a constant angular velocity, the net torque on it is zero.
i.e. $\tau =0$
$\left(F-\frac{B^2{a}^3\omega }{2R}+mg\sin \theta \right)\frac{\mathrm{OA}}{2}=0$
∴ $F=\frac{B^2{a}^3\omega }{2R}-mg\sin \theta$