Question

Figure shows a situation similar to the previous problem. All parameters are the same except that a battery of emf ε and a variable resistance R are connected between O and C. Neglect the resistance of the connecting wires. Let θ be the angle made by the rod from the horizontal position (show in the figure), measured in the clockwise direction. During the part of the motion 0 < θ < π/4 the only forces acting on the rod are gravity and the forces exerted by the magnetic field and the pivot. However, during the part of the motion, the resistance R is varied in such a way that the rod continues to rotate with a constant angular velocity ω. Find the value of R in terms of the given quantities.

Answer 1

It is given that the rod is rotated with angular speed in clockwise direction.
The emf induced in the rod (e) is $\frac{\mathit{B}\mathit{\omega }{\mathit{a}}^{\mathit{2}}}{\mathit{2}}$, with O at the lower potential and A at the higher potential.
The equivalent circuit can be drawn as:

$$\begin{gathered} i=\frac{e+E}{R}=\frac{{\displaystyle \frac{1}{2}}B\omega a^2+E}{R} \\ =\frac{B\omega a^2+2E}{2R} \end{gathered}$$
Because the rod rotates with uniform angular velocity, the net torque about point O is zero.
Now,
Net force on the rod, F_net = mg cos θ $-$ ilB
Net torque, τ = (mg cos θ $-$ ilB).(r/2) = 0
∴ mg cos θ = ilB
$$\begin{gathered} R=\frac{(B\omega a^2+2E)}{2\mathrm{R}}(a\mathit{\times }B) \\ \Rightarrow R=\frac{(B\omega a^2+2E)aB}{2mg\cos \theta } \end{gathered}$$