A metal rod of mass $m$ and length $a$ can rotate about a horizontal axis $O$, sliding along a circu

Question

A metal rod of mass $m$ and length $a$ can rotate about a horizontal axis $O$, sliding along a circular conductor of radius $a$. The arrangement is located in a uniform magnetic field of induction $B$ directed perpendicularly into the plane of the ring. The axis and the ring are connected to a voltage source to form a circuit of total resistance $R$. Neglecting friction, circuit inductance, and the resistance of the ring itself, determine the law according to which the source voltage $\mathcal{E}(t)$ must vary to make the rod rotate with a constant angular velocity $\omega$. The rod starts from the uppermost vertical position at $t=0$. (Assume $\mathcal{E}(t)$ is defined as positive when it tends to drive current radially outward from the center $O$ to the ring).

Professor PC Thomas Classes · Answer

Let the rod be at an angle $	heta = \omega t$ from the upward vertical. Gravity produces a clockwise torque: $	au_g = mg \left(\frac{a}{2}ight) \sin(\omega t)$. To maintain a constant angular velocity $\omega$, the net torque must be zero. Therefore, the magnetic force must provide an equal and opposite counter-clockwise torque $	au_m = 	au_g$. The magnetic torque on a rod carrying current $I$ from center to rim is $	au_m = \int_0^a r(I dr B) = \frac{1}{2} I B a^2$. Equating the torques: $\frac{1}{2} I B a^2 = \frac{1}{2} m g a \sin(\omega t) \implies I = \frac{mg \sin(\omega t)}{Ba}$. Due to its rotation, a motional emf is induced in the rod: $\mathcal{E}_{ind} = \frac{1}{2} B \omega a^2$. Using the right-hand rule for $\vec{v} 	imes \vec{B}$, the center $O$ is at a higher potential, meaning this induced emf opposes a current flowing from $O$ to the rim. Applying Kirchhoff's voltage law for the circuit: $\mathcal{E}(t) - \mathcal{E}_{ind} = I R$. Substituting the expressions for $\mathcal{E}_{ind}$ and $I$: $\mathcal{E}(t) = \frac{1}{2} B \omega a^2 + \left[ \frac{mg \sin(\omega t)}{Ba} ight] R = \frac{1}{2} B \omega a^2 + \frac{mg R}{Ba} \sin(\omega t)$.