Shouldn't this be a mechanics problem??

If we consider a small element of the ring of angular width $\delta\theta$ a a point which makes an angle $\theta$ with the upwards vertical, then there is charge $\tfrac{\delta\theta}{2\pi}C$ (if the total charge is $C$ ) in that segment, moving with velocity $\mathbf{v} = r\omega(\cos\theta\mathbf{i} - \sin\theta\mathbf{k})$ . Since the ambient magnetic field is $\mathbf{B} = B\mathbf{k}$ , this segment of the ring experiences a force $\delta\mathbf{F} \; = \; \tfrac{\delta\theta}{2\pi}C \mathbf{v} \wedge \mathbf{B} \; = \; -\tfrac{1}{2\pi}CBr\omega \cos\theta\,\delta\theta\mathbf{j}$ Thus the resultant force on the ring due to the magnetic field is $\mathbf{F} \; = \; -\tfrac{1}{2\pi}\int_0^{2\pi} CBr\omega \cos\theta\,d\theta\,\mathbf{j} \; = \; \mathbf{0}$

On the other hand, the force on the ring segment produces a moment $\delta\mathbf{Q} \; = \; r\big(\sin\theta\mathbf{i} + \cos\theta\mathbf{k}\big) \wedge \delta\mathbf{F} \; = \; \tfrac{1}{2\pi}CBr^2\omega\big(\cos^2\theta\mathbf{i} - \sin\theta\cos\theta\mathbf{k}\big)\delta\theta$ about the centre $O$ , and so the magnetic field $\mathbf{B}$ generates a torque $\mathbf{Q} \; = \; \tfrac{1}{2\pi} \int_0^{2\pi}CBr^2\omega\big(\cos^2\theta\mathbf{i} - \sin\theta\cos\theta\mathbf{k}\big)\,d\theta \; = \; \tfrac12CBr^2\omega\mathbf{i}$ about the centre $O$ .

We know that the weight of the rod plus ring is $W = 2T_0$ . Thus we know that $T_1 + T_2 = 2T_0 \hspace{2cm} \tfrac12DT_1 + Q \; = \; \tfrac12DT_2$ where $D$ is the length of the rod, and hence $T_1 \; = \; T_0 - D^{-1}Q \hspace{2cm} T_2 \; = \; T_0 + D^{-1}Q$ and hence we must have $D^{-1}Q \le \tfrac12T_0$ , and so $\omega \; \le \; \frac{DT_0}{CBr^2}$ With $D=5$ , $T_0=40$ , $C=20$ , $B=10$ and $r=1$ , we deduce that $\omega \le \boxed{1}\; \mathrm{s}^{-1}$ .

The maximum angular velocity is given by Omega= DT0/QBR^2