Physics - EMI/AC Question with Solution | TestHub

The magnetic flux through the ring is ${\Phi }_B=B(t)\cdot \pi R^2=(0.2-0.1t)\pi R^2$. The induced EMF around the ring is $\mathcal{E}=-\frac{d{\Phi }_B}{dt}=-\frac{d}{dt}[(0.2-0.1t)\pi R^2]=0.1\pi R^2$. Due to symmetry, the induced electric field $E$ is tangential and uniform along the ring: $\mathcal{E}=E(2\pi R)$. So, $E=\frac{0.1\pi R^2}{2\pi R}=\frac{0.1R}{2}$. The force on the charged ring is $F=QE=Q\frac{0.1R}{2}$. The torque on the ring is $\tau =FR=Q\frac{0.1R^2}{2}$. The moment of inertia of the ring is $I=M{R}^2$. The angular acceleration is $\alpha =\frac{\tau }{I}=\frac{Q\frac{0.1R^2}{2}}{M{R}^2}=\frac{0.1Q}{2M}$. Substituting the given values $Q=4\text{C}$ and $M=2\text{kg}$: $\alpha =\frac{0.1\times 4}{2\times 2}=\frac{0.4}{4}=0.1{\text{rad/s}}^2$.