Physics - EMI/AC Question with Solution | TestHub

The emf induced in a differential element on the ring is

$${\mathrm{dE}}_{\text{induced }}=(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}})\cdot \mathrm{d}\vec{\ell }$$

Note that vector $\vec{v}\times \vec{B}$ is parallel to vector $\mathrm{d}\vec{\ell }$ and vectors v and B are perpendicular to each other. Hence ${\mathrm{dE}}_{\text{induced }}=\mathrm{vB}\oint \mathrm{d}\ell =2\pi \mathrm{rBv}$ If A is the cross-sectional area of the ring and $\rho$ the resistivity, the resistance of the ring is

$$\mathrm{R}=\frac{\rho (2\pi \mathrm{r})}{\mathrm{A}}$$

The current in the ring is $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{2\pi \mathrm{rBvA}}{\rho 2\pi \mathrm{r}}=\frac{\mathrm{BvA}}{\rho }$ The induced electric field in the ring is anticlockwise, so is the current in the ring. From Fleming's left hand rule, the magnetic force is upward.

$${\mathrm{F}}_{\mathrm{B}}=\mathrm{I}\ell \mathrm{B}=\mathrm{B}\left(\frac{\mathrm{ABv}}{\rho }\right)2\pi \mathrm{r}=\frac{2\pi {\mathrm{AB}}^2\mathrm{vr}}{\rho }$$

According to Newton's second law, the equation of motion of the ring is

$$\mathrm{mg}=\frac{2\pi {\mathrm{rAB}}^2\mathrm{r}}{\rho }\mathrm{v}=\mathrm{ma}$$

The terminal velocity is reached when the magnetic force balances the weight of the ring. Terminal velocity is thus given by

$$\begin{array}{rlr}& \frac{2\pi A{B}^2{\mathrm{rv}}_{\mathrm{T}}}{\rho }=\mathrm{mg}& {\mathrm{v}}_{\mathrm{T}}=\frac{\mathrm{mg}\rho }{2\pi {\mathrm{AB}}^2\mathrm{r}}\end{array}$$