Physics - EMI/AC Question with Solution | TestHub

The induced EMF across the rod is given by $E={\int }_0^L(\vec{v}\times \vec{B})\cdot d\vec{l}$. Here $\vec{v}=r\omega \hat{\theta }$ and $\vec{B}=B_0(r/L)\hat{k}$. Thus, $\vec{v}\times \vec{B}=(r\omega \hat{\theta })\times (B_0(r/L)\hat{k})=-\frac{B_0\omega r^2}{L}\hat{r}$. Since $d\vec{l}=dr\hat{r}$, $E={\int }_0^L-\frac{B_0\omega r^2}{L}dr=-\frac{B_0\omega }{L}{\left[\frac{r^3}{3}\right]}_0^L=-\frac{1}{3}{B}_0\omega L^2$. The magnitude is $\frac{1}{3}{B}_0\omega L^2$. So, option A is correct. The current flowing through the resistor is $I=E/R=\frac{B_0\omega L^2}{3R}$. So, option B is correct. The power dissipated in the resistor is $P=I^{2}R={\left(\frac{B_0\omega L^2}{3R}\right)}^{2}R=\frac{B_0^2{\omega }^2{L}^4}{9R}$. So, option D is correct. Since $\vec{v}\times \vec{B}$ is in the $-\hat{r}$ direction, positive charges are pushed towards $O$. Thus, $O$ is at a higher potential. The current in the rod flows from the outer end (lower potential) to $O$ (higher potential). So, option C is incorrect.