Physics - EMI/AC Question with Solution | TestHub

The magnetic flux through the ring is ${\Phi }_B={\int }_0^R{B}_0\left(1-\frac{r^{\prime 2}}{R^2}\right)\cos (\omega t)(2\pi r^{\prime }d{r}^{\prime })=\frac{\pi B_0{R}^2}{2}\cos (\omega t)$. At $t=0$, ${\Phi }_B(0)=\frac{\pi B_0{R}^2}{2}$, so A is correct. The induced EMF is $\mathcal{E}=-\frac{d{\Phi }_B}{dt}=\frac{\pi B_0{R}^2\omega }{2}\sin (\omega t)$, giving peak EMF $\frac{\pi B_0{R}^2\omega }{2}$. Thus B is correct.

For $0<\omega t<\pi$, $\sin (\omega t)>0$. As flux in $+\hat{k}$ direction is decreasing for $0<\omega t<\pi$, Lenz's law dictates a counter-clockwise current. Thus C is incorrect. The average power is $P_{avg}=⟨I^{2}r⟩=\frac{1}{r}⟨{\mathcal{E}}^2⟩=\frac{1}{r}\frac{(\text{Peak EMF}{)}^2}{2}=\frac{{\pi }^2{B}_0^2{R}^4{\omega }^2}{8r}$, so D is correct.