Physics - EMI/AC Question with Solution | TestHub

Key Idea According to Faraday's second law of electromagnetic induction the induced emf is given by rate of change of magnetic flux linked with the circuit. Here, $B=0.04\mathrm{T}$ and $\frac{-\mathrm{dr}}{\mathrm{dt}}=2{\mathrm{mms}}^{-1}$ Induced emf, $\mathrm{e}=\frac{-\mathrm{d}\varphi }{\mathrm{dt}}=\frac{-\mathrm{BdA}}{\mathrm{dt}}=-\mathrm{B}\frac{\mathrm{d}\left(\pi {\mathrm{r}}^2\right)}{\mathrm{dt}}$ $$=-\mathrm{B}\pi 2\mathrm{r}\frac{\mathrm{dr}}{\mathrm{dt}}$$ Now, if $\mathrm{r}=2\mathrm{cm}$ $$\begin{array}{rl}\mathrm{e}& =-0.04\times \pi \times 2\times 2\times 10^{-2}\times 2\times 10^{-3}\\ & =3.2\pi \mu \mathrm{V}\end{array}$$