A conducting loop of radius 10 π cm is placed perpendicular to a...

As \left.\varepsilon\right|_{t=0.5 \mathrm{sec}}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}$<br />$=-\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}} \quad\left[\because \theta=0^{\circ} \Rightarrow \cos \theta=1\right]$<br />$=-\pi \times\left(\frac{10}{\sqrt{\pi}}\right)^2 \times 10^{-4} \times \frac{0-0.5}{0.5}=10^{-2} \mathrm{~V}=10 \mathrm{mV}$<br />As$\frac{d B}{d t}=$constant$\Rightarrow Induced emf will not change with time. So, \left.\mathrm{e}\right|_{0,5 \mathrm{sec}}=\left.\mathrm{e}\right|_{0.25 \mathrm{sec}}=10 \mathrm{mV}