A conducting loop of radius 10 π   cm is placed perpendicular to a uniform magnetic field of 0 . 5…

Given: $B=0.5 \mathrm{T} \text{at} t=0 \& B=0$, at $t=0.5 \mathrm{s}$.

Magnetic field will decrease linearly as given below. We can say, value of magnetic field at $t=0.25 \text{will be} B=0.25$.

Therefore,

${\epsilon }_{\text{ind }}=\left|\frac{\Delta \varphi }{\Delta t}\right|=\left|\frac{\Delta \left(BA\right)}{\Delta t}\right|=A\left|\frac{\Delta \left(B\right)}{\Delta t}\right|$

$=\pi \times {\left(\frac{10}{\sqrt{\pi }}\right)}^2\times {10}^{-4}\times \left(\frac{0.25}{0.25}\right)$

$={10}^{-2}\times 1$

$$\begin{gathered} =0.01 \mathrm{V} \\ =10 \mathrm{mV} \end{gathered}$$