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

Given: $B = 0 . 5T\text{at}t = 0 \text{ and } B = 0$, at $t = 0 . 5s$.

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,

${ε}_{\text{ind}} = \left|\dfrac{\Deltaϕ}{\Deltat}\right| = \left|\dfrac{\Delta\left(BA\right)}{\Deltat}\right| = A\left|\dfrac{\Delta\left(B\right)}{\Deltat}\right|$

$= \pi \times {\left(\dfrac{10}{\sqrt{\pi}}\right)}^{2} \times {10}^{- 4} \times \left(\dfrac{0 . 25}{0 . 25}\right)$

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

$= 0 . 01V = 10mV$