JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 24)
A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B = 0.8 T. When released the radius of the loop starts shrinking at a constant rate of 2 cms$$^{-1}$$. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be __________ mV.
Answer
10
Explanation
$E M F=\frac{d}{d t}\left(B \pi r^{2}\right)$
$=2 \mathrm{~B} \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=2 \times \pi \times 0.1 \times 0.8 \times 2 \times 10^{-2}$
$=2 \pi \times 1.6=\mathbf{1 0 . 0 6}~[$ rounding off $\mathbf{1 0 . 0 6}=\mathbf{1 0}]$
$=2 \mathrm{~B} \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=2 \times \pi \times 0.1 \times 0.8 \times 2 \times 10^{-2}$
$=2 \pi \times 1.6=\mathbf{1 0 . 0 6}~[$ rounding off $\mathbf{1 0 . 0 6}=\mathbf{1 0}]$
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