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QJEEM23Ph72c0: JEE Main 2023 Physics - Class 12 JEEM23Ph72c0 Solution: A certain elastic conducting material is stretched into a…

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^{-1} . The induced emf in the loop at an instant when the radius of the loop is 10 cm will be __________ mV.

Step-by-Step Verified Solution

EMF=ddt(Bπr2)E M F=\frac{d}{d t}\left(B \pi r^{2}\right)

=2 Bπrdrdt=2×π×0.1×0.8×2×10−2=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}

=2π×1.6=10.06 [=2 \pi \times 1.6=\mathbf{1 0 . 0 6}~[ rounding off 10.06=10]\mathbf{1 0 . 0 6}=\mathbf{1 0}]

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