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Q. 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 $2 \,cm ^{-1}$. The induced emf in the loop at an instant when the radius of the loop is $10\, cm$ will be___ $mV$.

JEE MainJEE Main 2023Electromagnetic Induction

Solution:

$ \text { EMF }=\frac{ d }{ dt }\left( B \pi r ^2\right) $
$ =2 B \pi r \frac{ dr }{ dt }=2 \times \pi \times 0.1 \times 0.8 \times 2 \times 10^{-2}$
$ =2 \pi \times 1.6= 1 0 . 0 6 \text { [round off } 1 0 . 0 6 = 1 0 \text { ] }$