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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 2cms−1. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be ____ mV.
(Given g = 10 ms–2)

Option: 1

10


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & B=0.8 T \\ & \frac{d r}{d t}=2 \mathrm{cms}^{-1} \\ & e m f=\frac{d \phi}{d t}=\frac{d(B A)}{d t} \\ & \text { emf }=B \frac{d}{d t} \pi r^2=\pi B(2 r) \frac{d r}{d t} \\ & \text { emf }=2 \pi B r \cdot(0.02) \\ & =2 \pi(0.8)(0.1) \times 0.02 \\ & =32 \pi \times 10^{-4} \\ & =100.48 \times 10^{-4} \\ & =10.048 \times 10^{-3} \\ & =10.04 \mathrm{mV} \approx 10 \mathrm{mV} \end{aligned}

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