A current carrying circular loop is perpendicular to a magnetic field of induction $10^{-4} \mathrm{~T}$. If…

A current carrying circular loop is perpendicular to a magnetic field of induction $10^{-4} \mathrm{~T}$. If the radius of the loop starts shrinking at a uniform rate of $2 \mathrm{mms}^{-1}$, then the emf induced in the loop at the instant, when its radius is $20 \mathrm{~cm}$ will be
  1. $0.02 \pi \mu \mathrm{V}$
  2. $0.08 \pi \mu \mathrm{V}$
  3. $0.03 \pi \mu \mathrm{V}$
  4. $0.05 \pi \mu \mathrm{V}$

Solution

Given, $B=10^{-4} \mathrm{~T}, r=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$ $ \frac{d r}{d t}=-\frac{2 \mathrm{~mm}}{\mathrm{~s}}=-2 \times 10^{-3} \frac{\mathrm{m}}{\mathrm{s}} $ Induced emf, $\varepsilon=-\frac{d \phi}{d t} \Rightarrow \varepsilon=-\frac{d}{d t}\left(B \cdot \pi r^2\right)$ $ \begin{aligned} & \varepsilon=-B \pi \cdot \frac{d r^2}{d t}=-B \pi\left(2 r \cdot \frac{d r}{d t}\right) \\ & \varepsilon=10^{-4} \times \pi \times 2 \times 20 \times 10^{-2} \times 2 \times 10^{-3} \\ & \varepsilon=8 \pi \times 10^{-8}=0.08 \pi(\mu \mathrm{V}) \end{aligned} $

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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