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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Physics
Chapter
Electromagnetic Induction

BITSAT 2016 · Physics · Electromagnetic Induction

A conducting circular loop is placed in a uniform magnetic field of...

A conducting circular loop is placed in a uniform magnetic field of 0.04T0.04 \mathrm{ T} with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 2mm/s2 \mathrm{ mm} / \mathrm{s}. The induced emf in the loop when the radius is 2cm2 \mathrm{ cm} is

Options

  1. A4.8πμV4.8 \pi \mu \mathrm{V}
  2. B0.8πμV0.8 \pi \mu \mathrm{V}
  3. C1.6πμV1.6 \pi \mu \mathrm{V}
  4. D3.2πμV3.2 \pi \mu \mathrm{V}

Correct answer

D. 3.2πμV3.2 \pi \mu \mathrm{V}

Step-by-step solution

Induced emf in the loop is given by e=−B⋅dAdt\mathrm{e}=-\mathrm{B} \cdot \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}} where A\mathrm{A} is the area of the loop. e=−B⋅ddt(π\mathrmr2)=−Bπ2rdrdtr=2cm=2×10−2mdr=2mm=2×10−3mdt=1se=−0.04×3.14×2×2×10−2×2×10−31V=0.32π×10−5V=3.2π×10−6V=3.2πμV\begin{array}{l}\mathrm{e}=-\mathrm{B} \cdot \frac{\mathrm{d}}{\mathrm{dt}}\left(\pi \mathrmr^{2}\right)=-\mathrm{B} \pi2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}\\ \mathrm{r}=2 \mathrm{ cm}=2 \times10^{-2} \mathrm{ m}\\ \mathrm{dr}=2 \mathrm{ mm}=2 \times10^{-3} \mathrm{ m}\\ \mathrm{dt}=1 \mathrm{ s}\\ \mathrm{e}=-0.04 \times3.14 \times2 \times2 \times10^{-2} \times \frac{2 \times 10^{-3}}{1} \mathrm{ V}\\ =0.32 \pi \times10^{-5} \mathrm{ V}\\ =3.2 \pi \times10^{-6} \mathrm{ V}\\ =3.2 \pi \mu \mathrm{V} \end{array}
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