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JEE Main 20137 April
Exam
JEE Main
Year
2013
Date
7 April
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2013 · 7 April · Physics · Electromagnetic Induction

A circular loop of radius 0 . 3cm lies parallel to a much bigger...

A circular loop of radius 0.3cm0 . 3cm lies parallel to a much bigger circular loop of radius 20cm20cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is1515 cmcm. If a current of 2.0A2 . 0A flows through the smaller loop, then the flux linked with a bigger loop is:

Options

  1. A3.3×10−11weber3 .3 \times 1{0}^{- 11}weber
  2. B6.6×10−9weber6 .6 \times {10}^{- 9}weber
  3. C9.1×10−11weber9 .1 \times {10}^{- 11}weber
  4. D6×10−11weber6 \times {10}^{- 11}weber

Correct answer

C. 9.1×10−11weber9 .1 \times {10}^{- 11}weber

Step-by-step solution

ϕ2=B1×A2=μ0IR122(R12+x2)32×πR22\begin{array}{ll}{ϕ}_{2} = {B}_{1} \times {A}_{2} \\ = \dfrac{{\mu}_{0}I{R}_{1}^{2}}{2( {R}_{1}^{2} + {x}^{2} )\dfrac{3}{2}} \times \pi{R}_{2}^{2}\end{array}

= μ0(2)(20×10−2)22[(0.2)2+(0.15)2]32×π(0.3×10−2)\dfrac{{\mu}_{0}( 2 ){( 20 \times {10}^{- 2} )}^{2}}{2[ {( 0.2 )}^{2} + {( 0.15 )}^{2} ]\dfrac{3}{2}} \times \pi( 0.3 \times {10}^{- 2} )

On solving

=9.216×10−11= \text{9.216} \times 1 {0}^{- 1 1}

≈9.2×10−11weber\approx \text{9.2} \times 1 {0}^{- 1 1} \text{weber}
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