Electromagnetic Induction

JEE Main level, single correct, medium

42 of 153

Question 42 of 153

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:

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

ϕ2=B1×A2=μ0IR122(R12+x2)32×πR22\begin{matrix}{\phi }_{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{matrix}

= 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}^{-11}

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

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