Electromagnetic Induction

JEE Main level, numerical answer, medium

79 of 153

Question 79 of 153

A square loop PQRS having 10 turns, area 3.6×10−3 m23.6 \times 10^{-3} \mathrm{~m}^2 and resistance 100Ω100 \Omega is slowly and uniformly being pulled out of a uniform magnetic field of magnitude B=0.5 T\mathrm{B}=0.5 \mathrm{~T} as shown. Work done in pulling the loop out of the field in 1.0 s1.0 \mathrm{~s} is ______ ×10−6 J\times 10^{-6} \mathrm{~J}.

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Answer: 3

ϵ=NBℓvi=ϵR=NBℓvRF=N(iℓB)=N2B2ℓ2vR\begin{aligned} & \epsilon=N B \ell v \\ & i=\frac{\epsilon}{R}=\frac{N B \ell v}{R} \\ & F=N(i \ell B)=\frac{N^2 B^2 \ell^2 v}{R}\end{aligned}
W=F×ℓ=N2 B2ℓ3R(ℓt)A=ℓ2 W=(10×10)(0.5)2×(3.6×10−3)2100×1 W=3.24×10−6 J\begin{aligned} & \mathrm{W}=\mathrm{F} \times \ell=\frac{\mathrm{N}^2 \mathrm{~B}^2 \ell^3}{\mathrm{R}}\left(\frac{\ell}{\mathrm{t}}\right) \\ & \mathrm{A}=\ell^2 \\ & \mathrm{~W}=\frac{(10 \times 10)(0.5)^2 \times\left(3.6 \times 10^{-3}\right)^2}{100 \times 1} \\ & \mathrm{~W}=3.24 \times 10^{-6} \mathrm{~J}\end{aligned}

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