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A square loop of area 25 \mathrm{~cm}^2 has a resistance of 10 \Omega . The loop is placed in uniform magnetic field of magnitude 40.0 \mathrm{~T}. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 \mathrm{sec}, will be

Option: 1

1.0 \times 10^{-3} \mathrm{~J}


Option: 2

2.5 \times 10^{-3} \mathrm{~J}


Option: 3

5 \times 10^{-3} \mathrm{~J}


Option: 4

1.0 \times 10^{-4} \mathrm{~J}


Answers (1)

best_answer

\begin{aligned} & \mathrm{l}=5 \mathrm{~cm} \\ & \mathrm{t}=1 \mathrm{sec} \\ & \mathrm{V}=\frac{0.05}{1}=0.05 \mathrm{~ms}^{-1} \\ & \mathrm{I}=\frac{40 \times 0.05 \times 0.05}{10}=\frac{\mathrm{BLV}}{\mathrm{R}}=0.01 \mathrm{~A} \\ & \mathrm{~F}=\mathrm{BIL}=40 \times 0.010 .05=0.02 \mathrm{~N} \\ & \mathrm{~W}=\mathrm{F} \ell=0.02 \times 0.05=1 \times 10^{-3} \mathrm{~J} \end{aligned}

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shivangi.shekhar

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