A square loop of area 25 cm² has a resistance of 10 $Ω$ . The loop is placed in uniform magnetic field of magnitude 40.0 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 sec, will be

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$5 \times 10^{- 3} J$

$1.0 \times 10^{- 4} J$

$2.5 \times 10^{- 3} J$

$1.0 \times 10^{- 3} J$

answer is C.

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Detailed Solution

Work done by agent
$\begin{array}{ll} = Power \times t = i^{2} R \times t \\ = {(\frac{BlV}{R})}^{2} Rt = \frac{B^{2} l^{2} V^{2} t}{R} = 1 \times 10^{- 3} J \end{array}$

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