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A square loop of side 1 m being moved towards right at a constant speed of 1m/s. At t = 0, the front edge of loop enters the 3m wide uniform magnetic field B=2 Tesla. The resistance of square loop is $2 Ω$ . The total heat produced in the loop during the interval 0 to 5sec is

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1 Joule

4 Joule

2 Joule

8 Joule

answer is A.

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Detailed Solution

$H e a t = {(\frac{B l v}{R})}^{2} \times R \times 2 = \frac{2 B^{2} l^{2} v^{2}}{R} = \frac{2 \times 4 \times 1 \times 1}{2} = 4 J$

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