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Question

A square loop of side 20 cm is placed in a magnetic field B=(5+2t2); such that, the magnetic field is perpendicular to the plane of the loop. If resistance of loop is 2Ω, heat generated in 5 sec in joules is (write until two digits after the decimal point)

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Solution

ε=dϕdt=ddtBA=ddt(5+2t2)a2
ε=4a2t=i=R=4a2t2=2a2t
Heat generated =50i2Rdt
= 504a4t2Rdt=4a4R(t33)50
Heat =4×0.24×2×533=0.533J

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