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in Electromagnetic Induction and Alternating Current by (48.6k points)
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A rectangular loop of sides 8 cm and 2 cm is lying in a uniform magnetic field of magnitude 0.5 T with its plane normal to the field. The field is now gradually reduced at the rate of 0.02 T/s. If the resistance of the loop is 1.6 Ω, then find the power dissipated by the loop as heat.

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Induced emf, |ε| = \(\frac{dΦ}{dt} = A\frac{dB}{dt}\) = 8 x 2 x 10-4 x 0.02

ε = 3.2 × 10-5 V

Induced current, I = \(\frac{ε}{R}\) = 2 × 10-5 A

Power loss = I2R = 4 × 10-10 × 1.6 = 6.4 × 10-10 W

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