A square loop of side 10 cm and resistance 0.5 ohm is moved out of a uniform magnetic field of 1 Tesla with a constant velocity of 2 meters per second. What is the power dissipated as heat in the loop?

B. 0.08 W — The induced EMF is epsilon = B * v * a. The power dissipated is P = epsilon squared / R. Substituting the values, we get (1 * 2 * 0.1) squared / 0.5 = 0.08 W.

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$A square loop of side a = 0.1 m and resistance R = 0.5 Ω is moved out of a uniform magnetic field B = 1 T with a constant velocity v = 2 m/s . What is the power P dissipated as heat in the loop?$

A. $0.04 W$
B. $0.08 W$
C. $0.16 W$
D. $0.02 W$

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Correct answer: B — $0.08 W$

Solution

The induced EMF is ε = B v a = 1 \times 2 \times 0.1 = 0.2 V . The power dissipated is P = \frac{ε ^{2}}{R} = \frac{( 0.2 ) ^{2}}{0.5} = \frac{0.04}{0.5} = 0.08 W .

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A square loop of side a=0.1 m and resistance R=0.5Ω is moved out of a uniform magnetic field B=1 T with a constant velocity v=2 m/s. What is the power P dissipated as heat in the loop?

Solution

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More Electromagnetic Induction practice questions

$A square loop of side a = 0.1 m and resistance R = 0.5 Ω is moved out of a uniform magnetic field B = 1 T with a constant velocity v = 2 m/s . What is the power P dissipated as heat in the loop?$