A square loop of side length Land resistance R is placed in a time-varying magnetic field B = B0 sin(omega t) directed perpendicular to the plane of the loop. What is the maximum power dissipated in the loop?

A. B_0^2 ^2 L^42R — The induced emf is given by Faraday's law as epsilon = -d(phi)/dt. Calculating the derivative of the flux phi = B*L^2 gives epsilon = B0*omega*L^2*cos(omega*t), and the power is P = epsilon^2/R, which has a maximum value when cos^2(omega*t) = 1.

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$A square loop of side length L and resistance R is placed in a time-varying magnetic field B = B_{0} sin (ω t) directed perpendicular to the plane of the loop. What is the maximum power dissipated in the loop?$

A. $\frac{B _{0}^{2} ω ^{2} L ^{4}}{2 R}$
B. $\frac{B _{0}^{2} ω ^{2} L ^{4}}{R}$
C. $\frac{B _{0}^{2} ω ^{2} L ^{2}}{2 R}$
D. $\frac{B _{0}^{2} ω L ^{4}}{2 R}$

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Correct answer: A — $\frac{B _{0}^{2} ω ^{2} L ^{4}}{2 R}$

Solution

The induced emf is given by Faraday's law as ϵ = - \frac{d ϕ}{d t} . Calculating the derivative of the flux ϕ = B L^{2} gives ϵ = B_{0} ω L^{2} cos (ω t), and the power is P = \frac{ϵ ^{2}}{R}, which has a maximum value when cos^{2} (ω t) = 1 .

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A square loop of side length L and resistance R is placed in a time-varying magnetic field B=B0​sin(ωt) directed perpendicular to the plane of the loop. What is the maximum power dissipated in the loop?

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More Electromagnetic Induction practice questions

$A square loop of side length L and resistance R is placed in a time-varying magnetic field B = B_{0} sin (ω t) directed perpendicular to the plane of the loop. What is the maximum power dissipated in the loop?$