A flat loop of wire consisting of a single turn of cross-sectional area 7.10 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 t to 2.00 t in 0.93 s. what is the magnitude of the resulting induced current if the loop has a resistance of 2.00?

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Physics

Rory Bowman

25 November, 22:19

A flat loop of wire consisting of a single turn of cross-sectional area 7.10 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 t to 2.00 t in 0.93 s. what is the magnitude of the resulting induced current if the loop has a resistance of 2.00?

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Nunez · Answer 1

From Ohm's law. V = IR. Where R = 2 Ohms. To calculate the induced current I; We need to calculate the electromotive force or voltage, V. From Faraday's law induced EMF = (The rate of change of magnetic flux density x Area) / (changein time). Or EMF = BA/t. Where B = Bf - Bi. And BA = Bf * A - Bi * A. Bf = 2.00 and Bi = 0.500 and t = 0.93s and the area, A = 7.1 cm^2 is 0.000071 m^2. 2 So Emf = 2.00 (0.000071) - 0.500 (0.000071) / (0.93) = 1.0654 * 10^ (-4) / 0.93 = 1.1415 * 10^ (-4). Substituting into ohms law, we have, I = (1.1415 * 10^ (-4)) / 2 = 0.57075 * 10^ (-4)