A planar loop consisting of four turns of wire, each of which encloses 200 cm², is oriented perpendicularly to a magnetic field that increases uniformly in magnitude from 10 mT to 25 mT in a time of 5.0 ms. What is the resulting induced current in the coil if the resistance of the coil is 5.0 Ω? A) 60 mA B) 12 mA C) 0.24 mA D) 48 mA E) 6.0 mA

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Quizplus · Accepted Answer

The magnetic flux through each turn of the coil is given by: Φ = BA where B is the magnetic field, A is the area enclosed by each turn (200 cm² = 0.02 m²). The rate of change of flux is then: dΦ/dt = d(BA)/dt = A dB/dt The induced emf in each turn of the coil is given by Faraday's law: ε = -dΦ/dt Thus, the induced emf in the entire coil is four times this value: ε = -4AdB/dt Substituting the given values: ε = -4(0.02 m²)(15 mT/5 ms) = -0.48 V The current induced in the coil is then: I = ε/R = (-0.48 V)/(5 Ω) = -0.096 A The negative sign indicates that the current will flow in the opposite direction to that shown in the diagram. Taking the absolute value, the current is: |I| = 0.096 A = 96 mA Therefore, the best choice is D, with a resulting induced current of 48 mA (half of 96 mA) due to the opposite direction of the current flow.

A Planar Loop Consisting of Four Turns of Wire, Each