A 5-turn square loop (10 cm along a side, resistance = 4.0 Ω) is placed in a magnetic field that makes an angle of 30° with the plane of the loop. The magnitude of this field varies with time according to B = 0.50t², where t is measured in s and B in T. What is the induced current in the coil at t = 4.0 s? A) 25 mA B) 5.0 mA C) 13 mA D) 43 mA E) 50 mA

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Quizplus · Accepted Answer

The induced emf is given by:emf = -dΦ/dt = -d(BAcosθ)/dt = -BAsinθ(dB/dt)= -(0.50t^2)(0.10 m)^2(sin 30°)(2t)= -0.0050t^3 VThe induced current is then:I = emf/R = -0.0050t^3 V / 4.0 Ω= -1.25 x 10^-6 t^3 AAt t = 4.0 s, the induced current is:I = -1.25 x 10^-6 (4.0 s)^3 A= -25 mAThe correct answer is A.

A 5-Turn Square Loop (10 Cm Along a Side, Resistance