A solenoid has N = 490 loops of wire and a cross-sectional area A = 15 cm².It is moved from a field-free region to one where the magnetic field B = 0.8 T points directly along the axis of the solenoid.If the motion proceeds with constant velocity and takes t = 15 seconds,what is the magnitude of the average induced EMF? A)0.039 V B)1.25 V C)12.0 V D)125 V E)4900 V

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Quizplus · Accepted Answer

The average induced EMF (ε) can be calculated using Faraday's law of electromagnetic induction: ε = -N(ΔΦ/Δt), where ΔΦ is the change in magnetic flux. Here, ΔΦ = B*A = 0.8 T * 15 cm² = 0.8 T * 0.0015 m² = 0.0012 Wb. Thus, ε = -490 * (0.0012 Wb / 15 s) = 0.0392 V.

A Solenoid Has N = 490 Loops of Wire and a Cross-Sectional