# Part AWhat is the magnitude of the emf induced in the secondary winding at the instant that the current inthe solenoid is 3.2 A ?Express your answer with the appropriate units.?HAValueEl =UnitsRequest AnswerSubmitProvide FeedbackNext>

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A very long, straight solenoid with a cross sectional area of 1.92 cmis wound with 92.1 turns of wire per centimeter. Starting at t=0, the current in the solenoid is increasing according to i(t) = (0.180 A/s2)t2. A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross sectional area as the solenoid.

What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2 A ?

|E| = ________ help_outlineImage TranscriptionclosePart A What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2 A ? Express your answer with the appropriate units. ? HA Value El = Units Request Answer Submit Provide Feedback Next> fullscreen
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Step 1 help_outlineImage TranscriptioncloseThe equation for the magnitude of the emf induced in the secondary winding according to Faraday's law is do dt Here, N2 is the number of turns in the secondary winding and ø is the magnetic flux associated with the secondary winding The equation for the magnetic flux is given by = BA Here, B is the magnetic field around the winding which is produced by the solenoid and A is the cross-sectional area of the winding The equation for the magnitude of the magnetic field produced by a solenoid at its core is given by B = ni Here, 4is the absolute permeability and n is the mumber of turns per unit length of the solenoid. Thus ф- (цт) А = 4Ani fullscreen
Step 2 help_outlineImage TranscriptioncloseAnd, N2 ni(t) dt di (t) = 46,N,An dt d(0.180 A/s) = 4N2A dt (0.180 A/s2)4,N,An(2t) - 2 (0.180 A/s2)4N,Ant (1) Rewrite the given equation of i (t) for time. (t) (0.180 A/s2) i(t) t2 0.180 A/s2 (t) 0.180 A/s2 fullscreen

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