A very long, straight solenoid with a cross-sectional area of 1.89cm 2 is wound

Question

A very long, straight solenoid with a cross-sectional area of 1.89cm 2 is wound with 93.4 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i(t) = ( 0.168A/s 2 )t 2. 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. Part A What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A ? Express your answer with the appropriate units.

Explanation / Answer

This should be e=N*A*(dB/dt)

n = 93.4 turns/cm = 9340 turns/m

The field inside the long solenoid is given by B = uo*ni

B = 4pix10^-7 x 9340 x 0..168t^2 = 1.97x10^-3 t^2

dB/dt = 3.9436x10^-3*t

A = 1.89cm^2 = 1.89x10^-4 m^2

|Emf| = rate of change of flux linkage

|Emf| = d(NAB)/dt = NA dB/dt

= 5 x 1.89x10^-4 x 3.9436x10^-3*t

= 3.72673x10^-6 t

If T is the time at which the current = 3.2A

3.2 = 0.168T^2

T = 4.364 s

|emf| = 3.72673x10^-6 x 4.364

= 1.626x10^-5 V

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