Question

A very long, straight solenoid with a cross-sectional area of 1.97 cm2 is wound with 94.6 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i(t)= ( 0.160 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. 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.

Answer 1

EMF equation e=N*A*(dB/dt)
________________________________

n = 94.6 turns/cm = 9460 turns/metre

The field inside the long solenoid is given by B = ??ni
B = 4?x10?? x 9460 x 0.161t² = 1.9020x10?³ t²

dB/dt = 3.8040x10?³ t
________________________________

A = 1.97cm² = 1.97x10?? m²

|Emf| = rate of change of flux linkage

|Emf| = d(NAB)/dt = NA dB/dt
= 5 x 1.97x10?? x 3.8040x10?³ t
= 3.7470x10?? t
_______________________________

If T is the time at which the current = 3.2A
3.2 = 0.161T²
T = 4.458s
_________________________________

|emf| = 3.7470x10?? T
= 3.7470 x x10?? x4.458
= 1.670x10?? V

Hope it will help you.

Please rate the answer.