A very long, straight solenoid with a diameter of 3.00 cm is wound with 40 turns

Question

A very long, straight solenoid with a diameter of 3.00 cm is wound with 40 turns of wire per centimeter, and the windings carry a current of 0.250 A. A second coil having N turns and a larger diameter is slipped over the solenoid so that the two are coaxial. The current in the solenoid is ramped down to zero over a period of 0.5 s.

A) What average emf is induced in the second coil if it has a diameter of 4.30 cm and N = 38? Express your answer in microvolts to two significant figures.

B) What is the induced emf if the diameter is 8.60 cm and N = 76? Express your answer in volts to two significant figures.

Explanation / Answer

A)

magnetic field of the solenoid B = uo*n*I

n = 40 /cm = 4000 turns/m

flux = N*B*A = B*pi*r^2

induced emf e = rate of change in flux

e = N*uo*n*I*pi*r^2/t

e = (38*4*pi*10^-7*4000*0.25*pi*0.015^2)/(0.5)

e = 6.6*10^-5 V

B)

magnetic field of the solenoid B = uo*n*I

n = 40 /cm = 4000 turns/m

flux = N*B*A = B*pi*r^2

induced emf e = rate of change in flux

e = N*uo*n*I*pi*r^2/t

e = (76*4*pi*10^-7*4000*0.25*pi*0.015^2)/(0.5)

e = 1.4*10^-4V

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