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 0.270 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 0.4 s.

Part A:What average emf is induced in the second coil if it has a diameter of 3.8 cm and N = 9? Express your answer in microvolts to two significant figures.

Part B:What is the induced emf if the diameter is 7.6 cm and N = 18? Express your answer in microvolts to two significant figures.

Please explain all steps.

Explanation / Answer

a)

Magnetic field due to solenoid is given by

B=uonI =(4pi*10-7)(40*102)*0.27=1.357*10-3 T

Area of coil

A=pi*0.0382/4=1.134*10-3 m2

Average emf induced in second coil is

E=NAdB/dt =9*(1.134*10-3)*(1.357*10-3/0.4)

E=3.46*10-5 A or 34.6 uA

b)

A=pi*0.0762/4 =4.536*10-3 m2

E=18*(4.536*10-3)*(1.357*10-3)/0.4

E=2.77*10-4 A or 277 uA

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