Question

A solenoidal coil with 30 turns of wire is wound tightly around another coil with 310 turns. The inner solenoid is 20.0 cm long and has a diameter of 2.50 cm . At a certain time, the current in the inner solenoid is 0.150 A and is increasing at a rate of 1700 A/s .

Part A

For this time, calculate the average magnetic flux through each turn of the inner solenoid.   

Part B

For this time, calculate the mutual inductance of the two solenoids;   

Part C

For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.   

Answer 1

Number of turns in inner solenoid = N1 = 310

Number of turns in outer solenoid = N2 = 30

Length of inner solenoid = L=20.0 cm = 0.20 m

Radius of inner solenoid = r=diameter/2= 2.50/2=1.25 cm=0.0125 m

The current in the inner solenoid = i = 0.150 A

Rate of increasing of current =di/dt = 1700 A/s.

The average magnetic flux through each turn of the inner solenoid. =B.A=(mu not)(N/L)i.A

The average magnetic flux through each turn of the inner solenoid. =[4(pi)*10^-7][310/0.20][0.15](pi)(0.0125)

A)The average magnetic flux through each turn of the inner solenoid. =8.45966*10^-8 weber
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The mutual inductance of the two solenoids=N2* 8.45966*10^-8

(B) The mutual inductance of the two solenoids =2.53789*10^-6 weber/A
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(C) The emf induced in the outer solenoid by the changing current in the inner solenoid.=3.94548*10^-4 V

Calculations maybe wrong, logic is correct !!

Hope this helps! :)