Question

A solenoidal coil with 28 turns of wire is wound tightly around another coil with 320 turns. The inner solenoid is 21.0cm long and has a diameter of 2.20cm . At a certain time, the current in the inner solenoid is 0.150Aand is increasing at a rate of 1600A/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 = 320

Number of turns in outer solenoid = N2 = 28

Length of inner solenoid = L=21.0 cm = 0.21 m

Radius of inner solenoid = r=diameter/2= 2.20/2=1.10 cm=0.11 m

The current in the inner solenoid = i = 0.150 A

Rate of increasing of current =di/dt = 1600 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][320/0.21][0.15](pi)(0.11)...

A)The average magnetic flux through each turn of the inner solenoid. =9.26423*10^-8 weber
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The mutual inductance of the two solenoids=N2*9.26423*10^-8

(B) The mutual inductance of the two solenoids =2.779269*10^-6 weber/A
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(C) The emf induced in the outer solenoid by the changing current in the inner solenoid.=5.91822*10^-4 V
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