Constants ▼ Part A A solenoidal coil with 26 turns of wire is wound tightly around...

Part A -

As mentioned in the problem -

Number of turns in inner solenoid = N1 = 310  

Number of turns in outer solenoid = N2 = 26

Length of inner solenoid = L=25.0 cm = 0.25 m

Radius of inner solenoid = r = diameter/2= 2.00/2=1.00 cm = 0.01 m

Again -

The current in the inner solenoid = i = 0.150 A

Rate of increasing of current =di/dt = 1600 A/s.

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

= [4(pi)*10^-7][310/0.25][0.15](pi)(0.01)^2

= 0.734 x 10^-7 Wb

= 7.34 x 10^-8 Wb

Part B -

The mutual inductance of the two solenoids, M =(N2* )/i

= (26 * 7.34 x 10^-8) / (0.15) Wb/A

= 1.272 x 10^-5 Wb/A

Part C -

EMF induced in the outer solenoid, E2 = -M*di/dt = -(1.272 x 10^-5) * 1600 =  -18.75 x 10^-3 V (Answer)