Question

A solenoidal coil with 23 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 25.0 cm long and has a diameter of 2.40 cm . At a certain time, the current in the inner solenoid is 0.120 A and is increasing at a rate of 1600 A/s

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

Express your answer in webers.

B)For this time, calculate the mutual inductance of the two solenoids.

Express your answer in henries.

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

Express your answer in volts.

Answer 1

Solution:-

                  We have a solenoid with length of L = 25.0 cm (L = 0.250 m), diameter of D = 2.40 cm(r = 0.024 m) and N1 = 350 turns. We have a second solenoid with N2 = 23 turns which are wound around the solenoid at its centre, It is given that at a certain time the current in the inner solenoid is I = 0.120 A and increasing at a di/dt = 1600 A/s

a) Calculate the average magnetic flux

At this instant the magnetic field at the centre of the solenoid is

B = u0*N1*I = u0*N1*I / L

B = (4π*10^-7 Tm/A)*(350)*(0.120A) / 0.250 m

B = 2.11*10^-4 T

Therefore, the flux through each turns of the inner solenoid is,

ΦB = BA = Bπr^2 = ¼*BπD^2

       = ¼*(2.11*10^-4 T)*(3.14)*(0.024m) ^2

ΦB = 9.54 * 10^-8 Wb

The average magnetic flux through each turns of inner solenoid is, 9.54 * 10^-8 Wb

b) Calculate mutual inductance of the two solenoid                                                                                                                                                                                   Since both the solenoid are wound tightly over each other’s then the magnetic flux through each turns of both solenoid is the same therefore the mutual inductance is,

M = N2* ΦB / I

M = (23)*(9.54 * 10^-8 Wb) / 0.120 A

M = 1.82 * 10^-5 H

The mutual inductance two solenoid is, 1.82 * 10^-5 H

c) Calculate emf induced in the outer solenoid,

The emf induced in the outer solenoid due to the changing current in the inner solenoid is,

ϵ = -M*di/dt

ϵ = -1.82 * 10^-5 H / 1600 A/s

ϵ = - 0.02912 V

emf induced in the outer solenoid is, ϵ = - 0.02912 V