Question

A solenoid coil with 22 turns of wire is wound tightly around another coil with 340 turns. The inner solenoid is 25.0 cm long and has a diameter of 2.00 cm. At a certain time, the current in the inner solenoid is 0.100 AA and is increasing at a rate of 1700 A/s.
For this time, calculate:
(a) the average magnetic flux through each turn of the inner solenoid;
(b) the mutual inductance of the two solenoids;
(c) the emf induced in the outer solenoid by the changing current in the inner solenoid.

Answer 1

Answer:

a) 1.34*10^-8 W

b) 1.18*10^-5 H

c) 20mV

Explanation:

a) To find the average magnetic flux trough the inner solenoid you the following formula:

$\Phi_B=BA=\mu_oNIA$

mu_o: magnetic permeability of vacuum = 4pi*10^-7 T/A

N: turns of the solenoid = 340

I: current of the inner solenoid = 0.100A

A: area of the inner solenoid = pi*r^2

r: radius of the inner solenoid = 2.00cm/2=1.00cm=10^-2m

You calculate the area and then replace the values of N, I, mu_o and A to find the magnetic flux:

$A=\pi(10^{-2}m)^2=3.141510^{-4}m^2\\\Phi_B=(4\pi*10^{-7}T/A)(340)(0.100A)(3.1415*10^{-4}m^2)=1.34*10^{-8}W$

the magnetic flux is 1.34*10^{-8}W

b) the mutual inductance is given by:

$M=\mu_o N_1 N_2 \frac{A_2}{l}$

N1: turns of the outer solenoid = 22

N2: turns of the inner solenoid

A_2: area of the inner solenoid

l: length of the solenoids = 25.0cm=0.25m

by replacing all these values you obtain:

$M=(4\pi*10^{-7}T/A)(340)(22)\frac{3.14*10^{-4}m^2}{0.25m}=1.18*10^{-5}H$

the mutual inductance is 1.18*10^{-5}H

c) the emf induced can be computed by using the mutual inductance and the change in the current of the inner solenoid:

$\epsilon_1=M\frac{dI_2}{dt}$

by replacing you obtain:

$\epsilon_1=(1.18*10^{-5}H)(1700A/s)=0.02V=20mV$

the emf is 20mV

Answer 2

Answer:

Φ = 5.27*10^-8 Wb

Mi = Mo = 1.16*10^-5 H

emf = - 0.0197 V      

Explanation:

Given:-

- The number of turns of inner solenoid, Ni = 340 turns

- The number of turns of outer solenoid, No = 22 turns

- The length of the inner solenoid, Li = 25.0 cm

- The diameter of inner solenoid, di = 2.00 cm

- The current in the Inner solenoid, I = 0.10 A

- The rate of change of current, dI / dt = 1700 A/s

Find:-

(a) the average magnetic flux through each turn of the inner solenoid;

(b) the mutual inductance of the two solenoids;

(c) the emf induced in the outer solenoid by the changing current in the inner solenoid.

Solution:-

- At the instant given the magnetic field at the center of the inner solenoid, Bi is given by:

                          $Bi = u_o*\frac{N_i}{L_i}*I$

Where, uo = 4π * 10^-7 T. m/A  

                          $Bi = 4\pi *10^-^7*\frac{340}{0.25}*0.1 \\\\Bi = 0.00017 T$

- So the flux through each coil of the inner solenoid is:

                          Φ = Bi*A

- Where A : The area of the circular region of wounded coil:

                          A = π*di^2 / 4

                          A = π*0.02^2 / 4

                          A = 0.00031

- Then the flux ( Φ ) is:

                          Φ = 0.00017*0.00031

                          Φ = 5.27*10^-8 Wb

- Both coils are tightly wounded on each other then the magnetic flux ( Φ ) through each turn of both solenoids is the same. The mutual inductance ( M ) of both will be same can be determined now:

                          Mi = Mo = No*Φ / I

                          Mi = Mo = 22*5.27*10^-8 / 0.1

                          Mi = Mo = 1.16*10^-5 H

- The induced emf in the outer coil due to rate of change of current in the inner coil ( dI / dt ) is given by:

                          emf = -M*dI/dt

                          emf = - 1.16*10^-5 * 1700

                          emf = - 0.0197 V