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](https://tex.z-dn.net/?f=%5CPhi_B%3DBA%3D%5Cmu_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\\](https://tex.z-dn.net/?f=A%3D%5Cpi%2810%5E%7B-2%7Dm%29%5E2%3D3.141510%5E%7B-4%7Dm%5E2%5C%5C%5CPhi_B%3D%284%5Cpi%2A10%5E%7B-7%7DT%2FA%29%28340%29%280.100A%29%283.1415%2A10%5E%7B-4%7Dm%5E2%29%3D1.34%2A10%5E%7B-8%7DW%5C%5C)
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}](https://tex.z-dn.net/?f=M%3D%5Cmu_o%20N_1%20N_2%20%5Cfrac%7BA_2%7D%7Bl%7D)
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](https://tex.z-dn.net/?f=M%3D%284%5Cpi%2A10%5E%7B-7%7DT%2FA%29%28340%29%2822%29%5Cfrac%7B3.14%2A10%5E%7B-4%7Dm%5E2%7D%7B0.25m%7D%3D1.18%2A10%5E%7B-5%7DH)
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}](https://tex.z-dn.net/?f=%5Cepsilon_1%3DM%5Cfrac%7BdI_2%7D%7Bdt%7D)
by replacing you obtain:
![\epsilon_1=(1.18*10^{-5}H)(1700A/s)=0.02V=20mV](https://tex.z-dn.net/?f=%5Cepsilon_1%3D%281.18%2A10%5E%7B-5%7DH%29%281700A%2Fs%29%3D0.02V%3D20mV)
the emf is 20mV