Physics: 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.

aightExplanation:no sorry I got pinterest...

A solenoid coil with 22 turns of wire is wound tightly

haileysolis5

26.01.2022

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

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