PHYSICS HW HELP PLS
2 answers:
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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:
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:
the magnetic flux is 1.34*10^{-8}W
b) the mutual inductance is given by:
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:
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:
by replacing you obtain:
the emf is 20mV