The figure below shows a short coil which is coaxial with a very long solenoid (

Question

The figure below shows a short coil which is coaxial with a very long solenoid (you may approximate it as infinite.)

b.) If the current supplied to the solenoid drops at a constant rate from 1.5 A to zero in in a time interval t = 25 ms,

What are the magnitudes of the emf and current induced in the coil during this time interval?
= ____ mV
Iind =  ____ mA

Also, how large is the induced electric field in the coil?
Eind = ____   N/C
Does the induced electric field need to be inside the same region of space as the changing magnetic field that caused it?

c.) What is the mutual inductance of this solenoid/coil system?
M = ____   mH

d.) Suppose the external current supply were detached from the solenoid and then connected instead to the coil.

If this current were also dropped from 1.5 A to zero in 25 ms,

What would be the size of the emf induced in the solenoid?
= ____   mV

Explanation / Answer

Coil:

Nc=120

rc=0.018m

Rc=5.9

Solenoid:

Ns=23100 turns/m

rs=0.016m

I=1.5 A

t=25x10^-3s

b)

The magnitude of the magnetic field inside of the solenoid is B = µ0NsI = 4*pi*10^-7*23100*1.5=0.044 T

At the end of t, the flux is zero because the current in the solenoid is zero.

f(t=25ms)=0

At the beggining:

i(t=0)= B*As = 0.044*pi(0.016)2=3.5x10^-5 Wb

We apply Faraday’s law to obtain the magnitude of the emf induced in the coil:

ind = Nc/t = Nc(f – i)/t = 120*(0-3.5x10^-5)/25x10^-3s= -0.168 V

Ohm’s law leads to Iind = ind /Rc = -0.0285 A

c) Mutual inductance is given by µo*Nc*Ns*pi*Rs^2=4*pi^2*10^-7*120*23100*0.016m=0.1751 mH

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