Current in circular wire surrounding infinite solenoid in a circuit

  • #1
zenterix
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Homework Statement
Consider the circuit shown below consisting of a battery with emf ##\mathcal{E}##, a resistor with resistance ##R##, a switch ##S##, and a long solenoid of radius ##a##, height ##H## that has ##N## turns.

Coaxial with the solenoid at the center of the solenoid is a circular copper ring of wire of radius ##b## with ##b>a## and resistance ##R_1##.

At ##t=0##, the switch is closed.
Relevant Equations
What is the induced current in the copper ring at the instant the switch is closed?
1713982031661.png


Assume the solenoid is very long such that the magnetic field can be approximated as constant inside.

If there weren't any ring surrounding the solenoid, then when the switch were closed we would have the following equation from Faraday's law

$$\oint \vec{E}\cdot d\vec{l}=-\mathcal{E}+IR=-L\dot{I}$$

$$\dot{I}+\frac{R}{L}I=\frac{\mathcal{E}}{L}$$

$$I(t)=\frac{\mathcal{E}}{R}\left (1-e^{-\frac{Rt}{L}}\right )$$

$$\dot{I}(t)=\frac{\mathcal{E}}{L}e^{-\frac{Rt}{L}}$$

We see that the current would initially be zero and would increase to a maximum value of ##\frac{\mathcal{E}}{R}## as ##t\to\infty##.

But we do have the surrounding copper wire.

My question is about the calculations when we take this wire into account.

Qualitatively, it seems that when the switch is closed there is a large ##\dot{I}## due to the battery.

The magnetic flux through the solenoid is

$$\Phi_{sol}=N\Phi_{turn}=N\frac{\mu_0 NI\pi a^2}{H}=\frac{N^2\mu_0\pi a^2}{H}I$$

and the rate of change of this flux is

$$\dot{\Phi}_{sol}=\frac{N^2\mu_0\pi a^2}{H}\dot{I}$$

Thus, the self-inductance of the solenoid is

$$L=\frac{N^2\mu_0\pi a^2}{H}$$

The back emf due to the solenoid is at its strongest when the switch is just closed because ##\mathcal{E}_L=-\dot{\Phi}=-L\dot{I}##.

Now consider the emf induced in the outer ring.

The flux through the ring due to the solenoid is simply the flux through one turn of the solenoid.

$$\Phi_{sol,ring}=\frac{\mu_0 N\pi a^2}{H}I$$

Thus

$$\dot{\Phi}_{sol,ring}=\frac{\mu_0 N\pi a^2}{H}\dot{I}=M\dot{I}$$

where ##M## is the mutual inductance for the solenoid and the outer ring.

At this point, what I initially thought to do was compute the current in the ring as follows

$$\mathcal{E}_{ring}=RI_{ring}=-\dot{\Phi}_{sol,ring}=-M\dot{I}=-\frac{\mu_0 N\pi a^2}{H}\dot{I}$$

$$I_{ring}(t)=-\frac{\mu_0 N\pi a^2}{RH}\frac{\mathcal{E}}{L}e^{-Rt/L}$$

$$=-\frac{\mathcal{E}}{NR}e^{-\frac{Rt}{n^2 H\mu_0\pi a^2}}$$

Note that in these calculations I assuming the following for the circulation direction and am assuming that the solenoid magnetic field is pointing in the same direction as ##\hat{n}_{rhr}##.

1713983361831.png

Therefore, we have a counterclockwise current (due to the negative sign and our clockwise positive circulation direction).

This makes sense qualitatively, but doesn't seem to be correct quantitatively.

With this reasoning, the answer to the problem (induced ring current at ##t=0## is ##-\frac{\mathcal{E}}{NR}##.

I think there is something missing: the self-inductance of the ring. Is this so?
 
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  • #2
Why do you think that there is something missing?
 
  • #3
The main reason is that this is a problem from an automated grading system on MIT Open Library and my answer of ##-\frac{\epsilon}{NR}## is incorrect according to the system.

Either the system has a bug or I've missed something.

When we do the calculations, we consider the self-inductance of the solenoid as current flows through it.

For the ring, there is an induced emf due to the current through the solenoid and this produces a current. Is there not a back emf on the ring due to this induced current as well?
 
  • #4
Have you tried to input a positive answer? What is the meaning of a negative current?
 
  • #5
In my calculations I assumed that the vector normal to the circuit loop is pointing into the screen. The direction of circulation of positive current is then clockwise.

We don't know how the solenoid is wound exactly, but we do know that no matter how it is wound, if the current flows according to the right-hand rule then flux points in the direction of the normal vector and so flux (and the derivative of flux) are positive.

(I show the above in more detail in post #6 of this other post)

For example, suppose that the solenoid is such that the normal vector points down. Then we have a clockwise current in the solenoid (when we view from above the solenoid shown in the picture below).

For the ring, using this same normal vector, the flux due to the solenoid is also positive.

$$\Phi_{sol,ring}=(\mu_0 nI)\pi a^2 $$

where ##n=\frac{N}{h}## is the number of turns per length of the solenoid.

$$\dot{\Phi}_{sol,ring}=\mu_0 n\pi a^2 \dot{I}$$

And then if we compute the induced emf we get

$$\mathcal{E}_{ring}=-\dot{\Phi}_{sol,ring}=-\mu_0 n\pi a^2 \dot{I}$$

This negative emf (assuming that ##\dot{I}>0##) means the induced current goes in the opposite direction to the positive circulation direction. Recall that the normal vector points down, positive circulation is clockwise, and so the induced current in the ring is counterclockwise (and just to be clear, when I say cw or ccw I mean from the perspective we have when we are looking from above at the diagram shown below)

1714010009866.png
 
  • #6
So, have you tried a positive input?
 
  • #7
nasu said:
So, have you tried a positive input?
I have tried positive input but it also is flagged as incorrect.
 
  • #8
I figured out what the issue is.

The induced current I computed is correct.

The system was calling the resistance in the circular wire ##R_1## and I typed in just ##R##.

Interestingly, the system accepted ##\mathcal{E}/NR_1## but not the negative.

1714012202711.png
 
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