Magnetic energy density, self-inductance of coaxial cylindrical shells

  • #1
zenterix
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Homework Statement
An inductor consists of two very thin conducting cylindrical shells, one of radius ##a## and one of radius ##b##, both of length ##h##.

This is shown in the two figures below.

Note in the rhs figure, we have the view of a cross-section and we have current flowing out of the page in the inner shell and into the page in the outer shell.

Assume that the current is uniformly distributed in both conductors and that ##b-a<<h## so that we can neglect edge effects.
Relevant Equations
What is the magnetic energy density for ##a<r<b##?

Calculate the inductance of this long inductor.
1715356756619.png


For ##a<r<b## we can calculate the magnetic field at some radius ##r## with Ampere's law

$$\oint_C\vec{B}\cdot d\vec{l}=\mu_0 I_{enc}=\mu_0 I$$

$$\implies B=\frac{\mu_0 I}{2\pi r}$$

$$\vec{B}=\frac{\mu_0 I}{2\pi r}\hat{\theta}$$

At this point my questions arise.

I don't see a current loop in this setup.

Therefore, I don't see an enclosed surface through which to compute a magnetic flux to be used in Faraday's law.

As far as I know, an inductor is pretty much any current loop because any current loop will produce a back emf due to a changing current in the loop.

I am not seeing why these coaxial cables form an inductor.
 
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  • #2
The current is going up in the middle and down on the outer cylinder. There is your loop.

However, note that you do not need to use Faraday's law here. The first part of the question is highly relevant to the second. Do you perhaps know a relationship involving the energy stored in an inductor, the current, and the inductance?
 
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  • #3
Consider the current ##I## flowing through the inner shell.

If that current started at zero and was increased to ##I## by an external electromotive force, then this external emf had to overcome the induced emf from the inductor.

Thus, we have ##\mathcal{E}_{ext}=-\mathcal{E}_L=L\dot{I}##.

The rate of work (ie, power) done by the external emf is

$$P=I\mathcal{E}_{ext}=IL\dot{I}\tag{1}$$

To obtain the total work done, and the energy stored in the inductor, we integrate

$$W_{ext}=\int_0^t LI\dot{I} dt\tag{2}$$

$$=\int_0^I LI dI\tag{3}$$

$$=\frac{LI^2}{2}\tag{4}$$

$$=U_B\tag{5}$$

the total energy stored in the magnetic field.

Now, there is another relationship which is apparently true which is the following

$$u_B=\frac{U_B}{V}=\frac{B^2}{2\mu_0}\tag{6}$$

where ##V## is volume and ##u_B## is the magnetic energy density.

I saw a calculation that shows this for a long solenoid. There was a comment saying this is true even when the magnetic field is non-uniform.

I'm not sure how to show this for the current problem.

Since ##u_B## is energy per unit volume, then if we integrate it over the entire volume where there is magnetic field (ie, the volume between the shells) we obtain the total energy stored in the magnetic field.
 
  • #4
zenterix said:
I'm not sure how to show this for the current problem.
It follows generally from electromagnetic theory.

zenterix said:
From a video I saw about this problem, apparently we need to integrate (6) over the volume between the shells. I'm not sure why though.
If you have the energy density distributed in a volume, how would you get the total energy in that volume?
 
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  • #5
Okay, yes, I immediately noticed that integrating energy per volume gives total energy.

Thus

$$u_B=\frac{B^2}{2\mu_0}=\frac{1}{2\mu_0}\left ( \frac{\mu_0 I}{2\pi r} \right )^2$$

and using as infinitesimal volume element a cylindrical shell

$$dV=2\pi r dr h$$

we have

$$U_B=\int_V u_B dV=\frac{\mu_0I^2h}{4\pi}\ln{\left (\frac{b}{a}\right )}$$

which we can equate to our other expression for ##U_B##, namely ##\frac{LI^2}{2}##.

We solve for ##L## to obtain

$$L=\frac{\mu_0h}{2\pi}\ln{\left (\frac{b}{a}\right )}$$
 
  • #6
I did not check the computations, but yes, the method is correct.
 
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  • #7
But how do we obtain the relationship ##u_B=\frac{B^2}{2\mu_0}##?
 
  • #8
We have ##U_B=\frac{LI^2}{2}##.

The volume in which there is magnetic field is ##h\pi (b^2-a^2)##.

$$u_B=\frac{U_B}{h\pi (b^2-a^2)}$$

$$=\frac{LI^2}{2}\frac{1}{h\pi (b^2-a^2)}$$

$$=\frac{L}{2}\left (\frac{2\pi rB}{\mu_0}\right )^2\frac{1}{h\pi (b^2-a^2)}$$

$$=\frac{B^2}{2\mu_0}\frac{L4\pi r^2}{\mu_0h(b^2-a^2)}$$
 
  • #9
##U_B## is total energy which is a function of ##I##.

##B##, on the other hand, is a function of ##r## for ##a<r<b##. Ie, the magnetic field is not constant between the shells.

Thus, the calculations in post #8 do not make sense.

What is then the definition of energy density that allows us to obtain the expression ##u_B=\frac{B^2}{2\mu_0}##?
 
  • #10
zenterix said:
But how do we obtain the relationship ##u_B=\frac{B^2}{2\mu_0}##?
This should be covered in any introductory book on electromagnetism or special relativity. In short, you can start from the Lorentz force law and deduce a continuity equation with the corresponding source being the energy loss of the current being acted upon by the EM field and the density being ##\frac{\epsilon_0 E^2}{2} + \frac{B^2}{2\mu_0^2}##. (The corresponding energy current density of the continuity equation is essentially the Poynting vector.)

A derivation in SR tensor notation can be found in my special relativity lecture notes. (Note: download link!)

zenterix said:
We have ##U_B=\frac{LI^2}{2}##.

The volume in which there is magnetic field is ##h\pi (b^2-a^2)##.

$$u_B=\frac{U_B}{h\pi (b^2-a^2)}$$

No, you cannot make this inference. This would be true if the energy density was constant, but it is not, because the field is not.
 
Last edited:
  • #11
Orodruin said:
No, you cannot make this inference. This would be true if the energy density was constant, but it is not, because the field is not.
It seems that OP knew about this at some point in time and then somehow forgot. See post #5 where apparently OP does the volume integral to get the stored energy from the energy density.
 

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