Angular momentum of concentric spheres immersed in magnetic field

  • #1

Homework Statement



Two concentric spherical shells carry uniformly distributed charges +Q (at radius a) and -Q (at radius b>a). They are immersed in a uniform magnetic field [itex]\vec{B}=B_0\hat{z}[/itex].

(a) Find the angular momentum of the fields (with respect to the center).

(b) Now the magnetic field is gradually turned off. Find the torque on each sphere, and the resulting angular momentum of the system.


Homework Equations



[itex]\vec{g}=\epsilon_0 \vec{E}\times\vec{B}[/itex]

[itex]\vec{l}=\vec{r}\times\vec{g}[/itex]

[itex]\nabla\times\vec{E}=-\frac{\delta \vec{B}}{\delta t}[/itex]

The Attempt at a Solution



I calculated an initial value for the angular momentum carried by the fields of [itex]\vec{L}=\frac{QB_0(b^2-a^2)}{3}\hat{z}[/itex]. This seems to make sense. The final angular momentum should be in the z hat direction as well, since the spheres will be spinning about the z axis.

I am stuck trying to find the electric field induced by the changing magnetic field. From Faraday's law, the electric field should be swirling around the z-axis, implying it should be in the phi hat direction. The field will exert a force on each sphere since they carry a charge and will send them spinning in opposite directions. This makes sense considering my initial angular momentum expression. But what is the easiest way to find the induced electric field? I attempted to use Faraday's law in integral form but I am not sure what Amperian loop to use. For cylindrical symmetry, I would use a simple circle and calculate the path integral of the electric field and the change in magnetic flux through that loop, then write my electric field in terms of B. I'm not so sure I can do the same thing with spherical symmetry.
 

Answers and Replies

  • #2
If I did use a circular Amperian loop of radius r about the origin, the path integral of the electric field would simply equal [itex]E\cdot 2\pi r[/itex] and the change in flux would be [itex]\pi r^2 \frac{dB}{dt}[/itex] so my induced electric field magnitude would be [itex]E=\frac{r}{2}\frac{dB}{dt}[/itex]. This has the right units, but I do not think it satisfies conservation of angular momentum. What am I doing wrong?
 
  • #3
TSny
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Part (a) looks very good to me except you might want to check the overall sign of your answer. Is the angular momentum in the +z or -z direction?

For part (b), I think you have the right idea. But it would help if you could add a little more explanation of how you are choosing your Amperian loops. You said that r is the radius of the loop about the origin. Did you mean about the z-axis rather than about the origin? I'm not sure if by "origin" you mean the center of the Amperian loop or the center of the spheres.
 
  • #4
I will check part (a), thanks!

For part (b), I meant that my Amperian loop is a loop of radius r with its center at the origin of the coordinate grid I defined, which is also the center of both spherical shells.
 
  • #5
TSny
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For part (b), I meant that my Amperian loop is a loop of radius r with its center at the origin of the coordinate grid I defined, which is also the center of both spherical shells.

I don't understand how you could choose your loops this way and still find the induced E field at arbitrary points on the surfaces of the spheres. For example, how would you construct a loop so that you can find the E field on the surface of one of the spheres at a point located at "45 degrees latitude" on the sphere?
 
  • #6
Good point. I'm not really sure. I suppose I just choose an arbitrary loop that is centered on the z-axis? That's basically my question. I'm not sure how to approach this part.
 
  • #7
TSny
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Consider dividing the surface of a sphere into thin annular rings about the z axis. Let each ring be an Amperian path.
 
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  • #8
Hm. So, an integral with respect to the polar angle theta? If I consider a sphere whose radius is a<r<b, the radius of each thin annular ring is then [itex]R=r\sin\theta[/itex]? But how do I set this integral up with Ampere's law?
 
  • #9
TSny
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You are not going to use Ampere's law. So, we should probably avoid the phrase "Amperian path" for the chosen loop. (What's the name of the law that you are using to get the induced electric field?)

You need to find the induced E field at points on the surfaces of the spheres. So, when setting up the relation between magnetic flux and induced E, your expression ##E 2 \pi r## should refer to a loop that sits on a sphere.
 
  • #10
You're right. I was using Faraday's law, my mistake.

For a sphere of radius a, at any polar angle [itex]\theta[/itex], the radius of the loop around the sphere is [itex]R=a\sin\theta[/itex], thus the path integral of that loop is simply [itex]E2\pi a\sin\theta[/itex] since we know E is in the same direction as the path at all times. Then I integrate this from [itex]\theta=0[/itex] to [itex]\theta=\pi[/itex]?

Sorry for not picking this up quicker and thank you for still helping me, I appreciate it a lot.
 
  • #11
Oh, so the electric field at any arbitrary point on the sphere (in terms of [itex]\theta[/itex]) would be [itex]\vec{E}=\frac{a\sin\theta}{2}\frac{dB}{dt}\hat{\phi}[/itex] using Faraday's law, correct? Then to find the total torque on the sphere I would have to integrate this then use [itex]\tau=\frac{dL}{dt}[/itex] to solve for the angular momentum of the sphere. Then do it all again for the sphere of radius b and I should get a final angular momentum equal to my initial angular momentum. Is that how I go about it?
 
  • #12
TSny
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Oh, so the electric field at any arbitrary point on the sphere (in terms of [itex]\theta[/itex]) would be [itex]\vec{E}=\frac{a\sin\theta}{2}\frac{dB}{dt}\hat{\phi}[/itex] using Faraday's law, correct? Then to find the total torque on the sphere I would have to integrate this then use [itex]\tau=\frac{dL}{dt}[/itex] to solve for the angular momentum of the sphere. Then do it all again for the sphere of radius b and I should get a final angular momentum equal to my initial angular momentum. Is that how I go about it?

Yes. Sounds good.
 
  • #13
Okay, so I did that and ended up with a final angular momentum nearly exactly the same as my initial but missing that factor of three. The factor of three came from integrating a sine cubed due to having to break up the theta hat into its cartesian components. Do you know whether I should or shouldn't have that factor of three?
 
  • #14
TSny
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Yes, you should get the factor of 1/3 and it should come from integrating sine cubed.
 
  • #15
Thank you, TSny! I really appreciate you taking the time to help me out!
 
  • #16
TSny
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You are welcome. That was a nice problem.
 

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