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What is the size of the magnetic dipole moment $\vec m$ of a superconducting diamagnetic sphere $radius=R$ in a uniform magnetic field $\vec B_0$? Since there is no free current, we can solve for $\Phi_m$, the scalar potential of $\vec H$.

The boundary conditions that I see are $r\to 0 \Rightarrow \Phi_m \lt \infty$ and $r\to\infty\Rightarrow\Phi_m\to r\cos\theta$

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Getting all your assignments done here? :P –  Bogo May 28 '13 at 23:02
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@AmanAbhishek This is probably my problem. Honestly, I didn't think 4 questions (none of which were the entire problem I was actually asked, but rather similar enough for me to learn concepts and be generalized to be possibly useful to future visitors) was overdoing it, but if the community has policies or whatever please let me know. –  Art M May 28 '13 at 23:51
    
@ArtM There are some relevant policies, but you don't seem to be violating them at all. You seem to have a healthy attitude toward asking questions so that you'll understand things better, which the rest of us respect. I suspect Aman was just having fun, as suggested by the poky tongue smiley. :) –  Mike May 29 '13 at 1:05
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@Mike Thanks you've all been really helpful. I was serious before about not wanting to step on any toes, but I'm glad to see everyone is enjoying studying physics :) –  Art M May 29 '13 at 1:18
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Since the sphere is perfectly diamagnetic, we have the condition that $\vec{B} = 0$ inside the sphere. By the principle of superposition, this means that the sphere has a magnetization $\vec{M}$ that induces a magnetic field $-\vec{B}_0$ inside the sphere. It turns out, this is satisfied for a uniform $\vec{M}$ pointing in $-\hat{B}_0$ direction, if you find the right magnitude for $\vec{M}$. To do so, use bound currents with the Biot-Savart Law to solve for the magnitude $M$. You'll find that $\vec{M}=-\frac{3}{2\mu_0}\vec{B}_0$. Finally, you can find the magnetic moment $\vec{m}$ by integrating this magnetization over the sphere's volume, giving you $\vec{m} = \frac{4}{3}\pi R^3 \vec{M} = -\frac{2\pi}{\mu_0} R^3 \vec{B}_0$.

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I didn't say it was perfectly diamagnetic. I said it was diamagnetic. Using Laplace's equation I think I found $B_{inside} = {-3uB_0\over2\mu_0+\mu}\hat z$. How can I use this to find the induced surface current density? It must be some discontinuity that I'm missing. What about the magnetic dipole? –  Art M May 29 '13 at 2:44
    
Superconductors exhibit perfect diamagnetism. See here. –  Izzhov May 29 '13 at 3:04
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