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 $r\to\infty\Rightarrow\Phi_m\to r\cos\theta$
Best Answer
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 $\vec{m} = \frac{4}{3}\pi R^3 \vec{M} = -\frac{2\pi}{\mu_0} R^3 \vec{B}_0$.