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Discharging a magnetized sphere
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[QUOTE="ELB27, post: 4963862, member: 516388"] [h2]Homework Statement [/h2] Imagine an iron sphere of radius ##R## that carries a charge ##Q## and a uniform magnetization ##\vec{M} = M\hat{z}##. The sphere is initially at rest. (a) Compute the Angular momentum stored in the electromagnetic fields. (b) Suppose we discharge the sphere, by connecting a grounding wire to the north pole. Assume the current flows over the surface in such a way that the charge density remains uniform. Use the Lorentz force law to determine the torque on the sphere, and calculate the total angular momentum imparted to the sphere in the course of the discharge. (The magnetic field is discontinuous at the surface... Does this matter?) [[I]Answer:[/I] ##\frac{2}{9}\mu_0 QR^2##] [h2]Homework Equations[/h2] Angular momentum per unit volume stored in electromagnetic fields: ##\vec{l} = \epsilon_0\left[\vec{r}\times\left(\vec{E}\times\vec{B}\right)\right]##. Lorentz force law for surface current: ##\vec{F} = \int\left(\vec{K}\times\vec{B}\right)da## where ##\vec{K}## is the surface current density. [h2]The Attempt at a Solution[/h2] Since the sphere is made of iron (conductor), all the charge will reside uniformly on the surface. Thus, ##\vec{E}_{inside} = 0## while ##\vec{E}_{outside} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{r}##. The magnetic field of a uniformly magnetized sphere is uniform inside and that of a dipole with moment ##\vec{m} = \frac{4}{3}\pi R^3\vec{M}## outside. Thus, ##\vec{B}_{inside} = \frac{2}{3}\mu_0M\hat{z}## while ##\vec{B}_{outside} = \frac{\mu_0R^3M}{3r^3}(2\cos\theta\hat{r} + \sin\theta\hat{\theta})##. Here I use spherical coordinates throughout where ##\theta## designates the angle with the z-axis. The calculation of the total angular momentum is trivial - ##\vec{L} = \frac{2}{9}\mu_0 QR^2\hat{z}##. Now comes the hard part - (b). I know how to calculate the torque and the angular momentum. The problem I'm having is to determine the surface current. I think that it is supposed to flow (initially) in the ##-\hat{\theta}## direction, towards the north pole, and that I can safely ignore the effect of the spin of the sphere as this motion does not contribute to the z-angular momentum. I thought about using a continuity equation for surface current - something like ##\nabla\cdot\vec{K} = -\frac{\partial\sigma}{\partial t}##. Since, by the above assumption, ##\vec{K}## has only a ##\theta## component, this equation becomes: ##\frac{1}{R\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta K) = -\frac{1}{4\pi R^2}\frac{dQ}{dt}##. Solving for K, I get that ##\vec{K} = \frac{dQ}{dt}\frac{\cot\theta}{4\pi R}(-\hat{\theta})##. Now, this solution does give me the correct answer in the end for angular momentum, but I'm troubled by its physical interpretation: ##\cot\theta## is negative in the southern hemisphere, which means that the current there will flow towards the south pole, not the north pole! Clearly, this is incorrect. So, in spite of magically receiving the correct answer, I suspect that the method is wrong. Can anyone point my mistake or suggest a better way to find the current? Any comments/suggestions will be greatly appreciated! EDIT: By the way, to answer the question posed at the end of the problem - the discontinuity of the magnetic field doesn't matter because only the component perpendicular to the surface, to wit: ##\frac{2}{3}\mu_0 M\cos\theta\hat{r}##, exerts a torque on the surface current. This component is continuous across surface currents. This is also the main cause of the difficulty of this problem - in the southern hemisphere it points in opposite direction compared to the northern hemisphere. The current, however, flows (in my opinion) in the same direction throughout. Thus, in order for a nonzero torque to exist, I believe that the speed in the southern hemisphere must be smaller than in the northern (but not opposite in direction!). [/QUOTE]
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