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Discharging a magnetized sphere

  1. Jan 3, 2015 #1
    1. The problem statement, all variables and given/known data
    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?) [Answer: ##\frac{2}{9}\mu_0 QR^2##]

    2. Relevant equations
    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.

    3. The attempt at a solution
    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!).
     
    Last edited: Jan 3, 2015
  2. jcsd
  3. Jan 3, 2015 #2

    TSny

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    When integrating the differential equation, there will be a constant of integration that you will need to determine.

    Another approach for finding ##K(\theta)## is to pick an arbitrary latitude line on the sphere corresponding to some angle ##\theta##. If ##dQ## is the loss of charge on the total sphere during some time ##dt##, then you should be able to determine the loss of charge in the portion of the sphere between the south pole and this latitude line in terms of ##dQ## and the area of this portion of the sphere. Relate this to ##K(\theta)##.
     
  4. Jan 3, 2015 #3
    First of all, it was very silly of me to omit the constant of integration - thank you for noticing. The correct expression reads ##K = -\frac{dQ}{dt}\frac{\cos\theta + c}{4\pi R\sin\theta}## where ##c## is a constant and the minus sign comes from the fact that ##\frac{dQ}{dt}<0## but ##K## is a magnitude. This not only has a correct physical interpretation if ##c≥1## but also gives the correct answer regardless of the exact value of ##c##.

    Secondly, I would like to follow your approach for consistency. Call ##dq## the charge during time ##dt## lost between the south pole and the arbitrary latitude ##\theta## while the total lost charge be ##dQ##. Since we assume that the density will remain uniform throughout, ##\frac{dq}{dQ} = \frac{\sigma A}{\sigma A_{total}} = \frac{A}{A_{total}}## where ##A## is the area of the spherical surface from the latitude to the south pole and ##A_{total}## is the total surface area of the sphere. Using a simple surface integral: ##A = \int_0^{2\pi}\int_{\theta}^{\pi}R^2\sin\theta d\theta d\phi = 2\pi R^2(1+\cos\theta)##. The total area is, of course, ##A_{total} = 4\pi R^2##. Substituting, we get ##dq = \frac{1+\cos\theta}{2}dQ##. Let ##I## be the total current flowing from the south pole to the latitude. By definition, ##I = \frac{dq}{dt} = -\frac{dQ}{dt}\frac{1+\cos\theta}{2}##, the minus sign, again, comes from the fact that ##\frac{dQ}{dt}<0## but ##I## is a magnitude. Now, by definition, ##K = \frac{dI}{dl_{⊥}} = -\frac{dQ}{dt}\frac{1+\cos\theta}{2}\frac{1}{2\pi R\sin\theta} = -\frac{dQ}{dt}\frac{1+\cos\theta}{4\pi R\sin\theta}## where ##dl_{⊥}## is an infinitesimal width of the surface current perpendicular to the current direction. Evidently, ##c=1## in my original approach.

    I have just two more questions. First, is there any simple boundary condition I can use to determine ##c## right away? Secondly, I never understood the formal definition ##K = \frac{dI}{dl_{⊥}}##. It says to take a derivative, but I always end up dividing by the width instead of differentiating. Can you shed some light on this?

    Thank you very much for your answer! I highly appreciate it and your approach was quite clever.

    EDIT: Concerning my first question, would a zero current at the south pole be an appropriate condition? (I mean physically, mathematically it all turns out correctly anyway).
     
    Last edited: Jan 3, 2015
  5. Jan 3, 2015 #4

    TSny

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    Your work is very nicely done.

    The integration constant can be determined by considering what K must be at ##\theta = \pi/2##. Consider the net current flowing out of the southern hemisphere and the net current flowing out of the northern hemisphere. How should these compare?

    If you can see that K must go to zero at the south pole, then that would be an easy way to get the integration constant. My intuition wasn't good enough to be confident about dealing with this point of the sphere.

    When I see the surface current expressed as ##K = \frac{dI}{dl_\perp}##, I'm with you. I think of it as division rather than a derivative. To interpret it as a derivative you could do something like the following. Imagine a curve in the surface that’s drawn perpendicular to the current flow. Let the starting point of the curve be O. Let ##s## be arc length along the curve measured from O. Consider the function ##I(s)## that gives the total current crossing the curve between O and the point at arc distance ##s##. Then the surface current density at ##s## is ##K = \frac{dI(s)}{ds}##, where here we actually have a derivative of a function. But, I don’t think this is a common way of looking at it.
     
  6. Jan 4, 2015 #5
    I see - they must be equal to one another and to half the total current ##-\frac{dQ}{dt}## by conservation of charge. Since the net current is ##K(2\pi R\sin\theta) = -\frac{dQ}{dt}\frac{c+\cos\theta}{2}##, at ##\theta=\pi/2## we have ##-\frac{dQ}{dt}\frac{c+0}{2} = -\frac{dQ}{dt}\frac{1}{2}## leading to the conclusion that ##c=1##.

    Thank you for the explanation - it is a bit clearer now. I sometimes wonder what is the point of such almost-misleading definitions - a good notation should suggest the answer, not hide it.
     
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