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Determining the magnetic field on the axis of a uniformly magnetized cylinder

  1. Nov 6, 2014 #1
    1. The problem statement, all variables and given/known data
    A cylindrical volume of length L, radius a, is given a uniform magnetization M along its axis (call it the z axis). The question is to first find the bound currents, then use them to find the magnetic field B, as well as "the auxiliary field" H - but only along the axis of the cylinder.

    2. Relevant equations

    The bound volume current is [itex] \vec J_b = \nabla \times \vec M [/itex]

    The bound surface current is [itex] \vec K_b = \vec M \times \hat n [/itex]

    The magnetization is [itex] \vec M = M_o \hat z [/itex]

    The auxiliary field is defined as [itex] \vec H = \frac{1}{\mu_o}\vec B - \vec M [/itex]

    Possibly using Amperian loops: [itex] \oint B \cdot dl = \mu_o I_e [/itex] ([itex] I_e [/itex] is the enclosed current, both bound and free)

    Also possibly using similar loops for H: [itex] \oint \vec H \cdot \vec dl = I_f [/itex] ([itex] I_f [/itex] is the free, enclosed charge...which is zero in this case, as far as I can tell)

    The magnetic vector potential is: [itex] \vec A = \frac{u_o}{4\pi}\int_V \frac{\vec J_b\vec (r')}{r_s}\ dV' + \frac{u_o}{4\pi}\oint_S \frac{\vec K_b\vec (r')}{r_s}\ da' [/itex]

    And from magnetic vector potential, the magnetic field [itex] \vec B = \nabla \times \vec A [/itex] or the curl of [itex] \vec A [/itex].

    Here [itex] r_s [/itex] is the magnitude of the separation vector. So [itex] r_s = r - r' [/itex], where [itex] r' [/itex] is the distance from the origin to the source charge and r is the distance from the origin to the field. Similarly all primed coordinates represent the distances or vectors to the source charges/currents in the system.

    3. The attempt at a solution

    Using boldface as vector notation, since M is uniform, I found Jb to be zero, while Kb was zero at the end faces of the cylinder, but not along the side face, where the right-hand-rule determines that Kb = [itex] M \hat \phi [/itex] (using cylindrical coordinates).

    After a horribly long period of confusion trying to apply Amperian loops across infinitesimal distances, I gave up and tried to use the formula for the magnetic vector potential directly. Only the closed surface integral remains since J is 0. In a nutshell, I chose my origin to be the center of the cylinder. I wrote [itex] r = z \hat z [/itex] while the primed, source r' was: [itex] r' = a \hat s + z' \hat z[/itex].

    Then the magnitude, [itex] r_s [/itex] was found to be [itex] \sqrt{a^2 + (z-z')^2} [/itex]

    The primed area, da', was written (in cylindrical) to be: [itex] ad\phi' dz' [/itex]

    So here's all of it put together:

    [itex] \vec A = \frac{\mu_o M}{4\pi}\int_0^{2\pi} \int_\frac{-L}{2}^\frac{L}{2} \frac{1}{\sqrt{a^2 + (z-z')^2}}ad\phi'dz' \hat \phi [/itex]

    Anyway in the end I used the fact that [itex] \int \frac{1}{\sqrt{a^2 + x^2}}dx = ln(x + \sqrt{a^2 + x^2}) [/itex] to compute the integral, and it was purely in terms of z, L/2, a, and some other constants.

    Then I computed the curl (in cylindrical coordinates), but here was the big problem. In my textbook (Griffiths, electrodynamics), the one of the two non-zero terms that I had to compute was [itex] \frac{1}{s}[\frac{\partial}{\partial s}(sA_\phi)] \hat z [/itex].

    My magnetic vector potential has no "s" term, because I am only considering points along the axis. That's why I have r = [itex] z \hat z [/itex] and not r = [itex] s \hat s + z \hat z [/itex].

    So basically this requires me to multiply by s, take the partial of s, then divide by s. In the end I have a magnetic field with a [itex] \frac{1}{s} [/itex] term! But I want to find the field at s = 0...along the axis. So my magnetic field blows up everywhere along z!

    I am really stuck on this one, hopefully someone can respond soon so that I have time to study for my other two midterms, before I try to finish the other four problems :(.

    Also note I'm pretty sure you can't just use the amperian loops as before because its a *finite* cylinder so the old symmetrical tricks don't work anymore. Unless you consider them in a really small area, but I have no idea how that would work. i.e. there *is* a radial component now, it doesn't get cancelled out by some infinite length. And anyway even if there was a way, I am deeply confused as to why the equation they give me in the book isn't working. Where am I going wrong?
     
    Last edited: Nov 6, 2014
  2. jcsd
  3. Nov 7, 2014 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello, adrian52. Welcome to PF!

    In order to find the curl of a vector at a point, you need to know the vector in a neighborhood of the point. So, to find the curl of ##\vec{A}## on the axis, you would need to know ##\vec{A}## at points just off the axis. That looks too hard. (I think you should have found that ##\small \vec{A} = 0## on the axis.)

    You have the surface current source ##\vec{K}_b## that can be used to find ##\vec{B}##. As you noted, there is not enough symmetry to use Ampere's law. You should have covered another law that describes the magnetic field ##d\vec{B}## produced by an element of current ##d\vec{K}_b##.
     
  4. Nov 7, 2014 #3
    Thanks for the reply, I actually went to my teaching assistant at my university and basically he shared the same view. I know that my problem is equivalent to having a tightly-wound solenoid, since the magnetization produces a surface current along the azimuthal direction (counter clockwise if I have my orientation right).

    I didn't consider solving the equivalent problem because I thought it might be too hard. I suppose that in this case I might need to use the Biot-Savart Law which I think you're referring to...I'm going to get some inspiration from my earlier tutorial notes on that problem, and apply it to this one. Thanks for the help anyway, back to work for me!

    Also my TA said I can't take the [itex] \delta \phi' [/itex] out because the unit vector changes as I go over the surface of the sphere. Not sure if that affects my s-dependence, but its a possible source of error.
     
  5. Nov 8, 2014 #4
    Sorry if double-posting not allowed, I just don't see the edit function for my previous post.

    Anyway! It worked out well. The second part of the question was pretty straightforward as well. Guess this thread is resolved
     
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