1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cylinder with uniform magnetization

  1. Jun 10, 2012 #1
    A short circular cylinder carries a uniform magnetization M parallel to its axis.
    I want to know what the divergence of M is. Using divergence in cylindrical coordinates I get that ∇[itex]\bullet[/itex]M = 0.

    Now I was pretty sure of this result until I read the following in my book:

    "∇[itex]\bullet[/itex]H = -∇[itex]\bullet[/itex]M only when the divergence of M vanishes. If you think I'm being pedantic consider the example of a bar magnet - a short cylinder of iron that carries a permanent uniform magnetization M parallel to its axis. In this case there is no free current anywhere, which can lead you to use the above formula. This is however not correct since the divergence of M does not vanish."

    This is obviously in disagreement with what I found for the short cylinder - can anyone explain what is wrong?
  2. jcsd
  3. Jun 11, 2012 #2


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Let's start from the macroscopic Maxwell equations for magnetostatics. You always have (in Heaviside Lorentz units)

    [tex]\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{H}=\vec{j}.\qquad (1)[/tex]

    In our case we have [itex]\vec{j}=0[/itex]. The connection between [itex]\vec{B}[/itex] and [itex]\vec{H}[/itex] is given by

    [tex]\vec{B}=\vec{H}+\vec{M}. \qquad (2)[/tex]

    In your case you have

    \vec{M}_0=\text{const} & \text{inside the cylinder}\\
    0 & \text{outside the cylinder}

    From (1) and (2) you indeed get

    [tex]\vec{\nabla} \cdot \vec{H}=-\vec{\nabla} \cdot \vec{M}=0[/tex]

    inside the cylinder and outside the cylinder. However you have a discontinuity of the magnetization at the boundary of the cylinder. Thus, at the boundary you have to translate the source freeness of [itex]\vec{B}[/itex] to the integral form, leading to the boundary condition

    [tex]\vec{n} \cdot (\vec{B}_{>}-\vec{B}_{<} )=0[/tex]

    along the surface of the cylinder. Here [itex]\vec{n}[/itex] is the usual surface-normal vector pointing out of the cylinder, and [itex]\vec{B}_{>}[/itex] ([itex]\vec{B}_{<}[/itex]) is the limit of the field when going to the cylinder surface from outside (from inside).

    For [itex]\vec{H}[/itex], using (2) you get

    [tex]\vec{n} \cdot (\vec{H}_{>}-\vec{H}_{<} )=-\vec{n} \cdot (\vec{M}_{>}-\vec{M}_{<})=+\vec{n} \cdot \vec{M}_0.[/tex]

    Now you can solve for [itex]\vec{H}[/itex] using the scalar magnetic potential and using the Green's function of the Laplace opertor.

    A very good discussion of this problem can be found in

    Sommerfeld, Lectures on Theoretical Physics, Vol. 3.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook