I have a cylindrical permanent magnet with uniform magnetization $\mathbf{M}=\mathbf{a_z}M$, length $L$ and Diameter $D$. The magnet has its center in the origin. So there is a length $L/2$ on each side of the $xy$-plane.

enter image description here

In an example in my book featuring this scenario, the author states:

"As a consequence of $\displaystyle\oint \mathbf{B} \cdot d\mathbf{S}=0$ we have the following case:

  1. The $\mathbf{B}$-field has a negative $r$-component for $z<0$

  2. The $\mathbf{B}$-field has no $r$-component for $z=0$

  3. The $\mathbf{B}$-field has a positive $r$-component for $z>0$

Now, I understand that the magnetic field lines must always close upon themselves, that is, they form a closed loop. I also realize we have the following situation with regards to the magnetic field and the magnetization.

enter image description here

Clearly, from the pictures, the magnetic field has positive $r$-component for $z>0$, and negative for $z<0$, but I'm not sure why this is the case. What does $\displaystyle\oint \mathbf{B} \cdot d\mathbf{S}=0$ have to do with it?

I would really need some help with this.


1 Answer 1


The no-monopole, or solenoidal, law is a mathematical expression of the fact that magnetic field lines have no beginning or end, but must circulate and be continuous.

Given that, plus the constraint that the normal component of the B-field has to be continuous at an interface (also a consequence of the solenoidal law) how else could you arrange for the field lines to get from the top half of the magnet back to the bottom half?

  • $\begingroup$ I'm just not sure why $r=0$ at $z=0$. I get that for the field lines to change direction, there must be a point where $r=0$, but why does this happen exactly at the center of the magnet? $\endgroup$
    – Carl
    Nov 1, 2020 at 11:24
  • $\begingroup$ @Carl where else would it happen given the symmetry about the plane $z=0$? $\endgroup$
    – ProfRob
    Nov 1, 2020 at 13:11

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