I have been performing an experiment at school in which I test the force on an iron surface from the magnetic field of an electromagnet. The electromagnet has a rectangular iron core. The theory predicts that the force increases linear with the surface area of the iron plate. This is because the volume between the plate and the magnet contains a certain amount of energy, which is equal to the force exerted on the plate times the distance between the plate and the magnet.

I found that the force does not increase linear with the surface. This is because of the divergence of the magnetic field of a bar magnet, which is what the core of the electromagnet essentially is. Researching the magnetic field of a bar magnet, I discovered that there is a higher density of field lines at the edges of the poles, and thus a stronger force on the plate. There are some images on this website, such as the one below.

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I am curious as to why the field is stronger here. I know that the electric field is stronger at edges and corners because the electrons repel and end up at a higher concentration there, is it the same concept for magnetic fields?


2 Answers 2


The explanation is related to the edge effects, but is very different to the corresponding of electric field. Here, in the magnet we will suposse uniform magnetization. To clarify my answer I will enhance the effect by choosing a disk shaped magnet, the radius >> the height, in fact a 2D problem. We can evaluate the field B as the superposition of the contributions of all small aligned magnets present in the disk. For a point near the centre of the disk, the magnetic field is very small because the contributions of the nearest magnetic dipoles is parcially canceled by the rest of the dipoles because its contribution is antiparallel. Actually, for a great disk shaped magnet the value of B is negligible at the centre, and we can understand this by considering the magnet as a circular loop of finite current (the magnetization) and infinite radius. By contrary, near the edges the field is greater because there are a minor quantity of dipole magnets to oppose the conntribution of the nearest.

  • $\begingroup$ Thank you, that seems logical. So from this theory it follows that the field is stronger at edges of a magnet, and even stronger at a corner because there are less 'tiny magnets' to cancel out the field, right? $\endgroup$
    – Rubenknex
    Commented Jun 12, 2014 at 19:53
  • $\begingroup$ Yes, I agree. The magnet behaves like a polarized dielectric, like an electret, instead a conductor. $\endgroup$
    – user50378
    Commented Jun 12, 2014 at 20:20

If a magnet is fashioned out of a slab with constant width in the $z$ direction and has constant magnetization $M\hat{z}$, then the magnetic field has $\vec{B}=\mu_0 \vec{M}+\mu_0\vec{H}$. Take the divergence of both sides to get $\nabla \cdot \vec{H}=-\nabla \cdot\vec{M}$. Taking time independence and no free currents, $\nabla\times\vec{H}=0$. So $H$ is like an electric field produced by "magnetic charge" $-\nabla \cdot\vec{M}$. In this case our fictitious magnetic charge is constant with positive sign on the top of the magnet, and constant with negative sign on the bottom. This parallel plate arrangement produces $\approx0$ $H$-field outside the magnet if the plates are very large and close together.

So in a flat thin slab of magnet of infinite extent, we actually get zero magnetic field outside of the magnet. As the other answer suggests, this is explained intuitively by noting that the bound currents look like an infinitely large current loop.


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