# I Demagnetizing effect of toroids or long thin cylinders

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1. Oct 16, 2016

### arpon

I read in the book, "Experiments on paramagnetic materials are usually performed on samples in the form of cylinders or ellipsoids, not toroids. In these cases, the value of the magnetic field inside the material is somewhat smaller than the value of magnetic field generated by the current in the surrounding winding because of the demagnetizing effect of induced currents that form on the surfaces of samples. In longitudinal magnetic fields, the demagnetizing effect may be rendered negligible by using cylinders whose length is much larger than the diameter, or it may be corrected for. In transverse magnetic fields, a correction factor must be applied. We shall limit ourselves to toroids or to long, thin cylinders of paramagnets in uniform fields where the values of the magnetic field are the same inside and outside the sample."
Why demagnetizing effect of toroids or long thin cylinders is negligible?

2. Oct 16, 2016

### Charles Link

There are two ways that the magnetic field inside a magnetized solid can be computed. One is from the magnetic surface currents. The second is using the pole method and using $B=\mu_o H+M$. The pole method is perhaps the easiest way to see how the magnetic field $B$ inside the magnetized material with magnetization $M$ gets reduced because of geometry. The $H$ consists of contributions from any magnetic poles. The poles are computed as density of magnetic charge (fictitious), $\rho_m=-\nabla \cdot M$ and using inverse square law to compute the $H$. The results the pole method gives for the magnetic field $B$ are the same as you get by computing magnetic surface currents. For uniform magnetization $M$, this gives magnetic surface charge density $\sigma_m=M \cdot \hat{n}$. It can be readily seen that a torus has no magnetic poles so that $B=M$, and the only poles on a long uniformly magnetized cylinder are on the endfaces (on the long cylindrical surface, $M$ is perpendicular to $\hat{n}$ so there are no poles along the surface of the cylinder.) The longer the cylinder, the less effect of the $H$ of the poles (inverse square law) throughout the major part of the magnetized cylinder. $\\$ Incidentally for a uniformly magnetized solid, the magnetic surface currents per unit length $K_m=M \times \hat{n}/\mu_o$. (see e.g. Griffith's E&M textbook). The magnetic field $B$ can alternatively be computed from the magnetic surface current via Biot-Savart's law, but for this case, the pole method offers the simpler mathematics. $\\$ Additional note: The effect of the endfaces on a long uniformly magnetized cylinder is to reduce the magnetic field $B$ by a factor of 2 to $B=M/2$ near the endface.

Last edited: Oct 16, 2016
3. Oct 16, 2016

### arpon

Thanks for your reply. But I cannot understand why induced currents form on the surfaces of samples.

4. Oct 16, 2016

### Charles Link

Qualitatively the formation of the surface currents is quite simple. Consider a checkerboard and let an electron do a square counterclockwise orbit around each square of the checkerboard. The electron currents in adjacent squares will precisely cancel, and the net effect will be a current running around the outside of the checkerboard. The electron currents are atomic states so there are no ohmic losses from the net current that appears to run along the outer surface. $\\$ The equation $\nabla \times M=\mu_o J_m$ is quantitatively how the magnetic current density $J_m$ is computed. At a surface boundary, by Stokes theorem, the result is magnetic surface current density per unit length $K_m =M \times \hat{n}/\mu_o$. One additional item is a single magnetic moment is given by $m=I A$ where $I$ is the current and $A$ is the area of the loop. The magnetization $M$ is $\mu_o$ multiplied by the number of such loops per unit volume. The magnetization vector points perpendicular to the plane of the loops. Additional item is that when considering z-angular momentum, the computed total angular momentum from all the orbiting electrons will equal that of the surface current. For electron spin, it gets a little more complicated, because of the spin 1/2 and the gyromagnetic ratio of g=2, but in any case, the surface current per unit length $K_m=M \times \hat{n}/\mu_o$ equation still holds. Note also, for a uniformly magnetized solid, $\nabla \times M=0$ except at the surface boundaries. (One final note: You may see two different definitions of magnetization $M$ in the literature. One uses $B=\mu_o H +M$ and the other uses $B=\mu_o H +\mu_o M$. Both of these are considered SI units. )

5. Oct 16, 2016

### Charles Link

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