# I Stray field and demagnetizing field

1. Jan 13, 2018

### Sahar ali

As Hs (stray field)= -Hd (demagnetizing field), What will happen if we minimize the stray field? Does minimize stray field will also reduce the strength of demagnetizing field? If yes then how will we minimize stray field so that it does not affect the demagnetizing field?

here few lines about the stray field and demagnetizing field is taken from
J. St¨ohr H.C. Siegmann
Magnetism From Fundamentals to Nanoscale Dynamics

To understand the nature of the three magnetic vectors B, H, and M.
We use the example of a flat ferromagnetic disk that has been perpendicularly
magnetized, by an external magnetic field. After this
process, the external magnetic field has been turned off, so that we consider
only the field generated by the ferromagnetic disk itself
In the absence of an external magnetic field, a flat disk magnetized perpendicular
to the surface is characterized by the three magnetic vectors B, H, and
M, The magnetic field inside the magnetic material is called the demagnetizing
field Hd because it is oriented opposite to the magnetization M, thus
tending to destroy it. The field outside the material is called the stray field Hs and
it loops around in space.

2. Jan 13, 2018

This topic seems to often be written up in some of the textbooks in a somewhat confusing manner. One suggestion is to read through this thread: https://www.physicsforums.com/threads/magnetic-field-of-a-ferromagnetic-cylinder.863066/ See also the Insight article: https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/ $\\$ In the pole model, a magnetic surface charge density $\sigma_m$ resides on the faces of the magnetized sample with $\sigma_m=\vec{M} \cdot \hat{n}$, making for a plus magnetic charge on one face and minus charge on the other. In the pole model, $H$ (= $H_d$ ) is computed just like $E$ in electrostatics with $\epsilon_o$ replaced by $\mu_o$. The equation $B=\mu_o H+M$ always applies, but deriving this equation really requires a comparison of the pole model to the surface current model. $\\$ Here, $H_{total}=H_{applied}+H_d$.

Last edited: Jan 13, 2018
3. Jan 14, 2018

### Sahar ali

Thanks lemme read the material you have given me a link to.

4. Jan 14, 2018

### Meir Achuz

Many textbooks do confuse the issue. It is a misnomer to call H inside a magnet a demagnetizing field, because it is B, not H, which exerts a torque on magnetic dipoles to affect the magnetization. Since H_d = - H_s, the only way to reduce H_d is to weaken the magnet. There is no need to do this since H_d does not affect the magnetization.

5. Jan 14, 2018

$H_d$ is geometry dependent, and also depends upon the magnetization $M$. $\\$ For a sphere, the demagnetizing factor $D=\frac{1}{3}$, so that $H_d=-(\frac{1}{3})\frac{M}{\mu_o}$. For a flat disc, $D \approx 1$, so that $H_d \approx -\frac{M}{\mu_o}$ . (Note: I edited this. Originally I forgot the $\mu_o$ in the denominator). $\\$ I agree with everything except the statement $H_d=-H_s$. $\\$ Additional comment: The result for the thin disc, because $H_d=-\frac{M}{\mu_o }$ , is that $B$ in the material is very nearly equal to $B_{external}$, independent of $M$, because $H_{total}=H_s+H_d$ , and in the material, $B=\mu_o H_{total}+M=\mu_o H_s=B_{external}$.

Last edited: Jan 14, 2018
6. Jan 15, 2018

### Meir Achuz

H_d=-H_s at the surface of "a flat ferromagnetic disk".

7. Jan 15, 2018

Please elaborate on this. I don't think I agree with this result. Outside of the material, the overall result of a layer of $\sigma_m=+M$ and a second layer of $\sigma_m=-M$ is no effect, so that $H_{outside}=H_s$. Inside the material , $H_d=-\frac{M}{\mu_o}$, as a result of these two layers. The magnetic surface charge is a result of $-\nabla \cdot M=\rho_m$, so that $\sigma_m=M \cdot \hat{n}$. $\\$ Also, since $B_{outside}=B_s =\mu_o H_s$, and the lines of flux of $B$ are continuous , $B$ in the material is also equal to $B_s$, as previously computed, for a thin disc of uniform magnetization $M$, independent of the magnetization. $\\$ The above results are computing using the pole model. Alternatively, in a surface current model, there are very little surface currents for a thin disc, where the surface current density per unit length $K_m=(M \times \hat{n})/\mu_o$, because the outside cylindrical surface is very narrow, completely unlike the case of a long cylinder. Thereby, there isn't enough surface current to generate a significant magnetic field $B$ for the case of a thin disc.