TL;DR: How do you calculate the field at a given point when both an external field and a paramagnetic material are present?

My overarching question has to do with the effect an induced magnetization (volume), $\mathbf{M}$, has on the surrounding magnetic field in a magnetostatic formulation.

From Griffiths (4th ed, Eq. 5.89 on pg. 255), we know that an isolated magnetic dipole, $\mathbf{m}$, will produce a magnetic field, $\mathbf{B}$, at a distance from the magnetic dipole (I will call this arbitrary point $P$).

$$ \mathbf{B}_{\rm dip}(\mathbf{r}) = \frac{\mu_0}{4 \pi}\frac{1}{r^3}[3(\mathbf{m} \cdot \mathbf{\hat{r}})\mathbf{\hat{r}}-\mathbf{m}]$$

Now, if in addition to the presence of a dipole, I turn on a homogeneous external magnetic field, $\mathbf{B}_0$. After waiting for a sufficiently long time such that the dipole aligns with the external magnetic field, the field at point $P$ should be the superposition of field due to $\mathbf{B}_0$ and the field due to the dipole.

Now instead of the dipole, let's say that there is paramagnetic material present. Accordingly, if I flip on an external magnetic field, $\mathbf{B}_0$, the material polarizes with a magnetization, $\mathbf{M}_0$. However, at the point $P$ (which is outside the domain of the paramagnetic material), the classical formulation (I believe) is that the total magnetic field, $\mathbf{B}$ would not be dependent on the magnetization, $\mathbf{M}_0$, of the paramagnetic material. If we follow the definition in Eq. 6.18 of Griffiths ($\mathbf{H} \equiv \frac{1}{\mu_0}\mathbf{B} - \mathbf{M}$), then we would see that the $\mathbf{B}$ field at point $P$ would only be due to the free current that is causing $\mathbf{B}_0$ in the first place.

However, if I envision the paramagnetic material as a collection of many small magnetic dipoles, and the application of an external field aligns enough of the dipoles in the direction of the external field such that the net magnetization is assumed to be in the direction of the external field, how could I now say that the coordination of these tiny dipoles within the paramagnetic material does not contribute to the magnetic field, $\mathbf{B}$ at the point $P$?

How do I rectify the seeming contradiction I described above (or maybe I'm misguided and overlooked something)? What would be the effect of the paramagnetic material on the magnetic field, $\mathbf{B}$ outside its domain?

It seems from a footnote on pg. 5 of Blundell's Magnetism in Condensed Matter that there would be an effect. "A magnetized sample will also influence the magnetic field outside it, as well as inside it (considered here), as you may know from playing with a bar magnet and iron filings." How should I think about this?


1 Answer 1


The magnetization does produce a contribution to the field outside the magnetized region. That's all there is to say, barring the calculation which will depend on the geometry.

But there is one commonly realized geometry where the field owing to the magnetization stays inside the region: the long cylinder. It is like a solenoid.

In the formula $$ {\bf H} = \frac{1}{\mu_0} {\bf B} - {\bf M}, $$ $\bf M$ is the dipole moment per unit volume at some location, $\bf B$ is the total magnetic field at that location owing to everything (i.e. owing to all currents and magnetic dipoles), and $\bf H$ is the quantity defined by this equation.

Ordinarily all the above (${\bf M},\, {\bf B}, \, {\bf H}$) refer to values after spatial averaging over a region large enough to smooth over the atomic structure of any material which may be present.

Let's consider the case of a short magnetized bar. Suppose there are no free currents anywhere, ${\bf j}_{\rm f} = 0$ throughout all of space. All we have are the aligned magnetic dipoles which make up the magnetized bar. Now one might form the intuition, from Maxwell's equation $\nabla \times {\bf H} = {\bf j}_{\rm f}$ (for static problems) that ${\bf j}_{\rm f}$ is "the source'' of $\bf H$ and so if there is no ${\bf j}_{\rm f}$ anywhere then ${\bf H} = 0$. This intuition is wrong. For, consider the Maxwell equation $\nabla \cdot {\bf B} = 0$. This tells us that $$ \nabla \cdot {\bf H} = - \nabla \cdot {\bf M}. $$ But this means that $\nabla \cdot {\bf M}$ acts as a source of $\bf H$ just as surely as charge density acts as a source of $\bf E$. In the case of the magnetized bar, $\nabla \cdot {\bf M} \ne 0$ at the edges of the bar (and it might be non-zero inside the bar too, but for uniform magnetization this divergence will be zero inside the bar). So we do have a source of $\bf H$ in this example. It means $\bf H$ cannot be zero at the edges of the bar, and therefore it is not zero elsewhere because the field equations for empty space guarantee its continuity.

  • $\begingroup$ Thank you for the response! I'm left wondering... how does this fit into the canonical H = B/mu_0 - M formulation? $\endgroup$
    – jmd
    Feb 25 at 15:54
  • $\begingroup$ @jmd further para added to answer this $\endgroup$ Feb 25 at 16:03
  • $\begingroup$ Hi Prof. Steane! Thank you for the addition. I am close to accepting your answer, but I have one more question. I am under the impression that M is only nonzero within the volume bounded by the material. Is this a simplification or is it true? If true, then I'm confused how B takes into account the field from dipoles. More accurately, I'm under the impression that H only considers free currents (Griffiths, Eq. 6.19), so if we rearrange the equation in terms of B (B = mu_0 (M+H)), then where do the magnetic fields from dipoles fit into calculation of B? Thank you so much! $\endgroup$
    – jmd
    Feb 25 at 18:20
  • $\begingroup$ @jmd ok I added some more. But that's all your getting! $\endgroup$ Feb 25 at 23:24
  • $\begingroup$ :) thank you, Professor $\endgroup$
    – jmd
    Feb 26 at 0:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.