When a ferromagnetic material is magnetized (by application of an external magnetic field), the domains are aligned in the direction of the field, and the ferromagnetic material itself acts as magnet whose north pole is near the south pole(of the field) and vice versa.

  1. If this is so, don't the two fields (the external and the one produced internally by the ferromagnetic material) cancel each other out?

  2. Why is the field inside a ferromagnetic material which is given by $$B =μ_0H + \mu_0M$$ (where $\mu_0$ - permeability of free space; $H$ - external field; $M$ - magnetic field of ferromagnet), much larger than the external field? Am I misunderstanding the relation between these vectors?

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  • $\begingroup$ Hint: clarify (first of all to yourself) what you mean with "north pole of the field". $\endgroup$
    – fra_pero
    May 23, 2020 at 12:29
  • $\begingroup$ The north pole of the field is also the place where the field lines from the south pole meet up and continue. So, the field lines would be from south to north inside the ferromagnetic material? That's why the flux line concentrate? $\endgroup$ May 23, 2020 at 15:41

2 Answers 2


The magnetic field produced by the magnetization is the dipolar field. This dipolar field is found by solving the Maxwell equations. For a static case, this comes down to solving $$ \nabla \cdot \mathbf{B} = 0 \Rightarrow \nabla \cdot (\mathbf{H}+ \mathbf{M}) = 0\,.$$

The part of this field that is outside the magnet is typically called the stray field $H_s$ (the field drawn on your picture). The field in inside the magnet is typically called the demagnetization field $H_d$. This demagnetization field always opposes the magnetization (you can prove this by solving the Maxwell equations), and from there also its name.

The field that you apply to saturate the magnetization is the external field $H_e$. Like you already indicate, the general relation between the overall magnetic field and the magnetization is $$ \mathbf{B} = \mu_0(\mathbf{H}+ \mathbf{M}) = \mu_0(\mathbf{H}_e + \mathbf{H}_{dipolar} + \mathbf{M})\,. $$

Inside the magnet, we have to work with the demagnetization field which has the opposite direction to the external field and the magnetization. So, if we consider one direction, we get

$$ B =\mu_0(H_e - H_{d} + M)\,. $$

Your questions can be answered now:

  1. The demagnetization field indeed tries to cancel out the external field (opposite direction) but you also need to take into account the magnetization in the determination of the magnetic induction $B$. Hence, the magnetic induction $B$ will be in the same direction as the external field and the magnetization.
  2. The demagnetization field will take the same value as the magnetization (but in the opposite direction) for a large completely magnetized element. If the external field $H_e$ is much smaller than the magnetization, indeed the demagnetization field will be larger.
  • $\begingroup$ In almost 40 years of teaching basic college level engineering physics, none of the texts that I used mentioned, H, and I saw no need to introduce it. A current carrying wire or magnetization in a ferromagnetic material both produce, B, fields. The resultant field at any point is the vector sum of any such contributions. H is a fictional concept which serves only as a source of confusion. $\endgroup$
    – R.W. Bird
    May 24, 2020 at 18:44

You are misunderstanding the meaning of $\mathbf H$ and $\mathbf M$. Magnetic strength/intensity $\mathbf H$ is not "external field" and magnetisation $\mathbf M$ is not "field of the ferromagnet".

Magnetic intensity $\mathbf H$ is total H-field, a sum of the H-field due to external source and H-field due to the ferromagnet:

$$ \mathbf H = \mathbf H_{ext} + \mathbf H_{ferro} $$ Similarly, magnetic induction $\mathbf B$ is total B-field due to both external B-field and B-field of the ferromagnet:

$$ \mathbf B = \mathbf B_{ext} + \mathbf B_{ferro} $$

Magnetisation $\mathbf M$ is average dipole moment per unit volume, it is not called a magnetic field of the ferromagnet.

Now to your questions:

1) Depends on which field. Let us assume that a ferromagnetic cylinder is put near the north pole of a weak permanent magnet cylinder.

|         S          |       permanent magnet (external source of magnetic field)
|         N          |       inside: B points down, H_mag points up

          X                  in between the bodies (point X), total B-field points down, total H-field points also down

|         S          |       ferromagnetic material
|         N          |       B points down, H_ferro points up, H points up

Magnetic field $\mathbf B$ is then increased both between the two faces of the two bodies (that is what ferromagnetic means) and by necessity, also inside both bodies. Since the B-field force lines have a pattern of circular motion loops, inside any of the two bodies, magnetic field $\mathbf B$ points towards the N pole of the body (to maintain continuity of the B-field across the body face).

The H-field is different in that its lines of force do not have circulating pattern, but instead originate on north poles of the bodies and go out, eventually sinking to the south poles. Near its north pole face, the H field of the body points away from the north pole, both inside and outside the body. Inside the body, it points towards the S pole and thus it is opposite to magnetic induction $\mathbf B$ (and also to magnetisation $\mathbf M$).

Since the H-field due to the external source points away from the N pole of the external source, the two H fields are pointing in opposite directions inside each body. Thus inside the ferromagnet, the H-field of the external source is counteracted by the H-field of the ferromagnet. Since the latter is much stronger, the H-field inside the ferromagnet does not vanish but points from the N-pole towards the S-pole.

2) Because the external field is assumed to be weak, far from saturating the ferromagnet. Then the magnetisation term $\mu_0 \mathbf M$ in

$$ \mathbf B = \mu_0 \mathbf H + \mu_0 \mathbf M $$ is much larger than the external field $\mathbf B_{ext}$.


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