We know that for a current carrying loop magnetic moment can be calculated by $$\overrightarrow{M} = I\times \overrightarrow{A}$$ Where I is current in loop and A is the area enclosed by the loop

And magnetic moment for a bar magnet is $$\overrightarrow{M} = m \times \overrightarrow{l}$$ Where m is pole strength and l is the length between the poles of bar magnet

Here my question is that is there some kind of relation between them about why both of them is described about magnetic moment and how these two formulas are actually related, if their is some connection between these two terms/formulas?


2 Answers 2


The most general formula for the magnetic dipole moment of a volume current distribution is $$ \vec{\mu} = \frac{1}{2} \iiint \vec{r} \times \vec{J} \, d \tau. $$ This has obvious generalizations to a surface current or a line current: $$ \vec{\mu} \sim \frac{1}{2} \iint \vec{r} \times \vec{K} \, d a \sim \frac{1}{2} \int \vec{r} \times \vec{I} \, d \lambda. $$ (I will use $\vec{\mu}$ rather than $\vec{M}$ for the dipole moment because it is more common to use $\vec{M}$ for the magnetization of the medium; see below.)

Connection to current loop formula

We can connect the last of these to the formula you wrote above by noting that if we parametrize a current loop by its arc length $\lambda$, then $d\vec{r}/d\lambda$ is a unit vector pointing along the path of the loop. This means that $$ \vec{I} = I \frac{d\vec{r}}{d\lambda} $$ This then means that we have $$ \vec{\mu} = I \left[\frac{1}{2} \int \vec{r} \times d\vec{r} \right] $$ and it can be shown that the quantity in brackets is equal to the vector area $\vec{a}$ of the loop. Thus, the fundamental equations above reduce to the equation $\vec{\mu} = I \vec{a}$ for a current loop.

Connection to bar magnet formula

The magnetization $\vec{M}$ of a medium is related to the bound current densities in that medium by $$ \vec{J}_b = \vec{\nabla} \times \vec{M}, \qquad \vec{K}_b = \vec{M} \times \hat{n} $$ and so in the absence of free currents we have $$ \vec{\mu} = \frac{1}{2} \iiint \vec{r} \times (\vec{\nabla} \times \vec{m}) \, d \tau + \frac{1}{2} \iint \vec{r} \times (\vec{M} \times \hat{n}) \, d a $$ Via a significant amount of algebra (see Zangwill's Modern Electrodynamics, §13.2.4 for the gory details), it is possible to show that $$\vec{\mu} = \iiint \vec{M} \, d\tau.$$

For a bar magnet, we typically have a cylinder with cross-sectional area $A$ and length $\ell$, with a uniform magnetization $\vec{M}$ along its axis. Call this axis the $z$-axis for concreteness, so that $\vec{M} = M \hat{z}$. Then in this case we have $$ \vec{\mu} = A \ell M \hat{z} = (AM) \vec{\ell}. $$ So if we identify $AM$ (the magnetization times the pole area) as the "pole strength" $m$ of the magnet,1 we end up with $\vec{\mu} = m \vec{\ell}$, as expected.

1 This definition can be further motivated by defining a "fictitious magnetic charge" such that $\rho_B = - \vec{\nabla} \cdot \vec{M}$ and $\sigma_B = \vec{M} \cdot \hat{n}$. Under this definition, the magnetic field of any magnetized object in the region where $\vec{M} = 0$ acts just like an electric field created by these same charge distributions. Moreover, for bar magnet, we have $\rho_B = 0$ and a "magnetic charge" on each pole of $\sigma A = \pm M A = \pm m$; and if $A$ is "small", then the magnetic field looks like the electric field of a dipole with "charges" $\pm m$. See Zangwill §13.4 for details & motivation of this construction.


Welcome to Physics S.E.! That's not a good way to think about the dipole moment of a magnet (sorry if I disagree with your teacher or textbook!) both because magnetic poles do not exist, and the formula doesn't have any predictive power: if you are given a cylindrical piece of magnetized material how do you know where the poles are or how strong they are?

The magnetized material in the bar magnet is made up of atoms each of which is a little atomic-sized current loop. When it is magnetized all these loops line up so the normals are parallel. The effect is that internally every atomic current is adjacent to another current on the next atom flowing the opposite way, which cancels its magnetic effect. But at the surface this is not true, so the magnetic effect of the whole cylinder is the same as that of a sheet of current flowing round the surface. Thus it is in effect a coil of wire wrapped around the outside of the cylinder, and this provides a way of linking the two types of magnet.

For example if we have a short cylinder - in shape like a coin - then the current flows around the perimiter of the coin, just like a current loop. The strength of the current is the magnetization $\cal{M}$ times the thickness $t$, and the dipole moment $M$ is thus $\cal{M}tA$. (This is also the magnetization multiplied by the volume of the material, which is nice)

  • $\begingroup$ I want to thank you for responding to my question. I can now see how the magnetic moments of the current carrying loop and bar magnet are interwoven rather than separate! However, could you please elaborate on how the strength of the current is the magnetization $\cal{M}$ times the thickness t so that I (a high school student) can grasp without much higher mathematics. Thank you once again for all your efforts. $\endgroup$
    – enthusiasm
    Aug 19, 2022 at 9:36
  • $\begingroup$ I'm impressed that you are a high school student! That point was not explained in my answer! The surface current per unit length is $\cal M$ (if the direction of $\cal M$ lies in the surface) quite generally, but I can't give an elementary proof of that, only derive it for particular example. Suppose the atoms are cubes of side $a$, and $I$ flows around four sides form a magnetic dipole $m=Ia^2$. The magnetization is then defined to be $\cal{M}=nm$ for $n=1/a^3$ atoms per unit volume so $\cal{M}=I/a$. The surface current per unit length is then $I/a=\cal{M}$ $\endgroup$
    – CWPP
    Aug 20, 2022 at 10:48

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