Deriving the formula for a magnetic dipole field

[Guest432]

Hello PF!
I am trying to find out how the formula for the magnetic field of a dipole magnet is derived, so far I've delved into maxwell equations, but I am finding the math too complicated to read. My state's education system is very far behind, and I am not accustomed to characters and symbols that aren't x, y, integrals, dy/dx etc...

From wikipedia (External magnetic field produced by a magnetic dipole moment):
https://wikimedia.org/api/rest_v1/media/math/render/svg/480e19d59c139435ef3e37d6bf0145746e19fe8f

What does the bold m and r represent? And how has this formula been derived?

My attempts have found that this may be an integration of Coulomb's law.

 

[f95toli]

The variables in bold are vectors.

Are you familiar with vectors? If not, there is no way you can understand the ME and you should probably start by picking up a good book about vector algebra.

Note that the math needed to understand the ME is fairly advances, you would normally(?) only learn the necessary math during your first year at university.

 

[Guest432]

The variables in bold are vectors.

Are you familiar with vectors? If not, there is no way you can understand the ME and you should probably start by picking up a good book about vector algebra.

Note that the math needed to understand the ME is fairly advances, you would normally(?) only learn the necessary math during your first year at university.

That may be the case, even then, do you mind having a crack at explaining it to me?

 

[vanhees71]

The multipole expansion of the electromagnetic field indeed requires quite some vector calculus. As was stressed #2, you cannot do any physics without having some knowledge about vector algebra. For electrodynamics (even electro and magneto statics) you also need some basic vector calculus.

To give you a glimpse, let's first look at the more simple electrostatic case. You can think of an electrostatic field as produced by charged particles at rest in some inertial reference frame. If the charges are located at positions $\vec{x}_j$ and because for electromagnetic fields the superposition principle holds, the electrostatic field is the sum of the electrostatic fields produced by the single charges, i.e., (in Heaviside-Lorentz units)
$$\vec{E}(\vec{x})=\sum_{i} \frac{q_i}{4 \pi|\vec{x}-\vec{x}|^2} \frac{\vec{x}-\vec{x}_i}{|\vec{x}-\vec{x}_i|}.$$
To analyze this further, it is easier to note that the Coulomb fields are all derivable from a potential, i.e.,
$$\vec{E}(\vec{x})=-\vec{\nabla} \Phi(\vec{x}),$$
where
$$\Phi(\vec{x})=\sum_i \frac{q_i}{4 \pi |\vec{x}-\vec{x}_i|}.$$
Now the idea of the multipole expansion is that you usually have a charge distribution, which is somewhere located in a finite region, say within a sphere of radius $R$. Then you can look at the field from some distance, i.e., at a point $\vec{x}$ with $|\vec{x}| \gg R$. Then you can expand the Coulomb potential in powers of $1/r$ with $r=|\vec{x}|$ in the following way
$$G(\vec{x},\vec{x}')=\frac{1}{|\vec{x}-\vec{x}'|} = G(\vec{x},0)+ \vec{x}' [\cdot \vec{\nabla}' G(\vec{x},\vec{x}')]_{\vec{x}'=0}+\ldots = \frac{1}{|\vec{x}|} + \frac{\vec{x}' \cdot \vec{x}}{|\vec{x}|^3}=\frac{1}{r} + \frac{\vec{x}' \cdot \vec{x}}{r^3}.$$
Now apply this to the above potential
$$\Phi(\vec{x})=\sum_i \frac{q_i}{r}+\sum_i q_i \frac{\vec{x}_i \cdot \vec{x}}{4 \pi r^3}+\cdots$$
As you see in leading order you get a Coulomb potential as if the total charge
$$Q=\sum_i q_i$$
was sitting at the origin. This is called the monopole contribution.

The next term is already the dipole contribution. You define the dipole moment of the charge distribution as
$$\vec{p}=\sum_i q_i \vec{x}_i.$$
Then you get
$$\Phi(\vec{x})=\frac{Q}{4 \pi r} + \frac{\vec{p} \cdot \vec{x}}{4 \pi r^3} +\ldots.$$
You can go on with this expansion, getting the higher multipole moments (quardupole, octopole, and so on).

The magnetostatic field looks very similar. The only difference is that there is no monopole moment since as far as we know no magnetic charges exist, and the expansion thus starts with the dipole moment. Thus far away from any stationary current distribution or also from a permanent magnet the field looks almost like a dipole field.

 

[lychette]

The variables in bold are vectors.

Are you familiar with vectors? If not, there is no way you can understand the ME and you should probably start by picking up a good book about vector algebra.

Note that the math needed to understand the ME is fairly advances, you would normally(?) only learn the necessary math during your first year at university.

Your advice is very sound. If there is 'no way' to understand ME without vector algebra could you recommend what you consider to be a 'good' book about vector algebra?

 

[Guest432]

The multipole expansion of the electromagnetic field indeed requires quite some vector calculus. As was stressed #2, you cannot do any physics without having some knowledge about vector algebra. For electrodynamics (even electro and magneto statics) you also need some basic vector calculus.

To give you a glimpse, let's first look at the more simple electrostatic case. You can think of an electrostatic field as produced by charged particles at rest in some inertial reference frame. If the charges are located at positions $\vec{x}_j$ and because for electromagnetic fields the superposition principle holds, the electrostatic field is the sum of the electrostatic fields produced by the single charges, i.e., (in Heaviside-Lorentz units)
$$\vec{E}(\vec{x})=\sum_{i} \frac{q_i}{4 \pi|\vec{x}-\vec{x}|^2} \frac{\vec{x}-\vec{x}_i}{|\vec{x}-\vec{x}_i|}.$$
To analyze this further, it is easier to note that the Coulomb fields are all derivable from a potential, i.e.,
$$\vec{E}(\vec{x})=-\vec{\nabla} \Phi(\vec{x}),$$
where
$$\Phi(\vec{x})=\sum_i \frac{q_i}{4 \pi |\vec{x}-\vec{x}_i|}.$$
Now the idea of the multipole expansion is that you usually have a charge distribution, which is somewhere located in a finite region, say within a sphere of radius $R$. Then you can look at the field from some distance, i.e., at a point $\vec{x}$ with $|\vec{x}| \gg R$. Then you can expand the Coulomb potential in powers of $1/r$ with $r=|\vec{x}|$ in the following way
$$G(\vec{x},\vec{x}')=\frac{1}{|\vec{x}-\vec{x}'|} = G(\vec{x},0)+ \vec{x}' [\cdot \vec{\nabla}' G(\vec{x},\vec{x}')]_{\vec{x}'=0}+\ldots = \frac{1}{|\vec{x}|} + \frac{\vec{x}' \cdot \vec{x}}{|\vec{x}|^3}=\frac{1}{r} + \frac{\vec{x}' \cdot \vec{x}}{r^3}.$$
Now apply this to the above potential
$$\Phi(\vec{x})=\sum_i \frac{q_i}{r}+\sum_i q_i \frac{\vec{x}_i \cdot \vec{x}}{4 \pi r^3}+\cdots$$
As you see in leading order you get a Coulomb potential as if the total charge
$$Q=\sum_i q_i$$
was sitting at the origin. This is called the monopole contribution.

The next term is already the dipole contribution. You define the dipole moment of the charge distribution as
$$\vec{p}=\sum_i q_i \vec{x}_i.$$
Then you get
$$\Phi(\vec{x})=\frac{Q}{4 \pi r} + \frac{\vec{p} \cdot \vec{x}}{4 \pi r^3} +\ldots.$$
You can go on with this expansion, getting the higher multipole moments (quardupole, octopole, and so on).

The magnetostatic field looks very similar. The only difference is that there is no monopole moment since as far as we know no magnetic charges exist, and the expansion thus starts with the dipole moment. Thus far away from any stationary current distribution or also from a permanent magnet the field looks almost like a dipole field.

Thank you for trying to explain that to me! Unfortunately you are correct in saying I need to learn vector algebra as this math is very confusing.

In my experiment, I get a magnetic field sensor and point it at a a small neodymium magnet and increase its distance, recording the magnetic field at each point up till 10 cm. It certainly does follow an inverse cubic relationship but I really feel need to explain the reason this occurs or at least explain the formula without just quoting it. Just quoting a formula without any intuition annoys me haha. Anyways, I was thinking about coulombs law which describes the electric field of a charge, and wouldn't integrating that formula with respect to V give a inverse cubic relationship formula as described? Is that not the basis behind the formula? Additionally, I want to relate the formula back to real world applications. I've looked at MRI machines and magnetic separation, but all I find is descriptions and complex explanations. Is there any field you can suggest that heavily relies on the inverse cubic relationship with magents to be effective?

 

[vanhees71]

The point is that you can expand your magnetic field (I've only used the electrostatic field, because it's simpler to explain; for the magnetostatic field it's similar, and you can also generalize the math to arbitrary electromagnetic fields) in a series in $(1/r)$, where $r$ is the distance of your point of observation from the source of the field. For the magnetic field this expansion starts with a term going like $(1/r^3)$ from the leading dipole term, i.e., if you are far away from the source (in your case the magnetization of your ferromagnet) the other terms are going faster to 0 than this leadin dipole term, and thus what you measure is a field falling like $(1/r^3)$ with distance. Note that there is also a angular dependence, i.e., you field also varies with the angle of the position vector of your point of ovservation with respect to the direction of the dipole moment of the magnet. I don't know, how to calculate this different from the way I've shown above for the electric field.

Of course, you are on the right track to ask, where this behavior comes from, and this should encourage you to learn math. Although often you hear the opposite, that's a lot of fun in itself! Many textbooks on electromagnetism have an introduction to vector calculus in the beginning. My favorite in this respect is the old book by Richard Becker. There's an English translation of it by Dover.

 

[Guest432]

the electrostatic field is the sum of the electrostatic fields produced by the single charges, i.e., (in Heaviside-Lorentz units)
$$\vec{E}(\vec{x})=\sum_{i} \frac{q_i}{4 \pi|\vec{x}-\vec{x}|^2} \frac{\vec{x}-\vec{x}_i}{|\vec{x}-\vec{x}_i|}.$$

I'm going through step by step trying to understand your proof, and I am going well, however I am curious as to where you got this information from?

 

[vanhees71]

You find this in many textbooks on electromagnetism. It's also in Griffths's undergraduate book, which also has a nice introduction to vector calculus. There the calculation, is however done in spherical coordinates rather than Cartesian. The result is of course the same, and the spherical coordinates have their own merit in making the connection to the spherical-wave expansion of electric and magnetic fields (i.e., in terms of angular-momentum eigenstates of the field, using spherical harmonics).

This is, however, not the most simple way. I think the approach with Cartesian coordinates, which leads to the equivalent way to expand the field, but it's written in irreducible Cartesian tensor fields. So for this approach you don't need the more complicated theory of spherical harmonics.

 

[Charles Link]

The electrostatic dipole field actually has the same geometric shape (inverse cube with the same angular shape) as the magnetic dipole even though the two are usually computed with different equations. The magnetic dipole is a microscopic loop of current I and area A with the magnetic field B computed from Biot-Savart's equation, while the electric dipole is a microscopic charge pair of charges +Q and -Q separated by a distance d and the electric field from the charges is computed from Coulomb's law which is inverse square for each charge. An alternative treatment of the magnetic dipole allows for fictitious magnetic charges of $+Q_m$ and $-Q_m$ separated a distance d, with the inverse square law applying like Coulombs law for each of the magnetic charges. This will in fact give the identical inverse cube relationship found by the Biot-Savart calculation for the microscopic current loop.

 

[Guest432]

The electrostatic dipole field actually has the same geometric shape (inverse cube with the same angular shape) as the magnetic dipole even though the two are usually computed with different equations. The magnetic dipole is a microscopic loop of current I and area A with the magnetic field B computed from Biot-Savart's equation, while the electric dipole is a microscopic charge pair of charges +Q and -Q separated by a distance d and the electric field from the charges is computed from Coulomb's law which is inverse square for each charge. An alternative treatment of the magnetic dipole allows for fictitious magnetic charges of $+Q_m$ and $-Q_m$ separated a distance d, with the inverse square law applying like Coulombs law for each of the magnetic charges. This will in fact give the identical inverse cube relationship found by the Biot-Savart calculation for the microscopic current loop.

Researching Biot-Savart's equation, I have come to this equation

Just a few questions, why do they leave the intergration sign there? Can't they integrate with respect to r? Does the unit vector limit the integration until the variables are known? Additionally, Vernier physics models this equation as

My limited knowledge on integration leads me to say that 'completing' the integration for the first formula would not yield an inverse cubic formula. Excuse my ignorance, I am attempting to learn work far beyond what is currently taught to me here at high school.

 

[Charles Link]

The integral for B for a magnetic dipole goes around a small microscopic current loop. Much of it gets canceled out just like for a plus and minus charge, much of it cancels. That's why you get an inverse cube. One suggestion is to work through the charged dipole case of +Q and -Q separated by a very small distance d. Pick an observation point at some arbitrary angle at some distance r and assume d<<r. The electric field E points away from the plus charge and towards the minus charge, each with an inverse square law. Add the two contributions to get the total electric field E. You just need to keep the linear term involving d (or (d/r)). With a little effort, I think you can show an inverse cube behavior, i.e. E is proportional to +Qd/r^3.