Atomic Structure of Magnets

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Claimed by Sarah Ghalayini (Fall 2017)

Logan Vaupel (Fall 2016)

Magnets generate their magnetic fields at the atomic level.

The Main Idea

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Figure 1: Two ways that a magnetic field can be generated. 1) An electron orbiting around the atomic nucleus. 2) An electron rotating around its axis.

The magnetic field produced by a magnet is the sum of the magnetic dipole moments generated by each individual atom. These very small magnetic fields are generated much like those of circular current loops; however instead of being generated by electrons or charges flowing through a wire, the field in each individual atom is produced in one three different ways:

  1. An electron orbiting around the atomic nucleus. (See Figure 1)
  2. An electron rotating around its axis. (See Figure 1)
  3. The rotation of protons and neutrons within the nucleus of the atom.

All three of these situations produce a magnetic dipole proportional to the angular momentum. Together, the magnetic dipoles of all the atoms in the magnet sum to give the total magnetic dipole of the magnet. The magnetic field at an observation location can then be found from this dipole.

A non-magnetic material (left) and a magnetic material (right). In the non-magnetic material, the atomic magnetic dipoles are random, and average out to zero. In the magnetic material, the magnetic field from the atomic magnetic dipoles are aligned, which allow for a nonzero magnetic field. Image retrieved from Hyperphysics]], free for educational use.

Although all atoms have electrons orbiting their nuclei, most materials are not magnetic. Each atom in these materials has a small magnetic dipole, however these dipoles are generally disordered and therefore the orbital and spin motions do not line up. Thus, the net magnetic field of each region throughout the material sums to zero. In magnetic materials, however, regions of magnetic dipoles line up enough to produce a significant magnetic field. Although some of these regions cancel other regions out, enough regions align to produce a nonzero magnet field. Such materials are called "ferromagnetic." This is allowed by interactions between atoms in certain elements (usually iron, nickel, cobalt, or alloys of these metals).

A material such as iron can be seen as a collage of different regions of nearly perfect alignment and net magnetic field. Normally, these regions are oriented at random relative to one another and do not produce a net magnetic field. But in the presence of a strong external magnetic field, such as one from a solenoid, the regions within the material that are already aligned with that external field will multiply in strength, creating a much larger net magnetic dipole moment within the formerly non-magnetic piece of iron. When the external field is removed, however, the iron returns to approximately the same net magnetic dipole as before the field was applied. But in the presence of a large enough external field, within certain non-pure alloys of iron, the internal regions can rotate and align enough to create a more permanent change in the material that remains after the external field is removed. This is called a permanent magnet, which you can probably find holding up your finger-paintings on your refrigerator at home. However, beating up the magnet with some hard blows or jolts can cause the regions to misalign, destroying the magnetic dipole. Heating the material above a certain temperature can produce the same effect.

To better understand why some materials better align to form magnets while others do not requires more in-depth knowledge of quantum mechanics, which is certainly interesting but not discussed here (so that your head doesn't explode from the concentrated sheer mind-blowing awesomeness of pure physics). Essentially, the alignment comes from electric interactions between atoms, not so much the weaker magnetic ones. For more information, see "Further Reading" below.

A Mathematical Model

We can model the magnetic dipole moment of a magnet by comparing the movement of electrons about the nucleus to the movement of current in a loop. In a current loop [math]\displaystyle{ {\mu = I \pi R^2} }[/math] when [math]\displaystyle{ R }[/math] is the radius of the loop. Since the units of [math]\displaystyle{ I }[/math] are [math]\displaystyle{ \frac{charge}{time} }[/math], the charge of an electron is [math]\displaystyle{ -e }[/math], and the period for one orbit around the nuclues is [math]\displaystyle{ t = \frac{2 \pi R}{v} }[/math] where [math]\displaystyle{ v }[/math] is the speed of the the electron, the magnetic dipole for one atom in a magnet simplifies to [math]\displaystyle{ \mu = \frac{e R v}{2} }[/math] where [math]\displaystyle{ R }[/math] is now the radius of the orbit.

The magnetic dipole is proportional to the angular momentum, [math]\displaystyle{ L }[/math] of the electron orbiting the nucleus. Assuming a circular orbit and assuming the speed of the electron is much less than the speed of light, [math]\displaystyle{ L = R p = R m v }[/math]. Multiplying the magnetic dipole by [math]\displaystyle{ \frac{m}{m} }[/math] reveals the proportionality of magnetic dipole and angular momentum [math]\displaystyle{ \mu = \frac{m}{m} \frac{e R v}{2} \Rightarrow \mu = \frac{e}{2 m} (R m v) = \frac{e}{2 m} L }[/math] From this equation the charge and mass of an electron and a proton can be plugged in to be compared. Because an electron weighs so much less than a proton, the magnetic dipole from the orbit of an electron is [math]\displaystyle{ 10^4 }[/math] times bigger than the magnetic dipole from the rotation of a proton or neutron in the nucleus, allowing the contributions from the protons and neutrons to be neglected.

For the purpose of calculating the magnetic dipole, it can be assumed that [math]\displaystyle{ L }[/math] is equal to Planck's Constant, [math]\displaystyle{ \hbar = 6.626 \times 10^{-34} \, J \cdot s }[/math]. Plugging in the charge and mass of an electron gives [math]\displaystyle{ \mu \approx 5.81 \times 10^{-23} \, A \cdot m^2 \, per \, atom }[/math].

Finally, if [math]\displaystyle{ \mu = N \mu_{atom} }[/math] where [math]\displaystyle{ N }[/math] is the number of atoms, this [math]\displaystyle{ \mu }[/math] is the magnetic dipole of the magnet.

However, it is worth noting that we made several simplifying assumptions to create this model. First, we assumed that the movement of electrons in their orbitals were moving in a circular path, creating angular momentum. In reality, atomic orbitals are much more complex probability distributions for finding an electron at a specific location. The different energy levels of electron orbits form different shapes with different angular momentums. The "s" orbital is a sphere with a net zero angular momentum, while the "p, d, and f" orbitals are asymmetrical and thus produce an angular momentum, which can produce a magnetic dipole moment. Also, we assumed that there was only one unpaired electron moving within its orbit per atom, when in reality, many materials, such as iron, have two or more unpaired electrons that contribute to the net magnetic dipole moment. A more accurate model would also take into account not just the magnetic contribution of individual atoms, but also how they impact one another. We also assumed that the atoms were perfectly aligned, when this would most certainly never happen in reality. Our modern understanding of magnets also suggests that the axial spinning of electrons contributes a great deal to the overall magnetic dipole moment, in comparison to the orbital angular momentum. Thus, a more accurate model would produce far more complex calculations. Nevertheless, this simpler model still gives useful insight into how the movement of electrons within atomic orbitals within a material can create a macro magnetic dipole effect.

A Computational Model

Here is a link that to glowscript showing a model of an electron orbiting a nucleus, which is how the magnetic dipole in a magnet is formed.



A bar magnet made from iron has a mass of 72 g. What is the magnetic dipole of the bar magnet?

[math]\displaystyle{ N = \left (\frac{72 \, g \, iron}{56 \, \frac{g \, iron}{mol \, iron}} \right ) \left (6.022 \times 10^{23} \frac{atoms}{mol} \right ) = 7.74 \times 10^{23} \, atoms }[/math]

[math]\displaystyle{ \mu = (7.74 \times 10^{23} \, atoms) \left (5.81 \times 10^{-23} \, \frac{A \cdot m^2}{atom} \right ) = 44.97 \, A \cdot m^2 }[/math]


A bar magnet that is 15% mass iron and 85% mass nickel has a mass of 126 g. What is the magnetic dipole of the bar magnet?

[math]\displaystyle{ 126 \, g \cdot 0.15 = 18.9 \, g \, iron }[/math]

[math]\displaystyle{ 126 \, g \cdot 0.85 = 107.1 \, g \, nickel }[/math]

[math]\displaystyle{ N = \left (\frac{18.9 \, g \, iron}{56 \, \frac{g \, iron}{mol \, iron}} \right ) \left (6.022 \times 10^{23} \frac{atoms}{mol} \right ) = 2.03 \times 10^{23} \, atoms }[/math]

[math]\displaystyle{ N = \left (\frac{107.1 \, g \, nickel}{59 \, \frac{g \, nickel}{mol \, iron}} \right ) \left (6.022 \times 10^{23} \frac{atoms}{mol} \right ) = 1.09 \times 10^{24} \, atoms }[/math]

[math]\displaystyle{ 2.03 \times 10^{23} \, atoms + 1.09 \times 10^{24} \, atoms = 1.29 \times 10^{24} \, atoms }[/math]

[math]\displaystyle{ \mu = (1.29 \times 10^{24} \, atoms) \left (5.81 \times 10^{-23} \, \frac{A \cdot m^2}{atom} \right ) = 74.95 \, A \cdot m^2 }[/math]


A compass originally points north. A bar magnet made of iron is placed [math]\displaystyle{ 20 \, cm }[/math] west of the compass on axis, with the north end of the magnet pointing towards the compass. The compass deflects [math]\displaystyle{ 55^\circ }[/math]. What is the mass of the magnet?

[math]\displaystyle{ \overrightarrow{B}_{net} = \overrightarrow{B}_{Earth} + \overrightarrow{B}_{magnet} }[/math]

[math]\displaystyle{ B_{magnet} = (2 \times 10^{-5} \,T) \tan{55^\circ} }[/math]

[math]\displaystyle{ B_{magnet} = 2.86 \times 10^{-5} \,T }[/math]

[math]\displaystyle{ B_{magnet} = \frac{\mu_0}{4 \pi} \frac{2 \mu}{r^3} }[/math]

[math]\displaystyle{ \mu = \frac{r^3 B_{magnet}}{2} \frac{4 \pi}{\mu_0} }[/math]

[math]\displaystyle{ \mu = \frac{(0.20 \, m)^3 (2.86 \times 10^{-5} \,T)}{2} \left (1 \times 10^7 \, T \cdot \frac{m}{A} \right) }[/math]

[math]\displaystyle{ \mu = 1.14 \, A \cdot m^2 }[/math]

[math]\displaystyle{ number \, of \, atoms = \frac{1.14 \, A \cdot m^2}{5.81 \times 10^{-23} \, \frac{A \cdot m^2}{atom}} }[/math]

[math]\displaystyle{ number \, of \, atoms = 1.96 \times 10^{22} \, atoms }[/math]

[math]\displaystyle{ mass \, of \, bar = (1.96 \times 10^{22} \, atoms) \left ( \frac{56 \, \frac{g \, iron}{mol \, iron}}{6.022 \times 10^{23} \frac{atoms}{mol}} \right ) }[/math]

[math]\displaystyle{ mass \, of \, bar = 1.82 \, g \, iron }[/math]


  1. How is this topic connected to something that you are interested in?
    1. Magnets are a classic toy given to children as an introduction to science, and I was no exception. I have fond memories making magnet models and playing with buckyballs, and remember how fascinated I was with how the magnets interacted with one another. What made them attract and repel one another at such far distances? This makes it all the more satisfying to learn more about how they work on a fundamental level.
  2. How is it connected to your major?
    1. A solid understanding of the structure of an atom and the resulting physical properties of a material are central to chemistry. Potential applications are endless and applicable to many disciplines including materials science where one might design an alloy for a product that is paramagnetic and can be used with a permanent magnet to design a product, such as a purse latch.
  3. Is there an interesting industrial application?
    1. Most certainly. If you understand how certain materials act in the presence of a magnetic field, you can incorporate that the into product design, as well as the processing stage when preparing that material. This is obvious in the automotive industry, but even in medicine and bioengineering an understanding of biomaterials is very helpful. This is especially helpful in designing prosthetics where it will be inconvenient and impractical if the product is magnetic.


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A lodestone. Image retrieved from Wikipedia

The "first" magnets were called lodestones and are just naturally magnetic pieces of metal. There are records of lodestones dating up to 2500 years ago from all over the world. By the 12th century, humans discovered they could suspend small pieces of lodestones which would allow them to rotate, and then used them for navigation. We still use compasses to this day!

Cool Applicaiton

In World War II, United States ships were equipped with magnetic field detection devices to scan for enemy submarines in the Atlantic. They discovered alternating bands of N-S and S-N oriented magnetic dipoles that ran parallel to the long ridges along the seafloor. This led physicists, geologists, and another science geeks to determine that as the iron-rich molten rock that comes up from the mantle during seafloor spreading of tectonic plates cools, iron fragments align with the earth's magnetic field, creating a magnetic dipole. The alternating bands indicates that the orientation of the earth's magnetic field reverses approximately 4 to 5 times per 100 million years. This gives valuable insight into the speed of seafloor spreading, as well as the long term history and behavior of plate tectonic interactions. If you don't believe me, check out this short video: Magnetic Mineral Alignment and WWII

See also

Magnetic Dipole Moment

The Angular Momentum Principle

Magnetic Field

Further reading


VIDEO: Magnetism and Quantum Mechanics

External links

[1] [2] [3]


Chabay, R., & Sherwood, B. (2015). Magnetic Field. In Matter & interactions (4th ed., Vol. 2, pp. 693-698). Hoboken, NJ: Wiley.

Ferromagnetism. (n.d.). Retrieved December 5, 2015, from

Lodestone. (n.d.). Retrieved December 5, 2015, from

Origin of Magnetism. (n.d.). Retrieved December 5, 2015, from