Atomic Structure of Magnets: Difference between revisions
Line 20: | Line 20: | ||
The magnetic dipole in a magnet is analogous to the magnetic dipole in a current loop. In a current loop <math>{\mu = I \pi R^2}</math> when <math>R</math> is the radius of the loop. Since the units of <math>I</math> are <math>\frac{charge}{time}</math>, the charge of an electron is <math>-e</math>, and the period for one orbit around the nuclues is <math>t = \frac{2 \pi R}{v}</math> where <math>v</math> is the speed of the the electron, the magnetic dipole for one atom in a magnet simplifies to <math>\mu = \frac{e R v}{2}</math> where <math>R</math> is now the radius of the orbit. | The magnetic dipole in a magnet is analogous to the magnetic dipole in a current loop. In a current loop <math>{\mu = I \pi R^2}</math> when <math>R</math> is the radius of the loop. Since the units of <math>I</math> are <math>\frac{charge}{time}</math>, the charge of an electron is <math>-e</math>, and the period for one orbit around the nuclues is <math>t = \frac{2 \pi R}{v}</math> where <math>v</math> is the speed of the the electron, the magnetic dipole for one atom in a magnet simplifies to <math>\mu = \frac{e R v}{2}</math> where <math>R</math> is now the radius of the orbit. | ||
The magnetic dipole is proportional to the [[The Angular Momentum Principle|angular momentum]], <math>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>L = R p = R m v</math>. Multiplying the magnetic dipole by <math>\frac{m}{m}</math> reveals the proportionality of magnetic dipole and angular momentum <math>\mu = \frac{m}{m} \frac{e R v}{2} \Rightarrow \mu = \frac{e}{2 m} (R m v) = \frac{e}{2 m} L </math> | The magnetic dipole is proportional to the [[The Angular Momentum Principle|angular momentum]], <math>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>L = R p = R m v</math>. Multiplying the magnetic dipole by <math>\frac{m}{m}</math> reveals the proportionality of magnetic dipole and angular momentum <math>\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>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. | ||
===A Computational Model=== | ===A Computational Model=== |
Revision as of 00:15, 5 December 2015
Austin Bryan
Short Description of Topic
The Main Idea
The magnetic field produced by a magnet is the sum of the magnetic fields 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 flowing through a wire, the field in each individual atom is produced in one three different ways:
- An electron orbiting around the atomic nucleus.
- An electron rotating around its axis.
- 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.
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 unaligned and disordered and therefore usually sum to zero. In magnetic materials, regions of magnetic dipoles line up. Although some of these regions cancel other regions out, enough regions align to produce a nonzero magnet field. This is allowed by interactions between atoms in certain elements (usually iron, nickel, cobalt, or alloys of these metals).
A Mathematical Model
The magnetic dipole in a magnet is analogous to the magnetic dipole in a current 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.
A Computational Model
How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript
Examples
Be sure to show all steps in your solution and include diagrams whenever possible
Simple
Middling
Difficult
Connectedness
- How is this topic connected to something that you are interested in?
- How is it connected to your major?
- Is there an interesting industrial application?
History
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
See also
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?
Further reading
Books, Articles or other print media on this topic
External links
References
This section contains the the references you used while writing this page