Magnetic Field of a Disk: Difference between revisions
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<math> B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}</math> | <math> B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}</math> | ||
<math> B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2 | <math> B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right]</math> | ||
==Examples== | ==Examples== | ||
Revision as of 13:19, 24 February 2018
Claimed by Andy Stevens (Spring 2018)
The Main Idea
Through this page, you will understand how to solve for the magnetic field produced by a moving charged, circular disk.
First, let us start with the basics. We know that moving charges spread out over the surface of an object will produce a magnetic field. This is similar to the concept of how charges spread out over an object allowed them to produce unique electric fields.
In order to figure out this magnetic field, we will start from the fundamental principles that we have learned already with regards to how magnetic fields are produced. We will then build on that and include the geometry of the object in question, in this a circular disk, in order to solve for the magnetic field produced by this disk.
A Mathematical Model
Say we have a flat circular disk with radius [math]\displaystyle{ R }[/math] and carrying a uniform surface charge density [math]\displaystyle{ \sigma }[/math]. It rotates with an angular velocity [math]\displaystyle{ \omega }[/math] about the z-axis.
To find the magnetic field [math]\displaystyle{ B }[/math] at any point [math]\displaystyle{ z }[/math] along the rotation axis, we can consider the disk to be a collection of concentric current loops. Breaking the disk into a series of loops with infinitesimal width [math]\displaystyle{ dr }[/math] and with radius [math]\displaystyle{ r }[/math], we will have a total area of
[math]\displaystyle{ dA = 2\pi r dr }[/math]
And therefore have a total charge of
[math]\displaystyle{ dQ = 2\pi \sigma rdr }[/math]
This loop is rotating at an angular velocity [math]\displaystyle{ \omega }[/math], which means the charge [math]\displaystyle{ dQ }[/math] on our loop makes a full rotation around the axis every period,
[math]\displaystyle{ T = \frac{2\pi}{\omega} }[/math] seconds.
Since a charge [math]\displaystyle{ dQ }[/math] makes a full rotation every [math]\displaystyle{ T }[/math] seconds, our infinitesimally thing ring is a current loop:
[math]\displaystyle{ dI = \frac{dQ}{T} = \frac{2\pi \sigma rdr}{2\pi / \omega} = \sigma\omega rdr }[/math]
For this single loop, equivalent to a current [math]\displaystyle{ dI }[/math], we can calculate the field a distance [math]\displaystyle{ z }[/math] above the axis. Using this proof from the Magnetic Field of a Loop page, we can get
[math]\displaystyle{ dB = \frac{\mu_0 r^2 dI}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} }[/math]
Now to find the total field for all of the concentric rings, we must integrate from [math]\displaystyle{ r = 0 }[/math] to [math]\displaystyle{ r = R }[/math]:
[math]\displaystyle{ B = \int_{0}^{R} \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}} }[/math]
[math]\displaystyle{ B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R} }[/math]
[math]\displaystyle{ B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right] }[/math]