Biot-Savart Law for Currents: Difference between revisions
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====A Mathematical Model==== | ====A Mathematical Model==== | ||
First We start off with the original version of the Biot-Savart Law. | |||
<math>\vec B=\frac{\mu_0}{4 \pi } \frac{q\vec v\times\hat r}{r^2}.</math> | |||
Because we are dealing with a portion of wire <math>\mathrm{d}\boldsymbol{\ell}</math> long with an Area A containing n moving particles with charge q, we find that the total number of moving charges is equal to |q|(nAv) which is also equal to I, the current in the wire. | |||
<math>B = \frac{\mu_0I}{4\pi}\frac{\mathrm{d}\boldsymbol{\ell} \times \mathbf{\hat r}}{r^2},</math> | |||
Because the shape of the current carrying wire can vary from a straight wire to a loop, we must integrate over the region of the wire. | |||
<math>B = \frac{\mu_0I}{4\pi}\int_{\mathrm{wire}}\frac{\mathrm{d}\boldsymbol{\ell} \times \mathbf{\hat r}}{r^2},</math> | |||
====A Computational Model==== | ====A Computational Model==== | ||
Revision as of 15:38, 29 November 2015
Claimed by David Medrano
Biot-Savart Law for Currents
The Biot-Savart Law can be used for more than just single moving charges; it can also be used to calculate the magnetic field for a large number of charges. One notable reason to do so is to find the magnetic field of a portion of a wire where there can be many moving charges. When we use Biot-Savart Law to find the magnetic field of a short wire, we can apply it to a variety of shapes.
A Mathematical Model
First We start off with the original version of the Biot-Savart Law. [math]\displaystyle{ \vec B=\frac{\mu_0}{4 \pi } \frac{q\vec v\times\hat r}{r^2}. }[/math]
Because we are dealing with a portion of wire [math]\displaystyle{ \mathrm{d}\boldsymbol{\ell} }[/math] long with an Area A containing n moving particles with charge q, we find that the total number of moving charges is equal to |q|(nAv) which is also equal to I, the current in the wire. [math]\displaystyle{ B = \frac{\mu_0I}{4\pi}\frac{\mathrm{d}\boldsymbol{\ell} \times \mathbf{\hat r}}{r^2}, }[/math]
Because the shape of the current carrying wire can vary from a straight wire to a loop, we must integrate over the region of the wire.
[math]\displaystyle{ B = \frac{\mu_0I}{4\pi}\int_{\mathrm{wire}}\frac{\mathrm{d}\boldsymbol{\ell} \times \mathbf{\hat r}}{r^2}, }[/math]
A Computational Model
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https://www.grc.nasa.gov/www/k-12/airplane/thermo0.html http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html https://www.grc.nasa.gov/www/k-12/airplane/thermo2.html http://www.phys.nthu.edu.tw/~thschang/notes/GP21.pdf http://www.eoearth.org/view/article/153532/