Moving Point Charge: Difference between revisions
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Crossing these two, we get < 0, 0, 49200> | Crossing these two, we get < 0, 0, 49200> | ||
3. The magnetic field will be this cross product multiplied by the charge of the proton <math> 1.6*10^{-19} </math> and divided by the magnitude of r squared. | 3. The magnetic field will be this cross product multiplied by the charge of the proton <math> 1.6*10^{-19} </math> and divided by the magnitude of r squared. Don't forget to also multiply this by <math> \mu_0 or 10^{-7}<math/>. | ||
The final answer will be <math> <0, 0, 1.75*10^{-19}> </math> | The final answer will be <math> <0, 0, 1.75*10^{-19}> </math> T | ||
=== | ===Medium=== | ||
The electron in the figure below is traveling with a speed of | The electron in the figure below is traveling with a speed of | ||
<math> v = | <math> v = 4*10^6 </math>m/s. What is the magnitude of the magnetic field at location A if r = <math> 7*10^{-10}</math>m and <math>\theta=57 </math>degrees | ||
[[File:Example2vB.png|300x400px]] | [[File:Example2vB.png|300x400px]] | ||
'''Solution:''' | |||
1. First split up the velocity in to its x and y components by multiplying the given velocity by cos(57) and sin(57) for x and y respectively. | |||
2. Find r hat and take the cross product of your new velocity vector with r hat. | |||
3. Multiply this by the magnitude of the charge for an electron, as well as by <math> \mu_0 <math/> and then divide this by the <math> r^2 </math> | |||
The final answer will be 0.11 T | |||
===Difficult=== | ===Difficult=== |
Revision as of 16:03, 1 December 2015
Page claimed by James Moroz Jmoroz3 (talk) 11:47, 20 November 2015 (EST)
This page covers the method of calculating the magnetic field from a moving point charge, derived from the Biot-Savart Law for magnetic fields.
The Main Idea
A Mathematical Model
The magnetic field of a moving point charge can be found using a derivation of the Biot-Savart Law for magnetic fields.
With this equation for the magnetic field given some current carrying object, we can rewrited Idl in terms of velocity in order to relate the velocity of the moving particle to the magnetic field at an observation location a distance r from this particle.
With this substitution, the final formula comes out to be:
where q is the charge of the particle, v is the velocity of the moving particle, and r is the distance from the observation location to the moving particle.
Examples
Be sure to show all steps in your solution and include diagrams whenever possible
Simple
At a particular instant, a proton at the origin has velocity < 4e4, -3e4, 0> m/s. Calculate the magnetic field at location < 0.03, 0.06, 0 > m, due to the moving proton.
Solution:
1. The first we need to do is find r hat. Given the vector <0.03, 0.06, 0>, we can calculate the normalized r hat vector to be < 0.447, 0.894, 0 >.
2. Once we have both the velocity and r hat vectors, we can take the cross product of these two as the equation tells us to do. Crossing these two, we get < 0, 0, 49200>
3. The magnetic field will be this cross product multiplied by the charge of the proton [math]\displaystyle{ 1.6*10^{-19} }[/math] and divided by the magnitude of r squared. Don't forget to also multiply this by [math]\displaystyle{ \mu_0 or 10^{-7}\lt math/\gt . The final answer will be \lt math\gt \lt 0, 0, 1.75*10^{-19}\gt }[/math] T
Medium
The electron in the figure below is traveling with a speed of [math]\displaystyle{ v = 4*10^6 }[/math]m/s. What is the magnitude of the magnetic field at location A if r = [math]\displaystyle{ 7*10^{-10} }[/math]m and [math]\displaystyle{ \theta=57 }[/math]degrees
Solution:
1. First split up the velocity in to its x and y components by multiplying the given velocity by cos(57) and sin(57) for x and y respectively.
2. Find r hat and take the cross product of your new velocity vector with r hat.
3. Multiply this by the magnitude of the charge for an electron, as well as by [math]\displaystyle{ \mu_0 \lt math/\gt and then divide this by the \lt math\gt r^2 }[/math]
The final answer will be 0.11 T
Difficult
An electron is moving horizontally to the right with speed [math]\displaystyle{ 5*10^6 }[/math] m/s. What is the magnetic field due to this moving electron at the indicated locations in the figure? Each location is d = 7 cm from the electron, and the angle θ = 35°. Give both magnitude and direction of the magnetic field at each location.
Connectedness
A single moving point charge represents the most simple situation of charges moving in space to produce a magnetic field. In reality, this situation rarely occurs, however understanding how a single moving point charge interacts to produce a field will allow you to understand how sets of moving charges produce a field in space as well.
History
The history of the Biot Savart law and its discovery can be found at the Biot-Savart Law.
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
Internet resources on this topic
References
http://maxwell.ucdavis.edu/~electro/magnetic_field/pointcharge.html This section contains the the references you used while writing this page