Physics Book - User contributions [en]

Moving Point Charge

2016-11-14T15:57:35Z

Dsmotrich3:

{{DISPLAYTITLE:Magnetic Field of a Moving Point Charge}}
Edited by Diem Tran Spring 2016.

The [[Biot-Savart Law]] for the magnetic field of a moving charge can be quite complex in calculations. The method of calculations can be broken down further into simpler steps.

==The Main Idea==

===A Mathematical Model===

A moving point charge not only has an electric field due to its possession of a charge but also has a magnetic field due to its velocity. The magnetic field of a moving point charge can be found using a derivation of the [[Biot-Savart Law]] for magnetic fields.

[[File:BiotSavartv.gif]]

where q is the scalar charge of the particle, v is the vector velocity of the moving particle, and r is the vector distance from the observation location to the position of the moving particle. The magnetic constant in the formula has the value of 1e-7 T(m/A). V × r is a cross product that can be computed by the formula:

[[File:CrossProduct.png]]

A trigonometric equivalence of the above formula is (p)(q)sin(θ)

Which can be written in terms of v and r as: (v)(r)sin(θ)
θ in this formula is the angle between the vectors v and r.

To determine the direction of the resulting vector from a cross product, we use the right hand rule where we align our right-hand fingers so that it can sweep from the vector v to the vector r. Where our thumb points as a result of this alignment is the direction of v × r.
The final direction of the magnetic field will depend on the charge of our particle. If our particle is positively charged, the magnetic field will point in the same direction as v × r. If the particle is negatively charged, the magnetic field will point in the opposite direction as v × r.

The unit of our result will be in Tesla, the unit for a magnetic field.

==Examples==

===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 [[File:BiotSavartv.gif|150x200px]] 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> 1.6*10^{-19} </math> and divided by the magnitude of r squared. Don't forget to also multiply this by <math> \mu_0 </math> or <math>10^{-7}</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
<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]]

'''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 <math> r^2 </math>

The final answer will be 0.11 T

===Complex===
An electron is moving horizontally to the right with speed <math> 6*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 = 8 cm from the electron, and the angle θ = 33°. Give both magnitude and direction of the magnetic field at locations 1, 2 and 3.

[[File:ExampleMCharge.png]]

'''Solution:'''

1. Find the r vector for each location, using <math> \theta </math> to calculate the x and y components

2. For each r vector, take the cross product v X r where the v is given.

3. Multiply each respective cross product by the magnitude of charge and <math> mu_0 </math>.

4. In order to find the direction of the magnetic fields, use the [[Right Hand Rule]]. One way to do this is to point your thumb in the direction of the velocity, your pointer finger in the direction of r hat, and look which way your palm is facing in order to find the direction of the magnetic field.

At location P2 and P5 the magnetic field will be zero. P1 will be into the page, P3 will be out of the page, P4 will be out of the page, and P6 will be into the page, all with a magnitude of <math> 8.17*10^{-18}</math> T.

==Important Applications==
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.

We can use the magnetic field of a single moving charge formula to derive a more practical formula of a short current carrying wire:

[[File:ILxR.png]]

Where dl is a vector pointing in the same direction as the conventional [[current]]
This formula is more applicable to many real life situations where we can have macroscopic objects that can be used to calculate the magnetic field, such as the [[Magnetic Field of a Long Straight Wire]] or a [[Magnetic Field of a Loop]].

We can apply the magnetic field formula to numerous situations by integration.

==History==

[[Oliver Heaviside]] first derived this relationship from Maxwell's Equations in 1888. The [[Biot-Savart Law]] was named after [[Jean-Baptiste Biot]] and [[Felix Savart]] who, in 1820, showed a needle deflection from a current carrying wire, thus relating electricity and magnetism.

== See also ==

[[Biot-Savart Law for Currents]]

[[Magnetic Force]]

[[Right-Hand Rule]]

==References==
http://maxwell.ucdavis.edu/~electro/magnetic_field/pointcharge.html
http://ruh.li/3DMathVectors.html

[[Moving Point Charge]]

Magnetic Field of a Long Straight Wire

2016-11-14T15:57:10Z

Dsmotrich3:

'''Claimed by Daniel Smotrich (Fall 2016)'''

In many cases, we are interested in calculating the electric field of a long, straight wire. Wires can create magnetic fields if they have a current flowing through them. If no current is flowing, then there will be no magnetic fields created. Below are steps that explain the derivation of the formula for calculating the magnetic field as well as how to calculate the direction of magnetic field.

==Calculation of Magnetic Field==

===Derivation===
Imagine centering a wire on the y-axis and having a current run through the wire in the +y direction. We are interested in finding the magnetic field at some point along the z axis, say <math> (0,0,z) </math>.

From here, it is an integral problem where you take an arbitrary piece of the rod and plug it into the generic formula for change in magnetic field: <math> \vec{B} =\frac{\mu_0}{4\pi} \frac{I(\vec{l} \times \hat{r})}{y^2+z^2} </math>

First, you can find the <math> \hat{r} </math>. The directional vector <math> \vec{r} </math> is equal to <math> (0,0,z) - (0,y,0) = (0,-y,z) </math>. You get this by doing final position - initial position.
Next, you can find the magnitude of r, and you will get <math> \sqrt{(z^2+y^2)} </math>. As a result, your <math> \hat{r} = \frac{(0,-y,z)}{\sqrt{(z^2+y^2)}} </math>

The last thing we need to calculate is the <math>\Delta \vec{L}</math>. This is nothing more than a unit vector that tells us what direction the current is flowing. Since we know that the current is flowing in the +y axis, our <math>\Delta \vec{L} = \Delta{y} (0,1,0) </math>.

Now that we have everything we need, we can plug it into the equation and evaluate the cross product. As a result we get <math> \Delta \vec{B} =\frac{\mu_0}{4\pi} \frac{I \Delta {y}}{(z^2+y^2)^{3/2}} (z,0,0) </math>

The final step is to integrate this. Since it is centered at the origin, we have to integrate from -L/2 to L/2. So our equation looks like <math> \int\limits_{-L/2}^{L/2}\ \Delta \vec{B} =\frac{\mu_0}{4\pi} \frac{I}{(z^2+y^2)^{3/2}} (z,0,0) \delta {y}</math>

Integrating this, we get the expression <math> B = \frac{\mu_0}{4\pi} \frac{LI}{z(\sqrt{z^2+(L/2)^2})} </math>

===Approximation===

If you know that L>>r, then you know that <math> \sqrt{r^2+(L/2)^2} = L/2 </math>. Therefore, if you have a really long wire and you are trying to find the magnetic field of a point relatively close to the rod, you can use the approximation <math> \frac{\mu_0}{4\pi} \frac{2I}{r} </math>.

==Direction of Magnetic Field==

If you are simply interested in finding the direction of the magnetic field, all you have to do is use the right hand rule. Point your right thumb in the direction of the current, and your hand will curl in the direction of the magnetic field. So, for this situation, we point our thumb in the y direction and, at a point on the +z axis, we can see that our fingers curl right, or towards the +x direction.

==References==
Matter and Interactions Vol. II

This page was created by Arjun Patra

Magnetic Field of a Long Straight Wire

2016-11-14T15:55:58Z

Dsmotrich3:

Moving Point Charge

2016-11-14T15:52:38Z

Dsmotrich3:

{{DISPLAYTITLE:Magnetic Field of a Moving Point Charge}}
Edited by Diem Tran Spring 2016. '''Claimed by Daniel Smotrich (Fall 2016)'''

The [[Biot-Savart Law]] for the magnetic field of a moving charge can be quite complex in calculations. The method of calculations can be broken down further into simpler steps.

==The Main Idea==

===A Mathematical Model===

A moving point charge not only has an electric field due to its possession of a charge but also has a magnetic field due to its velocity. The magnetic field of a moving point charge can be found using a derivation of the [[Biot-Savart Law]] for magnetic fields.

[[File:BiotSavartv.gif]]

where q is the scalar charge of the particle, v is the vector velocity of the moving particle, and r is the vector distance from the observation location to the position of the moving particle. The magnetic constant in the formula has the value of 1e-7 T(m/A). V × r is a cross product that can be computed by the formula:

[[File:CrossProduct.png]]

A trigonometric equivalence of the above formula is (p)(q)sin(θ)

Which can be written in terms of v and r as: (v)(r)sin(θ)
θ in this formula is the angle between the vectors v and r.

To determine the direction of the resulting vector from a cross product, we use the right hand rule where we align our right-hand fingers so that it can sweep from the vector v to the vector r. Where our thumb points as a result of this alignment is the direction of v × r.
The final direction of the magnetic field will depend on the charge of our particle. If our particle is positively charged, the magnetic field will point in the same direction as v × r. If the particle is negatively charged, the magnetic field will point in the opposite direction as v × r.

The unit of our result will be in Tesla, the unit for a magnetic field.

==Examples==

===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 [[File:BiotSavartv.gif|150x200px]] 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> 1.6*10^{-19} </math> and divided by the magnitude of r squared. Don't forget to also multiply this by <math> \mu_0 </math> or <math>10^{-7}</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
<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]]

'''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 <math> r^2 </math>

The final answer will be 0.11 T

===Complex===
An electron is moving horizontally to the right with speed <math> 6*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 = 8 cm from the electron, and the angle θ = 33°. Give both magnitude and direction of the magnetic field at locations 1, 2 and 3.

[[File:ExampleMCharge.png]]

'''Solution:'''

1. Find the r vector for each location, using <math> \theta </math> to calculate the x and y components

2. For each r vector, take the cross product v X r where the v is given.

3. Multiply each respective cross product by the magnitude of charge and <math> mu_0 </math>.

4. In order to find the direction of the magnetic fields, use the [[Right Hand Rule]]. One way to do this is to point your thumb in the direction of the velocity, your pointer finger in the direction of r hat, and look which way your palm is facing in order to find the direction of the magnetic field.

At location P2 and P5 the magnetic field will be zero. P1 will be into the page, P3 will be out of the page, P4 will be out of the page, and P6 will be into the page, all with a magnitude of <math> 8.17*10^{-18}</math> T.

==Important Applications==
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.

We can use the magnetic field of a single moving charge formula to derive a more practical formula of a short current carrying wire:

[[File:ILxR.png]]

Where dl is a vector pointing in the same direction as the conventional [[current]]
This formula is more applicable to many real life situations where we can have macroscopic objects that can be used to calculate the magnetic field, such as the [[Magnetic Field of a Long Straight Wire]] or a [[Magnetic Field of a Loop]].

We can apply the magnetic field formula to numerous situations by integration.

==History==

[[Oliver Heaviside]] first derived this relationship from Maxwell's Equations in 1888. The [[Biot-Savart Law]] was named after [[Jean-Baptiste Biot]] and [[Felix Savart]] who, in 1820, showed a needle deflection from a current carrying wire, thus relating electricity and magnetism.

== See also ==

[[Biot-Savart Law for Currents]]

[[Magnetic Force]]

[[Right-Hand Rule]]

==References==
http://maxwell.ucdavis.edu/~electro/magnetic_field/pointcharge.html
http://ruh.li/3DMathVectors.html

[[Moving Point Charge]]