Difference between revisions of "Forest UCM PnCP QubUniBfield"

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(Created page with " Consider a charged particle moving the x-y plane in the presence of a uniform magnetic field with field lines in the z-dierection. :<math>\vec{v} = v_x \hat i + v_y \hat j</ma…")
 
 
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=Charge in Bfield=
 
 
 
Consider a charged particle moving the x-y plane in the presence of a uniform magnetic field with field lines in the z-dierection.
 
Consider a charged particle moving the x-y plane in the presence of a uniform magnetic field with field lines in the z-dierection.
  
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;Motion in the z-direction has no acceleration and therefor constant (zero) velocity.
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;Motion in the z-direction has no acceleration and therefore constant (zero) velocity.
  
 
;Motion in the x-y plane is circular
 
;Motion in the x-y plane is circular
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The position is also composed of two oscillating components that are out of phase by 90 degrees
 
The position is also composed of two oscillating components that are out of phase by 90 degrees
  
:<math>x^* = x + i y= \frac{v_{\perp}}{i \omega} e^{-i\omega t} = -i\frac{v_{perp}}{\omega} \left ( \cos(\omega t) - \sin(\omega t) \right )</math>
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:<math>x^* = x + i y= \frac{v_{\perp}}{i \omega} e^{-i\omega t} = -i\frac{v_{\perp}}{\omega} \left ( \cos(\omega t) - \sin(\omega t) \right )</math>
  
 
The radius of the circular orbit is given by  
 
The radius of the circular orbit is given by  
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The momentum is proportional to the charge, magnetic field, and radius
 
The momentum is proportional to the charge, magnetic field, and radius
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=charge in B-field and E-field=
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Problem 2.53
  
  
  
 
[[Forest_UCM_PnCP#Charged_Particle_in_uniform_B-Field]]
 
[[Forest_UCM_PnCP#Charged_Particle_in_uniform_B-Field]]
 
http://hep.physics.wayne.edu/~harr/courses/5200/f07/lecture10.htm
 
  
  

Latest revision as of 14:44, 10 September 2014

Charge in Bfield

Consider a charged particle moving the x-y plane in the presence of a uniform magnetic field with field lines in the z-dierection.

[math]\vec{v} = v_x \hat i + v_y \hat j[/math]
[math]\vec{B} = B \hat k[/math]


Lorentz Force
[math]\vec{F} = q \vec{E} + q\vec{v} \times \vec{B}[/math]
Note
the work done by a magnetic field is zero if the particle's kinetic energy (mass and velocity) don't change.
[math]W = \Delta K.E.[/math]

No work is done on a charged particle forced to move in a fixed circular orbit by a magnetic field (cyclotron)


[math]\vec{F} = m \vec{a} = q \vec{v} \times \vec{B} = q\left ( \begin{matrix} \hat i & \hat j & \hat k \\ v_x & v_y &0 \\ 0 &0 & B \end{matrix} \right )[/math]
[math]\vec{F} = q \left (v_y B \hat i - v_x B \hat j \right )[/math]

Apply Newton's 2nd Law

[math]ma_x = qv_yB[/math]
[math]ma_y = -qv_x B[/math]
[math]ma_z = 0[/math]


Motion in the z-direction has no acceleration and therefore constant (zero) velocity.
Motion in the x-y plane is circular

Let

[math]\omega=\frac{qB}{m}[/math] = fundamental cyclotron frequency

Then we have two coupled equations

[math]\dot{v}_x = \omega v_y[/math]
[math]\dot{v}_y = - \omega v_x[/math]

determine the velocity as a function of time

let

[math]v^* = v_x + i v_y[/math] = complex variable used to change variables
[math]\dot{v}^* = \dot{v}_x + i \dot{v}_y[/math]
[math]= \omega v_y + i (-\omega v_x)[/math]
[math]= -i \omega \left ( \omega v_x +i\omega v_y \right )[/math]
[math]= -i \omega v^*[/math]
[math]\Rightarrow[/math]
[math]v^* = Ae^{-i\omega t}[/math]

the complex variable solution may be written in terms of [math]\sin[/math] and [math]\cos[/math]

[math]v_x +i v_y = A \left ( \cos(\omega t) - i \sin ( \omega t) \right )[/math]

The above expression indicates that [math]v_x[/math] and [math]v_y[/math] oscillate at the same frequency but are 90 degrees out of phase. This is characteristic of circular motion with a magnitude of [math]v_{\perp}[/math] such that

[math]v^* = v_{\perp}e^{-i\omega t}[/math]

Determine the position as a function of time

To determine the position as a function of time we need to integrate the solution above for the velocity as a function of time

[math]v^* = v_{\perp}e^{-i\omega t}[/math]

Using the same trick used to determine the velocity, define a position function using complex variable such that

[math]x^* = x + i y[/math]

Using the definitions of velocity

[math]x^* = \int v^* dt = \int v_{\perp}e^{-i\omega t} dt[/math]
[math]= \frac{v_{\perp}}{i \omega} e^{-i\omega t} [/math]

The position is also composed of two oscillating components that are out of phase by 90 degrees

[math]x^* = x + i y= \frac{v_{\perp}}{i \omega} e^{-i\omega t} = -i\frac{v_{\perp}}{\omega} \left ( \cos(\omega t) - \sin(\omega t) \right )[/math]

The radius of the circular orbit is given by

[math]r = \left | x^* \right | = \frac{v_{perp}}{\omega} = \frac{mv_{perp}}{qB}[/math]
[math]r = \frac{p}{qB}[/math]
[math]p=qBr[/math]

The momentum is proportional to the charge, magnetic field, and radius


charge in B-field and E-field

Problem 2.53


Forest_UCM_PnCP#Charged_Particle_in_uniform_B-Field


http://www.physics.sfsu.edu/~lea/courses/grad/motion.PDF

http://physics.ucsd.edu/students/courses/summer2009/session1/physics2b/CH29.pdf

http://cnx.org/contents/77faa148-866e-4e96-8d6e-1858487a520f@9