Forest UCM PnCP QubUniBfield

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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.

[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]
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


[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


[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]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]

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

charge in B-field and E-field

Problem 2.53