## Motion of a Charged Particle in a Constant Magnetic Field

The motion of a particle with charge $q$ and mass $m$ in a constant magnetic field aligned along the $z$-axis, $\vec{B}=B_z\,\hat{z}$, is described by,

 $y$ $x$
$$\vec{r} =\left( x_0+R\cos(\omega t + \phi)-R\cos(\phi)\right)\,\hat{x} + \left( y_0 + R\sin(\omega t + \phi)-R\sin(\phi)\right)\,\hat{y} + \left(z_0 +v_{z0}t \right)\,\hat{z},$$ $$\vec{v} = -\omega R\sin(\omega t + \phi)\,\hat{x}+ \omega R\cos(\omega t + \phi)\,\hat{y} + v_{z0}\,\hat{z},$$ $$\vec{a} = -\omega^2 R\cos(\omega t + \phi)\,\hat{x} - \omega^2 R\sin(\omega t + \phi)\,\hat{y} + 0\,\hat{z},$$ $$\vec{F} = m\vec{a} = -m\omega^2 R\cos(\omega t + \phi)\,\hat{x} - m\omega^2 R\sin(\omega t + \phi)\,\hat{y} + 0\,\hat{z},$$

where,

$$\omega = -\frac{qB_z}{m}, \qquad \phi = \text{atan}\left(-\frac{v_{x0}}{v_{y0}}\right),\qquad R=\frac{1}{|\omega|}\sqrt{v_{x0}^2+v_{y0}^2}.$$
 $x_0=0$ [m], $y_0=0$ [m], $v_{x0}=0$ [m/s],$m=1$ [kg], $q=1$ [C] $B_{z}=$ 0 [T] $v_{x0}=$ 4 [m/s] $v_{y0}=$ 4 [m/s]

The force on the particle is the Lorentz force,

$$\vec{F} =q\vec{v}\times\vec{B}.$$

Since this force is perpendicular to the velocity of the particle, the force does no work. This means that the kinetic energy of particle will not change and the magnitude of the velocity will not change. Because of the properties of the cross product, the magnitude of the force only depends on component of the velocity perpendicular to the magnetic field, $|q\vec{v}\times\vec{B}|=q|\vec{v}_{\perp}||\vec{B}|$. The magnitude of the force is constant and a force of constant magnitude that is always directed perpendicular to the velocity causes a particle to move in a circle in the plane perpendicular to the $\vec{B}$ vector. The position vector for a particle moving in a circle in the $x-y$ plane when there is no force applied in the $Z$-direction is,

$$\vec{r} =\left( x_0+R\cos(\omega t + \phi)-R\cos(\phi)\right)\,\hat{x} + \left( y_0 + R\sin(\omega t + \phi)-R\sin(\phi)\right)\,\hat{y} + \left(z_0 +v_{z0}t \right)\,\hat{z},$$

Here $R$ is the radius of the circle, $\omega$ is the angular frequency, and $\phi$ is a phase that will be specified below. Note that at $t=0$ the position of the particle is $(x_0,y_0,z_0)$. Differentiating to determine the velocity vector yields,

$$\vec{v} =\frac{d\vec{r}}{dt} = -\omega R\sin(\omega t + \phi)\,\hat{x}+ \omega R\cos(\omega t + \phi)\,\hat{y} + v_{z0}\,\hat{z}.$$

Evaluating this expression at $t=0$ results in the relations $v_{x0} = -\omega R\sin(\phi)$ and $v_{y0} = \omega R\cos(\phi)$ which can be used to determine $\phi$ and $R$,

$$\tan(\phi)= -\frac{v_{x0}}{v_{y0}}, \qquad R=\frac{1}{|\omega|}\sqrt{v_{x0}^2+v_{y0}^2}.$$

Differentiating the velocity yields the acceleration,

$$\vec{a} = \frac{d\vec{v}}{dt} = -\omega^2 R\cos(\omega t + \phi)\,\hat{x} - \omega^2 R\sin(\omega t + \phi)\,\hat{y} + 0\,\hat{z}.$$

Multiplying by the mass gives the force vector,

$$\vec{F} = m\vec{a} = -m\omega^2 R\cos(\omega t + \phi)\,\hat{x} - m\omega^2 R\sin(\omega t + \phi)\,\hat{y} + 0\,\hat{z}.$$

This expression for the force can be equated with $\vec{F} =q\vec{v}\times\vec{B}$, where in this case, $\vec{v}\times\vec{B}=v_yB_z\,\hat{x}-v_xB_z\,\hat{y}= \omega R\cos(\omega t + \phi)B_z\,\hat{x}+\omega R\sin(\omega t + \phi)B_z\,\hat{y}$.

$$-m\omega^2 R\cos(\omega t + \phi)\,\hat{x} - m\omega^2 R\sin(\omega t + \phi)\,\hat{y} = q\omega R\cos(\omega t + \phi)B_z\,\hat{x}+q\omega R\sin(\omega t + \phi)B_z\,\hat{y}.$$

Equating like terms we find that the angular frequency is,

$$\omega = -\frac{qB_z}{m}.$$

This is called the cyclotron frequency. A charged particle in a constant magnetic field moves in a circle with an angular frequency $\omega$ which only depends on the value of the magnetic field and the charge to mass ratio of the particle. The radius of the circle depends on the initial velocity of the particle. The larger the velocity, the larger the radius. If the initial velocity is zero, there is no force and the particle does not move. If there is an initial velocity in the $z$-direction, the particle will continue to move with this velocity in the $z$-direction since no force acts in this direction. In this case the particle spirals around the direction of the magnetic field. Energetic charged particles from the sun spiral along the magnetic field lines of the earth and are directed to the north and south magnetic poles where they put atoms and molecules in the atmosphere in excited states and cause the northern lights.

Using the relations $v_{x0} = -\omega R\sin(\phi)$ and $v_{y0} = \omega R\cos(\phi)$, the position vector can be rewritten in another form that is more convenient for plotting, $$\vec{r} =\left( x_0+\frac{v_{y0}}{\omega}\cos(\omega t)+\frac{v_{x0}}{\omega}\sin(\omega t)-\frac{v_{y0}}{\omega}\right)\,\hat{x} + \left( y_0 + \frac{v_{y0}}{\omega}\sin(\omega t)-\frac{v_{x0}}{\omega}\cos(\omega t)+\frac{v_{x0}}{\omega}\right)\,\hat{y} + \left(z_0 +v_{z0}t \right)\,\hat{z}.$$

This problem can also be solved numerically with a Numerical 6th order differential equation solver.

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