Gyration (helical rotation) of a particle of charge $q$ inside an electric ($\mathbf{E}$) and magnetic ($\mathbf{B}$) field is due to the combined effect of Coulomb force and Lorentz force.¹⁾ The equation of motion for this particle can be written as:

$$ m\frac{d\mathbf{v}}{dt} = q(\mathbf{E}+\mathbf{v}\times\mathbf{B}) $$

where $m$ is the mass of the particle and $\mathbf{v}$ is its velocity. If there is no electric field, this equation will contain only the Lorentz force part. Then taking the dot product of the velocity with the two sides gives:

$$ m\frac{d\mathbf{v}}{dt} \cdot \mathbf{v} = q (\mathbf{v}\times\mathbf{B}) \cdot \mathbf{v} \ \Rightarrow \ frac{d}{dt}\left(\frac{1}{2}mv^2\right) = 0 $$

Because $\mathbf{v}\cdot (\mathbf{v}\times\mathbf{B})=0$ and both sides are divided by two. That means the kinetic energy ($mv^2/2$) and the magnitude of the velocity (speed) of this particle are both constant. A static magnetic field can never change the kinetic energy of a particle.

In a constant magnetic field along the z-axis, B=Bhatk and the three components in the equation of motion are then

\begin{align*} & m\dot{v}_x = qBv_y \\ & m\dot{v}_y = - qBv_x \\ & m\dot{v}_z = 0 \end{align*}

where $\dot{v}_x=dv_x/dt$ is the first derivative; Same for three components. The last equation states that the $z$-component of the velocity in the direction parallel to the magnetic field is constant. Differentiating the first equation again gives the second derivative like this.

\begin{align*} & m\ddot{v}_x = qB\dot{v}_y = qB(-qBv_x/m) \\ & \Rightarrow \ddot{v}_x = -\left(\frac{qB}{m}\right)^2 v_x = -\omega_g^2 v_x \end{align*}

where $\omega_g=(qB/m$) is the gyrofrequency or cyclotron frequency, whose sign is opposite for positive and negative charges. Similarly it can be shown that $\ddot{v}_y = -\omega^2_g v_y$ for the other component. These two second derivative equations are actually harmonic oscillator equations whose solutions are:

\begin{align} & x - x_0 = r_g \sin\omega_g t \label{ho1} \\ & y - y_0 = r_g \cos\omega_g t \label{ho2} \end{align}

where the sign of the two components of displacement will be opposite for electrons and ions. and gyroradius

$$ r_g = \frac{v_\perp}{|\omega_g|} = \frac{mv_\perp}{|q|B} $$

where $v_\perp = (v_x^2+v_y^2)^{1/2}$ is the constant speed in the plane perpendicular to the magnetic field.

Now the trajectory obtained according to the equations $\ref{ho1}$ and $\ref{ho2}$ is actually a circular orbit. The center of this orbit $(x_0,y_0)$ is called the guiding center. Around this center there is a circular current along the orbit. The internal magnetic field created by this current acts in opposition to the external magnetic field that creates this current. Its name is diamagnetic effect.

If the particle’s velocity has no component parallel to the magnetic field, the particle will move in a circular path. But if there is a component in parallel ( $v_z$ or $v_\parallel$ ), then the particle trajectory will be twisted like a helix as shown below.

The distance from one orbit to another in the direction of the magnetic field is called the pitch, which depends on the pitch angle ($\alpha$).

$$ \alpha = \tan^{-1} \frac{v_\perp}{v_\parallel} $$

That is, the pitch angle depends on the ratio of the two components of the particle’s velocity.