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. The equation of motion of 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.