A charged particle will move in a circular path in a magnetic field if there is no component of its velocity along the field. If there is a component along the field, it results in helical motion, known as electromagnetic gyration. On top of this gyration, electric drift occurs if the electric field changes with time or is non-uniform in space. Due to the drift, the helical path of the gyrating particle shifts in a specific direction. We will discuss two types of drifts: E-cross-B drift (cross-field drift) and polarization drift.

Consider a charged particle gyrating in a magnetic field with an external electric field also present. The two components of the particle’s velocity equation can be written separately: one parallel to the magnetic field and one perpendicular to it. The parallel component is

$$m\dot{v}_\parallel = qE_\parallel$$

which results in acceleration along the magnetic field. However, in most plasmas of the solar system, it is difficult to maintain an electric field parallel to the magnetic field because fast-moving electrons along the magnetic field lines cancel out the electric field. If the perpendicular electric field to the B-field lies along the x-axis, the two perpendicular components of the velocity equation become

$$\begin{align*} & \dot{v}_x = \omega_g v_y + \frac{q}{m} E_x \\ & \dot{v}_y = - \omega_g v_x \end{align*}$$ where $\dot{v}_x = (qB/m)v_y+qE_x/m$ and $\omega_g = qB/m$. For the y-component, there is no electric field. By differentiating the two equations above, we get

$$\begin{align*} & \ddot{v}_x = - \omega_g^2 v_x \\ & \ddot{v}_y = - \omega_g^2 \left(v_y + \frac{E_x}{B}\right) \end{align*}$$ These equations resemble the [[em-gyration|gyration equations]] if we assume $v_y + E_x/B \equiv v_y’$. Thus, these two equations also describe a gyration, although in this case, the guiding center of the gyration drifts in the $-y$ direction. This drift of the guiding center is called E-cross-B drift, with velocity

$$\mathbf{v}_E = \frac{\mathbf{E} \times \mathbf{B}}{B^2}$$

which does not depend on the electric charge, so the drift is the same for all types of charges. The video below explains this:

Here, the E-field is perpendicular to the B-field, and the plane of gyration for both an ion and an electron is shown. As ions accelerate along the direction of the E-field and decelerate in the opposite direction, the opposite happens for electrons. However, since electrons gyrate in the opposite direction, both ions and electrons ultimately drift in the same direction. Using an ion as an example makes this clearer: when the ion moves along the E-field, its gyro-radius increases due to acceleration, and decreases when moving in the opposite direction. As a result of this alternating increase and decrease, the center of the gyro-orbit, i.e., the guiding center, starts to drift. The drift direction is along the cross product of the E and B fields, i.e., perpendicular to both fields.

The true explanation of the drift lies in the Lorentz transformation of the electric field in the frame of the moving charge. In the moving frame, the transformed field is

$$\mathbf{E}' = \mathbf{E} + \mathbf{v} \times \mathbf{B} = 0$$

because the charge is considered free. Therefore,

$$\mathbf{E} = -\mathbf{v} \times \mathbf{B}$$

Taking the cross product of both sides with the B-field yields the drift velocity.

Another drift relation can be derived directly from the equation of motion of a charge in the presence of E and B fields. Taking the cross product of both sides of the equation of motion with $\mathbf{B}/B^2$, we get

$$\begin{align*} & m\frac{d\mathbf{v}}{dt} \times \frac{\mathbf{B}}{B^2} = q\frac{\mathbf{E} \times \mathbf{B}}{B^2} + \frac{q}{B^2} (\mathbf{B} \times \mathbf{v} \times \mathbf{B}) \\ & \Rightarrow \frac{m}{q} \frac{d\mathbf{v}}{dt} \times \frac{\mathbf{B}}{B^2} = \frac{\mathbf{E} \times \mathbf{B}}{B^2} + \frac{q}{B^2}[(\mathbf{B} \cdot \mathbf{B})\mathbf{B} - (\mathbf{B} \cdot \mathbf{v})\mathbf{B}] \\ & \Rightarrow \mathbf{v} - \frac{\mathbf{B} (\mathbf{v} \cdot \mathbf{B})}{B^2} = \frac{\mathbf{E} \times \mathbf{B}}{B^2} - \frac{m}{q} \frac{d\mathbf{v}}{dt} \times \frac{\mathbf{B}}{B^2} \end{align*}$$ where the two terms on the left together represent a perpendicular velocity vector, and the first term on the right is the E-cross-B drift. If we assume that all temporal changes during a gyro-period are negligible, this perpendicular velocity can be thought of as the perpendicular drift velocity $\mathbf{v}_d$. Thus,

$$\mathbf{v}_d = \mathbf{v}_E - \frac{m}{qB^2} \frac{d}{dt}(\mathbf{v} \times \mathbf{B}) = \mathbf{v}_E + \frac{1}{\omega_g B} \frac{d\mathbf{E}_\perp}{dt}$$

Since we know from the Lorentz transformation that $\mathbf{v} \times \mathbf{B} = -\mathbf{E}$ and the gyro-frequency $\omega_g = qB/m$, the last term on the right side of the equation is known as polarization drift:

$$\mathbf{v}_p = \frac{1}{\omega_g B} \frac{d\mathbf{E}_\perp}{dt}$$

This drift differs from E-cross-B drift in at least two ways: it is proportional to the particle’s mass and also depends on its charge; both of these terms are related to the gyro-frequency.