Special theory of relativity > Motion of relativistic particles in electric and magnetic fields > Problem 14.4.28

Statement

$14.4.28.$
In a crossed electric field of intensity E and a magnetic field of induction B, a
relativistic charged particle ”drifts” across the fields. What is the drift velocity
of a particle?

Solution

Consider a particle of charge q moving in a region where a uniform electric field $\mathbf{E} $and a uniform magnetic field$ \mathbf{B} $coexist, with $\mathbf{E} \perp \mathbf{B}$

The total force on the particle, according to the Lorentz law (in its relativistic form, perfectly valid at any speed), is:

$\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$

The drift is defined as a straight-line uniform motion in which the acceleration is zero. If the particle moves with constant velocity $\mathbf{v}_d $ the net force must vanish:

$q(\mathbf{E} + \mathbf{v}_d \times \mathbf{B}) = 0 \quad\Rightarrow\quad \mathbf{E} + \mathbf{v}_d \times \mathbf{B} = 0$

To solve for $\mathbf{v}_d$, we take the cross product with \mathbf{B} on the right and use the vector triple product identity

$\mathbf{E} \times \mathbf{B} + (\mathbf{v}_d \times \mathbf{B}) \times \mathbf{B} = 0$

The second term expands as:

$(\mathbf{v}_d \times \mathbf{B}) \times \mathbf{B} = (\mathbf{v}_d \cdot \mathbf{B})\mathbf{B} - B^2\mathbf{v}_d$

Since $\mathbf{v}_d$ is perpendicular to $\mathbf{B}$ (the drift has no component along the magnetic field), $\mathbf{v}_d \cdot \mathbf{B} = 0$
This leaves:

$\mathbf{E} \times \mathbf{B} - B^2\mathbf{v}_d = 0 \quad\Rightarrow\quad \mathbf{v}_d = \frac{\mathbf{E} \times \mathbf{B}}{B^2}$

The magnitude of this velocity, given that the fields are perpendicular, is simply:

$\boxed{v_d = \frac{E}{B}}$

This result does not depend on the mass or the charge of the particle, and it is valid even in the relativistic regime, as long as the electric field is not too intense compared with the magnetic field.

Answer

$\boxed{v_d = \frac{E}{B}}$