Solution

Transformation to a system without an electric field

If $E < cB$ (that is, $k < 1$),
there exists an inertial reference frame moving with drift velocity

$\mathbf{v}_d$

relative to the laboratory, in which the electric field vanishes and only an effective magnetic field remains. The velocity of that frame is precisely the electric drift velocity:

$\mathbf{v}_d = \frac{\mathbf{E} \times \mathbf{B}}{B^2}, \qquad v_d = \frac{E}{B} = kc$

In that privileged system, the particle feels no electric force and moves only under the magnetic field, describing a uniform circular motion with a constant speed that we will call $\beta_1 c$

Relativistic velocity composition

Upon returning to the laboratory system, the velocity of the particle is obtained by combining the circular velocity in the moving system $(\beta_1 c)$ with the drift velocity of the system itself$(v_d = kc)$ Since both motions are collinear at certain instants (parallel or antiparallel), the relativistic addition formula gives the extreme values of the observed velocity:

Maximum velocity (when $\beta_1$and k point in the same direction):

$v_{\max} = \frac{\beta_1 c + kc}{1 + \beta_1 k} = c\,\frac{\beta_1 + k}{1 + \beta_1 k}$

Minimum velocity (when $\beta_1$ and k point in opposite directions):

$v_{\min} = \frac{\beta_1 c - kc}{1 - \beta_1 k} = c\,\frac{\beta_1 - k}{1 - \beta_1 k}$

The statement tells us that this minimum velocity is precisely $\beta c$ Therefore:

$\beta c = c\,\frac{\beta_1 - k}{1 - \beta_1 k} \quad\Rightarrow\quad \beta = \frac{\beta_1 - k}{1 - \beta_1 k}$

Expression for the maximum velocity in terms of$\beta and k$

From the previous relation we can solve for $\beta_1$

$\beta_1 = \frac{\beta + k}{1 + \beta k}$.

Substituting this expression into the formula for the maximum velocity we obtain:

$v_{\max} = c\,\frac{\dfrac{\beta + k}{1 + \beta k} + k}{1 + \dfrac{\beta + k}{1 + \beta k}\,k} = c\,\frac{(\beta + k) + k(1 + \beta k)}{(1 + \beta k) + (\beta + k)k}$

Simplifying the numerator and denominator:

$v_{\max} = c\,\frac{\beta + k + k + \beta k^2}{1 + \beta k + \beta k + k^2} = c\,\frac{\beta(1 + k^2) + 2k}{1 + k^2 + 2\beta k}$