Two charged particles traverse identical helical paths in an opposite sense in a uniform magnetic field $\vec B = {B_0}\hat k$.
A) They have equal z-components of momenta.
B) They must have equal charges.
C) They necessarily represent a particle-antiparticle pair.
D) The charge to mass ratio satisfies ${\left( {\dfrac{e}{m}} \right)_1} + {\left( {\dfrac{e}{m}} \right)_2} = 0$.

Hint: Identical helical path means that the radius of helical path and frequency of both the particles’ is the same. The force due to the magnetic field must be determined thoroughly to comment on the nature of the particle’s motion and its other aspects.

Complete step by step solution:
Total velocity of a particle in the magnetic field, $v = {v_ \bot } + {v_\parallel }$
Where,
${v_ \bot } = v\sin \theta =$ the perpendicular component of velocity which governs the circular motion of the particle
${v_\parallel } = v\cos \theta =$ the parallel component of velocity governs its straight (linear) motion of the particle
The force on the particle by the magnetic field, $\vec F = q\left( {\vec v \times \vec B} \right)$
$\Rightarrow F = qvB\sin \theta$
The magnetic force balances the centripetal force$\left( { = \dfrac{{mv_ \bot ^2}}{r}} \right)$ on a particle undergoing a circular motion,
$\Rightarrow q{v_ \bot }B\sin \theta = \dfrac{{mv_ \bot ^2}}{r}$
Since centripetal force is perpendicular to the motion of the particle,
$\therefore \theta = {90^ \circ }$
$\Rightarrow \sin \theta = \sin {90^ \circ }$
$\Rightarrow \sin \theta = 1$
$\Rightarrow q{v_ \bot }B = \dfrac{{mv_ \bot ^2}}{r}$
Canceling $v$ on both sides and rearranging the equation, we get
$\Rightarrow r = \dfrac{{m{v_ \bot }}}{{qB}}$
$\Rightarrow r = \dfrac{{mv}}{{qB}}$…………… equation (1)
Where, $r =$ the radius of the helical path
$m =$ mass of particle
$v =$ velocity of particle
$B =$ the magnitude of the magnetic field
$q =$ charge of particle
$\theta =$ the angle between the velocity of the particle and the direction of the magnetic field.
Also, Time period (of the helical path), $T = \dfrac{{2\pi }}{\omega }$
Here, $\omega = \dfrac{v}{r}$
$\Rightarrow T = \dfrac{{2\pi r}}{v}$
Substituting from equation (1):
$\Rightarrow T = \dfrac{{2\pi }}{v}.\left( {\dfrac{{mv}}{{qB}}} \right)$
Canceling $v,$ Therefore, $T = \dfrac{{2\pi m}}{{qB}}$
Now, frequency,$\nu = \dfrac{1}{T}$
$\therefore \nu = \dfrac{{qB}}{{2\pi m}}$
Let the frequency of the first particle be ${\nu _1}$ and that of the second particle be ${\nu _2}$.
Given that, ${\nu _1} = {\nu _2}$
$\Rightarrow \dfrac{{{q_1}B}}{{2\pi {m_1}}} = \dfrac{{{q_2}B}}{{2\pi {m_2}}}$
$\Rightarrow \dfrac{{{q_1}}}{{{m_1}}} = \dfrac{{{q_2}}}{{{m_2}}}$
$\therefore {\left( {\dfrac{q}{m}} \right)_1} = {\left( {\dfrac{q}{m}} \right)_2}$
$\therefore {\left( {\dfrac{q}{m}} \right)_1} - {\left( {\dfrac{q}{m}} \right)_2} = 0$
Hence, the two charged particles must have an equal charge to mass ratio, but they should be opposite, i.e., one should be positive, and the other should be negative.

Therefore, the correct answer is [D], ${\left( {\dfrac{e}{m}} \right)_1} - {\left( {\dfrac{e}{m}} \right)_2} = 0$.

Note: Lorentz force guides a particle’s movement in a region where both electric and magnetic fields are present. If the force due to the electric field overpowers, then the particle moves in a straight-line path. If the magnetic field overpowers, then the particle moves in a circular path within the region. Moreover, a helical path is followed when forces due to both the fields are acting significantly.