Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field 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 anti-particle pair (d) The charge to mass ratio satisfy ((e)/(m))+((e)/(m))=0

Dear Student,
Given,
Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field B given by:
$\overrightarrow{B}=B_o\hat{k}$
The charged particles curl due to the magnetic force which is given by:
$\overrightarrow{F_B}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$
Magnetic force always act perpendicular to the direction of $\overrightarrow{v} and \overrightarrow{B}$. If the velocity $\overrightarrow{v}$ is along the direction of B then no magnetic force acts on the charged particle as a result of which the charged particle does not form a helical path. Since, velocity is not along z-direction, hence its momentum must also not be along z direction. Hence, option (a) is FALSE.

Since, the helical paths of the two charged particles are of opposite sense, there charge cannot be same. Hence, option (b) is FALSE.

Since the charged particles traverse identical helical paths in completely opposite sense, they can represent a particle - antiparticle pair. But, it is not a necessary criteria. Hence, option (c) is FALSE.

$$\begin{gathered} Pitch of the particle, P = T\times v\cos \left(\theta \right) \\ where, \\ \theta is the angle between \overrightarrow{v} and \overrightarrow{B} \\ T is the time period, T=\frac{2\pi }{\omega }=\frac{2\pi m}{qB} \\ P=\frac{2\pi mv\cos \left(\theta \right)}{qB} \\ Since, the helix is identical, so the pitch, P must be a cons\tan t. \\ Hence \\ {\left(\frac{2\pi mv\cos \left(\theta \right)}{qB}\right)}_1={\left(\frac{2\pi mv\cos \left(\theta \right)}{qB}\right)}_2 \\ {\left(\frac{q}{m}\right)}_1={\left(\frac{q}{m}\right)}_2 \\ Hence, the charge to mass ratio must be same. \\ So, the CORRECT option is \left(d\right) \end{gathered}$$
Regards