Solution :

$a_E=-e \overrightarrow{E} lm$

This direction is opposite to the direction of $\overrightarrow{E}$ Acceleration of position due to electric field $\overrightarrow{a_E}=\large\frac{e \overrightarrow E}{m}$

This direction is along $\overrightarrow {E}$

From this it is clear that $ aE \neq a'E$

therefore option 'a' is not correct.

$\overrightarrow{F} = -e(\overrightarrow{u} \times \overrightarrow{B})$

$a_m = \large\frac{\overrightarrow{F}}{m}$

$\qquad= \large\frac{-e(\overrightarrow{u} \times \overrightarrow{B})}{m}$

Magnetic force on position is $ \overrightarrow{F'}=-e(-\overrightarrow{v} \times \overrightarrow{B})$

$a^1_m= \large\frac{\overrightarrow{F_1}}{m}=\large\frac{-e (\overrightarrow{v} \times \overrightarrow{B})}{m}$

From this it is clear that $a_m=a'_m$

The magnetic forces on both the particle cause equal accleration is correct

As both the particles (electron and position ) are of same mass and same changes in magnitude and this acceleration due to magnetic field is also equal.

As a result of this they gain or loose the energy at the same rate.

Therefore option both particles gain or loose energy at the rate is correct

Due to electric field , net electric force force on electron - position pair is $-e \overrightarrow{E} + e \overrightarrow{E}=0$

Net magnetic force on electron - position pair

$\quad= -e( \overrightarrow{v} \times \overrightarrow{B}) + [-e \overrightarrow{v} \times \overrightarrow{B})]=-2e ( \overrightarrow{v} \times \overrightarrow{B})$

This clearly states that the motion of center of mass (CM) is determined by magnetic field alone.

The motion of the center of mass(M) is determined by $\overrightarrow{B}$ alone is correct

Answer : $\text{all the above} $