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# A cubical region of space is filled with some uniform electric and magnetic fields . An electron enters the cube magnetic fields. An electron enters the cube across one of its faces with velocity $\overrightarrow {v}$ and a position enters via opposite face with velocity - $\overrightarrow v$ At this instant .

$\begin{array}{1 1} \text{the magnetic forces on both the particle cause equal accleration } \\ \text{both particles gain or loose energy at the rate} \\ \text{the motion of the center of mass(M) is determined by$\overrightarrow{B}$alone} \\ \text{all the above} \end{array}$

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A)
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 : $\text{all the above}$