A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity `vecv` and a positron enters via opposite face with velocity `-vecv`. At this instant,

Options

A. the electric forces on both the particles cause identical accelerations
B. the magnetic forces on both the particles cause equal acceleration
C. both particles gain or loose energy at the same rate
D. the motion of the centre of mass (CM) is determined by `vecB` alone

Answer

B, C, D

Explanation

Acceleration of electron due to electric field is `veca_E=-evecE//m`.
Its direction is opposite to the direction of `vecE`.
Acceleration of positron due to electric field is `vec(a_E^')=evec(E//)m`, Its direction is along the direction of `vecE`.
Thus `vec(a_E^')!=vec(a_E)`. Hence, option (a) is wrong.
Magnetic force on electron, `vecF=-e(vecvxxvecB)`
Acceleration of electron, `vec(a_m)=(vecF)/(m)=-(e(vecvxxvecB))/(m)`
Magnetic force on position, `vec(F^')=e(-vecvxxvecB)=-e(vecvxxvecB)`
Acceleration of position, `vec(a_m^')=(vec(F^'))/(m)=-(-e(vecvxxvecB))/(m) :. vec(a_m)=vec(a_m^')`. Hence, option (b) is true.
As both the particles (electron and positron) are of same mass and same charge in magnitude having same acceleration due to magnetic field, hence they gain or loose the energy at the same rate. Thus option (c) is true.
Due to electric field, net electric force on electron-position pair, `=-evecE+evecE=0`.
Net magnetic force on electron-position pair `=-e(vecvxxvecB)+[-e(vecvxxvecB)]=-2e(vecvxxvecB)`
Therefore, the motion of the centre of mass (CM) is determined by magnetic field alone. Thus option (d) is true.