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. `(a,b,c,)`
B. `(b,c,d)`
C. `(a,d)`
D. `(a,c,d)`

Answer

2

Explanation

Force on electron, `vec(F)_(e)=-evec(E)` Force on the positron, `vec(F)_(e)=evec(E)`
Since, `vec(F)_(e)=vec(F)_(e)` so electric forces on both the particles cause accelerations in opposite direction.
So, option `(a)` is incorrect.
Magnetic force on a moving charge particle,
`vec(F)_(m)=q(vec(upsilon)xxvec(B))`
Its acceleration, `vec(a)_(m)=(vec(F)_(m))/(m)=(q(vec(upsilon)xxvec(B)))/(m)`
Acceleration of the electron due to magnetic force,
`vec(a)_(m)=(-evec(upsilon)xxB)/(m)`
Acceration of the position due to magnetic force,
`vec(a)_(m)=(-e(vec(upsilon)xxB))/(m)" ":. vec(a)_(m)=vec(a)_(m)`
Here, `(m upsilon^(2))/(R)=eupsilonB,upsilon=(eBR)/(m)`
Maximum kinetic energy for both the particles,
`K_(max)=(1)/(2)m upsilon^(2)=(1)/(2)m(e^(2)B^(2)R^(2))/(m^(2))=(e^(2)B^(2)R^(2))/(2m)`
Net electric force on the electron positron pair,
`vec(F)_(m)=-evec(E)+evec(E)=vec(0)`
Net magnetic force on the electron `-` position pair,
`vec(F)_(m)=-e(vec(upsilon)xxvec(B))+e(-vec(upsilon)xxvec(B))`
`=-2e(vec(upsilon)xxvec(B))=2e(vec(B)xxvec(upsilon))`
Clearly, the motion of the centre of mass of electron`-` position pair is determined by `vec(B)` alon.
Hence, option `(b), (c )` and `(d)` are correct.