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

B

Explanation

Force on electron, `vecF_e = e vecE` Force on the positron, `vecF_e = evecE`
. Since, `vecF_e = vecF_e` , so electric forces on both the particles cuase accelerations in opposite direction. So, option(a) is incorrect. Magnetic force on a moving charge particle,
`vecF_m = q(vecv xx vecB)`
Its acceleration, `veca_m = (vecF_m)/(m) = (q(vecv xx vecB))/(m)`
Acceleration of the electron due to magnetic force,
`vec a_m = (-evecv xx vecB)/(m)`
Acceleration of the position due to magnetic force
` veca_m = (e(-vecv xx vecB) )/(m) therefore veca_m = veca_m`
Here , `(mv^2)/(R ) = ev B, v = (eBR)/( m)`
Maximum kinetic energy for both the particles ,
`K_(max) = 1/2 mv^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 - position pair
`vecF_m = - e vecE +e vecE = 0`
Net magnetic force on the electron - position pair
` vecF_m = -e (vecv xx vec B) + e (-vecv xx vecB)`
Clearly , the motion of the centre of mass of electron - position pair determined by `vecB` alone . hence , option b,c, and d are correct