Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field `vec(B)=B_(0)hat(K)`

Options

A. They have equal z-components of momenta
B. They must have equal charges.
C. They necessarily represent a particle-antiparticle pair.
D. The charge to mass ratio satisfies:
`(e/m)_(1) + (e/m)_(2) = 0 `.

Answer

d

Explanation

Let us consider a charged particle of charge q and mass m, moving with a velocity v inclined to the direction of magnetic field `vec B` at some angle `theta`.
We know that, due to perpendicular component of `vec v i.e. v sin theta`, the charged particle will move in a circular path under the influence of an external magnetic field.
`:. ` If r is the radius of circular path, then the necessary centripetal force will be provided by `vec B`.
`q(v sin theta) B = (m(v sin theta)^(2))/r = (mv sin theta)/(qB) ` ....(i)
Due to a parallel component of velocity i.e. `v cos theta`, the charged particle will move along a straight line path. Also, the linear distance travelled by the charged particle in one period of revolution of a circular path is pitch, given by:
`"Pitch " = v cos theta xx T`
` T = (2 pi m)/(qB) ` [From (i)]
Pitch ` = v cos theta xx (2 pi m)/(qB) = (2 pi m v cos theta)/(qB) ` ...(ii)
It is given that the two particles are in identical helical paths. So, their pitches and radii of the circular path will be same. So, from(i) :
`(m_(1) v sin theta)/(q_(1)B) = (m_(2)v sin theta )/(q_(2)B)`
`rArr " " m_(1)/q_(1) = m_(2)/q_(2) rArr |q_(1)/m_(1)| = |q_(2)/m_(2)|`
The same can be obtained from (ii).
Also, the charged particles are traversing in a completely opposite sense, which means these charges are opposite in nature.
`:. " "q_(1)/m_(1) = (-q_(2))/m_(2) rArr q_(1)/m_(1) + q_(2)/m_(2) = 0 `
Therefore, option (d) is correct.