Question

Question #21d0f

Answer 1

Only possible if velocity of the charged particle is`=1.907times 10^7 m//s`. Direction of `vec v` must be as assumed in the solution.
It is independent of `e/m`

Explanation:

This is a question related to Motion of a Charged Particle in Electric and Magnetic Field. Let us recall the applicable Lorentz Force equation

`vec F=q[vecE + (vecvxxvecB)]`
where `q` is the charge of the particle, `vecE` is the electric field, `vecv` is the velocity of the charged particle and `vecB` is the magnetic field.

Assumption of direction of three Vectors

Let the charged particle move in the direction given by `hat x`, the electric field be in the `haty` and magnetic field be in the `hatz` direction.

Electric field `vecE` will exert force on the charge in its direction, assuming the charge `q` to be positive.

The direction of the force exerted by the magnetic field is cross product of velocity and magnetic field vectors and is

`hat x times hat z=-haty`.
It is acting in a direction opposite to the electric field vector.
As it is required that the charge should pass through in a straight line, it implies that

Lorentz force must be zero.

`0=q[|vecE|hat y - |vecv|.|vecB|haty]`
`=>|vecE|=|vecv|.|vecB|`
`16728=0.000877times |vecv|`
`|vecv|approx1.907times 10^7 m//s`