Question #21d0f

1 Answer
Jan 30, 2016

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