Question

A charged particle is projected in a space where electric field
Ē = Eoi and Ē = Boj is present. If particle is projected with
velocity v = Vok(Eo = VoBo ), then
Particle move on a circular path
Particle move on a straight line path
Particle move on a parabolic path
Particle move on a helical path​

Answer 1

fm=0 , theta =180° , v= constant , path will be straight line

Answer 2

Given:- Ē = Eoi and Ē = Boj

To find:- Motion of the particle

Solution:-

The main equation required to express the electric field is given by the following equation:-

Fm = q*[E+[V X B]]

where it is a vector equation and V x B is a cross product.

It can be also expressed as,

Fm = q*[Eoi + [Vok x Boj]]

Solving it further we get,

Fm = q*[Eoi + VoBo(-i)]

as k x j = -i

Fm = q*[Eoi - VoBoi]

but Eo  VoBo

Therefore the equation becomes,

Fm = q*[VoBoi-VoBoj]

Fm= q*[0] = 0

Therefore we conclude that v is constant Fm= 0

and theta is 0 or 180 degrees hence the motion is in straight line.

Particle move on a straight line path is the answer.