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. A particle of mass m and charge q is moving in a region where uniform, constant electric and magnetic field E and B are present. E and B are parallel to each other, At time t = 0, the velocity v 0 of the particle is perpendicular to E. ( Assume that its speed is always 0 , E and B and their magnitude v 0 , E and B

. A particle of mass m and charge q is moving in a region where uniform, constant electric and magnetic field E and B are present. E and B are parallel to each other, At time t = 0, the velocity v0 of the particle is perpendicular to E. ( Assume that its speed is always 0,  E and B and their magnitude v0, E and B 

Grade:upto college level

1 Answers

Deepak Patra
askIITians Faculty 471 Points
7 years ago
Hello Student,
Please find the answer to your question
Because the forces due to parallel electric and magnetic fields on a charged particle moving perpendicular to the fields will be at right angles to each other, electric force being along the direction of while magnetic force perpendicular to the plane containing and , so magnetic force will not affect the motion of charged particle in the direction of electric field and vice-versa. So, the problem is equivalent to superposition of two independent motion as shown in the adjoining figures.
So, for motion of the particle under electric field alone,
y = qE / m i.e., dvy / dt = qE / m
or dt i.e., vy = qE / m t …(1)
While at the same instant, the charges particle under the action of magnetic field will describe a circle in the x –z plane with
r = mv0 / qB i.e., ω = v0 / r = qB / m
So, angular position of the particle at time t in the x-z plane will be given by
Θ = ωt = qB / m t
and therefore, in accordance with figure accordance with figure
vx = v0 cos θ = v0 cos ωt = v0 cos (qB / m t ) …(2)
and vz = v0 sin θ = v0 sin ωt = v0 sin (qB / m t) …(3)
Thanks
Deepak patra
askIITians Faculty

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