Direction of a particle in a uniform magnetic field.

[nvictor]

Homework Statement

a beam of particles with velocity v^vector enters a region that has a uniform magnetic field B^vector in the +x direction.

show that when the x component of the displacement of one of the particles is 2 * Pi * (m / qB) * vcos(theta), where theta is the angle between v^vector and B^vector, the velocity of the particle is in the same direction as it was when the particle entered the field

Homework Equations

the period of circular motion is T = 2 * Pi * m / q * B

The Attempt at a Solution

if I divide v = d * t = x * t

if I choose t to be T then I find v = v cos(theta).

then thing is, I don't really undertstand the question, and I want to know how to proceed from there.

 

[tiny-tim]

Hi nvictor!

(have a theta: θ and a pi: π and use bold for vectors )

show that when the x component of the displacement of one of the particles is 2 * Pi * (m / qB) * vcos(theta), where theta is the angle between v^vector and B^vector, the velocity of the particle is in the same direction as it was when the particle entered the field

The displacement in the x-direction is (vcosθ)t (because speed in that direction is constant), so the question is asking you to prove that the period is 2πm/qB

 

[nvictor]

thanks a lot for the reply.