Particle Focusing in a Uniform Magnetic Field

[bon]

Homework Statement

Particles with charge e and mass m are emitted with velocity v from a point source. Their directions of emission make a small angle with the direction of a uniform constant flux density B. Show that the particles are focussed to a point at a distance 2pi mv/Be from their source and at integral multiples of this distance.

Homework Equations

The Attempt at a Solution

Can't see how this will lead to focussing? solving F = e(vxB) i get that the x and y components of velocity will be constant while md^2 z/dt^2 = -Bevsin(theta) - which doesn't lead to focussing!

Anyone see how I can solve this?

Thanks!

 

[tiny-tim]

hi bon!

¬ solving F = e(vxB) i get that the x and y components of velocity will be constant while md^2 z/dt^2 = -Bevsin(theta) …

no, the z component is constant, and the x and y vary …

try dotting v' = (e/m) v x B with v or with B …

then concentrate just on the x and y components ​

 

[bon]

hi bon!


no, the z component is constant, and the x and y vary …

try dotting v' = (e/m) v x B with v or with B …

then concentrate just on the x and y components ​

Hi!

Thanks - i see where i went wrong now. But I am still getting the wrong answer...I get dist = v2pi/w rather than v2pi/w^2

I find that x = A sinwt, y = B sinwt i.e. both =0 where wt = 2pi etc.. but that is where t = 2pi/w

but z = vcostheta t

so if theta is small, z = vt, = v(2pi/w)

Where have i gone wrong?

Thanks!

 

[tiny-tim]

hi bon!

(have an omega: ω and a theta: θ and try using the X² icon just above the Reply box )

I find that x = A sinwt, y = B sinwt i.e. both =0 where wt = 2pi etc..

where do you get that from?

since the original equation is in v, it might be safer to start with an equation in x' and y' rather than x and y

 

[eczeno]

I find that x = A sinwt, y = B sinwt

these are not the results I get. how did you arrive at them?

 

[bon]

hi bon!

(have an omega: ω and a theta: θ and try using the X² icon just above the Reply box )


where do you get that from?

since the original equation is in v, it might be safer to start with an equation in x' and y' rather than x and y

Hi Tiny Tim

I got this from F = ma = q(vxB)

I got that x'' = -(eB/m)^2 x, which, together with the boundary conditions gives this solution...

I got x'' = -(eB/m)^2 x from the eom.

 

[tiny-tim]

hi bon!

I got this from F = ma = q(vxB)

I got that x'' = -(eB/m)^2 x

i don't understand how you got an equation in x from a (first-degree) equation in v

 

[bon]

Thanks Tiny Tim - I think I've solved this now :) THough am stuck on another question! Please could you see this post:

https://www.physicsforums.com/showthread.php?t=486385

Thanks!