Radial force on charged particle in beam of positive ions

[astrocytosis]

Homework Statement

Many experiments in physics call for a beam of charged particles. The stability and “optics” of charged-particle beams are influenced by the electric and magnetic forces that the individual charged particles in the beam exert on one another. Consider a beam of positively charged ions with kinetic energy E₀. At t = 0 the beam has a radius R₀, a uniform charge density ρ₀, and is traveling in the x-direction.

(a) Derive an expression for the radial force on a charged particle in the beam at some initial radial position r₀ (note that there will be electric and magnetic forces acting on the particle).

(b) Apply F = ma to determine the radial velocity of the charged particle when its radial position has increased from r₀ to 2r₀. You’ll need to make one assumption here – the charge density of the particle beam remains uniform (but not constant in time!) as the beam expands.

Homework Equations

(1) F_mag = Q(v x B)

(2) F_net = Q[E + (v x B)]

The Attempt at a Solution

[/B]
I am not sure how to treat this situation. Do the charges in the center of the beam act like a current-carrying wire of circular cross section? In this case the volume current density would be J = I/πR₀² at t=0 and the magnetic field generated by the beam at the position of the particle would be B = μ₀I/2πr₀. Would the electric field be that of a line charge, E = kρ₀/r₀? Can I plug these fields into (2) to get the radial force?

Are any of these trains of thought going in the right direction? Thank you.

 

[mfb]

Do the charges in the center of the beam act like a current-carrying wire of circular cross section?

Yes.
Note that you need the magnetic field within the beam, not the magnetic field outside.

Would the electric field be that of a line charge

Sure. Same thing here, you need it inside.

You can express I via the particle velocity and given constants.

Can I plug these fields into (2) to get the radial force?

Sure, once you have the correct fields.

 

[astrocytosis]

Ok, so would the magnetic field within the beam just be due to I_enc=ρvA, so B = μ₀ρ₀r₀/2? And E = kρ₀/r₀? Or am I missing a subtlety?

I'm also a bit unsure how to deal with the direction of the force; because the field is circulating and the velocity is in x, the direction of the magnetic component of the force is rotating?

 

[mfb]

Something is still wrong with the electric field.

The field is constant in time, there is nothing rotating. It is orthogonal to the beam direction so there is a force on the electron (which is also constant in time - assume the outwards motion is negligible).

 

[astrocytosis]

I applied Gauss's law to get the electric field: E(2πr₀L) = Qenc/ε₀ =ρ₀πr₀L/ε₀ → E = ρ₀r₀/2ε₀ (only inside the beam)

When I plug that into (2) I get F = ρ₀πr₀L(ρ₀r₀/2ε₀ + ½μ₀ρ_or₀v²) (where v x B is just vB since the fields are orthogonal)

but I don't think there should be a factor of length L since it wasn't given in the problem statement. And how does E₀ fit in?

EDIT: I haven't been sleeping enough lately... Q is just the charge on the particle itself.

 

[mfb]

Q is just the charge on the particle itself.

Right.

And how does E₀ fit in?

What is the relation between kinetic energy and speed?

You can simplify the expression for the force.

 

[astrocytosis]

I can say ½v² = E₀/m and simplify to F = qρ₀r₀(1/2ε₀ + E₀μ₀/m].

If I set F=ma and get a, v = √(2aΔx). Would 2r₀-r₀=r₀ be Δx in this case, since I'm looking for the radial velocity?

 

[mfb]

v = √(2aΔx)

Be careful with the different directions (along the beam and perpendicular) and your variables.

The beam will expand. How does that influence the force?

 

[astrocytosis]

The force when the particle is at 2r₀ is F = 2qρ₀r₀(1/2ε₀ + E₀μ₀/m], and v = √(2ar₀)?

 

[mfb]

What happens to the charge density if the beam spreads?

 

[astrocytosis]

Does the density go as

ρ'V' = ρ₀V
ρ'π(2r₀)²L=ρ₀πr₀²L
ρ' = ρ₀/4 ?

So the force is overall reduced by a factor of ½ at r=2r₀?

 

[mfb]

Right.

 

[astrocytosis]

Thank you! This was very helpful