Find the position of a proton in an E and B field

[zehkari]

Homework Statement

(See Attempt).

Relevant Equations

(See Attempt).

Hello all,

I have a question with the helix path of proton in a magnetic field that I am a bit stuck on.

Question:

Equations:
F = qv X B
F = mv^2/r
d=vt

My Attempt:

Think the graph drawn is good enough for questions (a). However, I am stuck on (b) and (c).
Firstly I am not entirely sure what question (b) is asking, the position where? At t = 0?
And then I can obtain an equation for the position of the proton after three complete revolutions (c), which is three 'pitches'. But I do not have the velocity in the z direction.

I also think I am overlooking the Electric field here. Which could play a role with the velocity in the z direction.

So I am not sure how to calculate the velocity in the z direction and what question (b) is asking.

Any help would be appreciated. Thank you for your time.

 

[kuruman]

I also think I am overlooking the Electric field here. Which could play a role with the velocity in the z direction.

What would be the motion of the proton if the magnetic field were not there? Would it move in a straight line or accelerate? In this case, you can simply superimpose the motion in the electric-field-only case and the motion in the magnetic-field-only case. (Why?)

 

[zehkari]

What would be the motion of the proton if the magnetic field were not there? Would it move in a straight line or accelerate? In this case, you can simply superimpose the motion in the electric-field-only case and the motion in the magnetic-field-only case. (Why?)

The motion of the proton in the z direction would be due to the electric field applying a linear acceleration?

So you could say the velocity in the z direction is:

$$v_z = u_z + at$$

So,

$$v_z = at $$

But as $a = \frac{Eq}{m}$, then,

$$v_z = \frac{Eq}{m}t$$

Displacement could be found with another kinematic equation as well,

$$z = \frac{1}{2}\frac{Eq}{m}t$$

Am I on the right track? I am still confused with adding a time variable to the problem.

Thank you for your time.

*Edit:

Also a quick question, would $v_z = \frac{E}{B}$ work for the velocity? As I thought this only worked when the electric field is perpendicular to the magnetic.

 

[kuruman]

The motion of the proton in the z direction would be due to the electric field applying a linear acceleration?

Yes.

So you could say the velocity in the z direction is:$$v_z = u_z + at$$So,$$v_z = at $$

Yes.

But as $a = \frac{Eq}{m}$, then,$$v_z = \frac{Eq}{m}t$$

Yes, except what you have above is $a_z$, not $a$.

Displacement could be found with another kinematic equation as well,$$z = \frac{1}{2}\frac{Eq}{m}t$$

No. The $z$-component of the position goes as $t^2.$

Am I on the right track? I am still confused with adding a time variable to the problem.

You are on the right track, but what do you find confusing about the time variable? In this case the motion in the $xy$-plane is independent of the motion in the $z$-direction.

Also a quick question, would $v_z = \frac{E}{B}$ work for the velocity? As I thought this only worked when the electric field is perpendicular to the magnetic.

It only works when you have crossed electric and magnetic fields and it is the condition for the particle to emerge move in a straight line through the field region when its velocity is perpendicular to both fields. It is irrelevant to the situation you have here.

 

[zehkari]

No. The $z$-component of the position goes as $t^2.$

Yeah, sorry didn't double check.

You are on the right track, but what do you find confusing about the time variable? In this case the motion in the $xy$-plane is independent of the motion in the $z$-direction.

Question (b) asks to find the position of the proton along the z direction.
Does this mean at $t = 0$? If so the equations found for velocity and accerlation on the z plane do not work? As I can't think of a way to avoid having both unkown variables of time and distance using kinematics.

It only works when you have crossed electric and magnetic fields and it is the condition for the particle to emerge move in a straight line through the field region when its velocity is perpendicular to both fields. It is irrelevant to the situation you have here.

Thank you.

 

[zehkari]

Hey, I think I have the correct method now:

As,
$$z = \frac{1}{2}\frac{Eq}{m}t^2$$

And the time would be the same as one pitch (T), then,
$$z = \frac{1}{2}\frac{Eq}{m}T^2$$

With substituion,
$$z = \frac{1}{2}\frac{Eq}{m}{(\frac{2m\pi}{bq})}^2$$

For part (c), 3 revolutions would be $(3T)^2$.

For (b) I obtain a position of 4 meters. Which is slightly smaller than the radius.

 

[kuruman]

For (b) you need to provide the $z$ coordinate as a function of time, that is for any time $t$. You have done that.
For part (c) you need to provide all three components of the position vector.

 

[zehkari]

Thank you.