Difference in Radius of Charged Particles in Magnetic Field

[tristanslater]

Homework Statement

Ions having equal charges but masses of M and 2M experience a constant electric field while they travel a fixed distance d and then enter a uniform magnetic field perpendicular to their path. If the heavier ions follow a circular arc of radius R, what is the radius of the arc followed by the lighter?

Homework Equations

R = mv / qB

The Attempt at a Solution

The wording of this question is a little ambiguous, but since no values are given for B or E, I figured that the magnetic field was the only one effecting the motion of the particles and that the electric field merely accelerates the particle to the initial velocity.

My thinking was that since the only force is magnetic force causing the circular motion:
F_c = F_B
mv^2 / R = qvB

Solving for R:
R = mv / qB (turns out this is a well established formula)

So, if R = 2Mv / qB (radius of heavy particle), then R / 2 = Mv / qB (radius of light particle in terms of radius of heavy particle.)

My answer: R / 2

However, the correct answer is (unless there is an error) R / sqrt(2).

Can someone explain why?

Thanks.

 

[TSny]

Welcome to PF!

Is the speed v the same for the two ions?

 

[tristanslater]

Thank you!

It doesn't say. I just copied and pasted the question exactly. What you see is what was provided.

I had thought that perhaps that was the issue, the electric field, if it was used to accelerate the two particles the lighter one would have been accelerated by more. Perhaps I need to think through that line of reasoning and see if that gets the correct answer.

 

[tristanslater]

Thank you!

It doesn't say. I just copied and pasted the question exactly. What you see is what was provided.

I had thought that perhaps that was the issue, the electric field, if it was used to accelerate the two particles the lighter one would have been accelerated by more. Perhaps I need to think through that line of reasoning and see if that gets the correct answer.

After thinking this through, I'm pretty sure that if that was the case, the velocity of the heavy object would be half of the light object, which would then result in the radii being the same! Still doesn't make sense.

 

[TSny]

After thinking this through, I'm pretty sure that if that was the case, the velocity of the heavy object would be half of the light object,

The heavier ion will not have half the speed of the lighter ion.

Hint: Compare the work done on each ion by the electric force. Work is related to energy.

Or, alternately, you can use a kinematics approach. Since the force is constant, what can you say about the acceleration?

 

[tristanslater]

The heavier ion will not have half the speed of the lighter ion.

Hint: Compare the work done on each ion by the electric force. Work is related to energy.

Or, alternately, you can use a kinematics approach. Since the force is constant, what can you say about the acceleration?

Thanks for your input. However, I already figured out that the velocity of the light would be twice the heavier. Unfortunately, this only makes my original problem worse.

I actually think the key is that the wording is very ambiguous. It doesn't really describe the situation well. What direction is the electric field in? Did either of the particles have velocity to begin with? Is the electric field still present when experiencing the magnetic field as in a mass spectrometer?

Can anyone get a clear picture of what is even happening?

Thanks.

 

[TSny]

Thanks for your input. However, I already figured out that the velocity of the light would be twice the heavier. Unfortunately, this only makes my original problem worse.

Can you show how you deduced that the lighter ion has twice the speed of the heavier ion?

(You are right that the problem statement should have specified that the ions are initially at rest.)

 

[tristanslater]

Actually, on second thought, that might not be the case. I was thinking they were accelerated for the same amount of time, but it says over a fixed distance.

So, starting with a = F_E / m (F_E is not dependent on mass, so should be the same for each), the acceleration of the light particle, a = F_E / m, and a_heavy = F_E / 2M => a_heavy = 1/2 * a_light.

Using d = 1/2at^2 (assuming starting from rest) for the light particle: d = 1/2at^2, and solving for t (again basing around light particle): t = sqrt(2d / a) => t_heavy = sqrt(2d / (1/2 * a)) = sqrt(4d / a) = sqrt(2) * sqrt(2d / a) = sqrt(2) t.

Now back to velocity: v = at (again for light), v_heavy = 1/2 * a * sqrt(2) * t = (sqrt(2) / 2) * v.

To summarize:
Light: m, a, t, v
Heavy: 2m, a/2, sqrt(2) * t, (sqrt(2) / 2) * v

Back to original: R_light = mv / qB, and R_heavy = R = 2m * ((sqrt(2) / 2) * v) / (qB). The 2's cancel: R = sqrt(2) * (mv / qb) = sqrt(2) * R_light. Rearranging: R_light = R / sqrt(2).

Yay!

Thanks so much for getting me on the right track!

 

[TSny]

OK, that looks good.

If you're familiar with the concepts of work and energy, then you can get to the answer much quicker. The work is W = Fd = qEd, which is the same for each ion. The work determines the kinetic energy, KE, of each ion. So, the final KE of each ion is the same. The ratio of the speeds can then be found from the ratio of the KE's.

 

[tristanslater]

OK, that looks good.

If you're familiar with the concepts of work and energy, then you can get to the answer much quicker. The work is W = Fd = qEd, which is the same for each ion. The work determines the kinetic energy, KE, of each ion. So, the final KE of each ion is the same. The ratio of the speeds can then be found from the ratio of the KE's.

I am familiar, I just didn't think of doing it that way. Maybe I'll try that for practice.

 

[TSny]

OK.

Also, for the kinematics approach, you can use the formula $v^2 = v_0^2 + 2 a x$ to relate the final velocity to the acceleration and distance. This avoids having to work with time.

 

[tristanslater]

OK.

Also, for the kinematics approach, you can use the formula $v^2 = v_0^2 + 2 a x$ to relate the final velocity to the acceleration and distance. This avoids having to work with time.

Thanks again! How do I mark this answered?

 

[tristanslater]

Thanks again! How do I mark this answered?

Nevermind. Found it.