Particle's path in magnetic field

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Homework Help Overview

The discussion revolves around a charge moving through a magnetic field, specifically focusing on the path it will follow and the calculations for the radius and period of that path. The parameters given include a velocity of 10 m/s, a charge of 3 C, and a magnetic field strength of 4 T.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the need for mass in the equations related to the motion of the charged particle in the magnetic field. There is uncertainty about whether to use the mass of a proton or another value, given the charge's magnitude. The equations for radius and period are mentioned, but participants express confusion about how to proceed without knowing the mass.

Discussion Status

Several participants have provided insights into the relevant equations, such as the Lorentz Force equation and the relationship between force, mass, and radius. There is an ongoing exploration of the implications of charge and mass, with some participants questioning the assumptions made about the mass of the charged particle. The discussion remains open, with no consensus reached on how to resolve the uncertainties regarding mass.

Contextual Notes

Participants note the unusual charge value and its implications for the mass of the particle, raising questions about the physical feasibility of such a scenario. The lack of explicit information about the mass is acknowledged as a significant constraint in solving the problem.

ivanwho49
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Homework Statement



A charge enters a magnetic field. Draw the path the charge will follow and find the radius and period of the particle's path.

v=10m/s (drawn as a straight line pointing right)
q=3c
B=4T (points outward)

Homework Equations



I'm not sure.

The Attempt at a Solution



The charge is drawn as positive. Wouldn't I need the mass for an equation? I don't know what the mass would be. Would it just be the mass of a proton?

Also, anyone know what equation I would use?
 
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I think the equation for radius is mv/(qB) but I still don't know if I should use 1.67*10^(-27) kg as the mass. Q = 3C so it seems like it would be a mistake to make the mass equal to that of a proton's mass when Q does not equal 1.6*10^(19) C...

Then the equation for the period is qB/m..?
 
ivanwho49 said:

Homework Statement



A charge enters a magnetic field. Draw the path the charge will follow and find the radius and period of the particle's path.

v=10m/s (drawn as a straight line pointing right)
q=3c
B=4T (points outward)


Homework Equations



I'm not sure.

The Attempt at a Solution



The charge is drawn as positive. Wouldn't I need the mass for an equation? I don't know what the mass would be. Would it just be the mass of a proton?

Also, anyone know what equation I would use?

ivanwho49 said:
I think the equation for radius is mv/(qB) but I still don't know if I should use 1.67*10^(-27) kg as the mass. Q = 3C so it seems like it would be a mistake to make the mass equal to that of a proton's mass when Q does not equal 1.6*10^(19) C...

Then the equation for the period is qB/m..?

Welcome to the PF.

The Relevant Equation would be the Lorentz Force equation. Can you show us what that is?

And I think you are right -- the Lorentz force just gives the force -- you need the mass to know what the radius of the resulting circular motion is. Have you written out the full question?
 
ivanwho49 said:
I think the equation for radius is mv/(qB) but I still don't know if I should use 1.67*10^(-27) kg as the mass. Q = 3C so it seems like it would be a mistake to make the mass equal to that of a proton's mass when Q does not equal 1.6*10^(19) C...

Then the equation for the period is qB/m..?

Ivan, your equation seems correct. As for the mass can you think of a reason why an object the size of a house could also have the charge given in the question? The mass might be able to be obtained from the equation you used to derive r=mv/qb. What equation was this?
 
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berkeman said:
Welcome to the PF.

The Relevant Equation would be the Lorentz Force equation. Can you show us what that is?

And I think you are right -- the Lorentz force just gives the force -- you need the mass to know what the radius of the resulting circular motion is. Have you written out the full question?

The Lorentz Force equation is F = q(E + v x B) but I don't know how to use it in this occasion.. I have the question written out exactly as our professor stated it but I'm assuming the mass should be stated because I don't know how to solve it otherwise.
 
kinematics said:
Ivan, your equation seems correct. As for the mass can you think of a reason why an object the size of a house could also have the charge given in the question? The mass might be able to be obtained from the equation you used to derive r=mv/qb. What equation was this?

So mass is independent of charge...? but I thought a single proton would always have charge +e. So in this case, the charge is 3C.. is it possible that it still has the same mass as a regular proton?

The equation used to derive r=mv/qb is F=qvB=m(v^2/R).

I can find out what the force is with Lorentz Force equation as berkeman said, I know what q, v, and B are, but that still leaves m and R. It seems to me as if both quantities are dependent of each other and I must know one to get the other. Is there a way around this?



Thank you both!
 
Your F = q(E + v x B) is spot on. There is only a B field, so F = q (v x B) . Right hand thumb to the right, index finger out of the paper pointing to your nose: middle finger points in the direction of F.

F is perpendicular to v, so no change in magnitude of v, only in direction. And this remains so: , |v| constant, |F| constant and ##\vec F \perp \vec v##. The sure sign of a path that is characterized by a radius (what a giveaway in the original problem tekst!), for which we know (right?) F = mv2/R. So there is your ## R={mv\over QB}##. As long as you have no clue about m, no value for R.

To collect 3 Coulomb of charge you need at least 6 x 1018 protons and a lot of energy to squeeze them together. That many protons still is only 10 nanograms, though. But to avoid instant discharge by a lightning flash you need a substantial charge carrier. A 1 meter conducting sphere has a capacitance of 1 ##\mu##F, with 3 C the surface potential is 3 x 109 Volt. 4 Tesla of nicely homogenetic magnetic field isn't all that trivial either and barely possible above the, say, 1 m2 scale. (Others will correct me if I am wrong).

I think this is just a physics thought exercise and somewhat not-good: good students get distracted by the missing of m and average students put m=1 kg or something arbitrary.

I was triggered by this thread because as a 17 year old I had an oral grammar school exam where they put a Teltron tube in front of me that I had never seen before. The thing measures e/m as a ratio, using your formula. But googling 'e over m experiment' shows that JJ Thomson stole the cake in 1897.
 
ivanwho49 said:
So mass is independent of charge...? but I thought a single proton would always have charge +e. So in this case, the charge is 3C.. is it possible that it still has the same mass as a regular proton?

The equation used to derive r=mv/qb is F=qvB=m(v^2/R).

I can find out what the force is with Lorentz Force equation as berkeman said, I know what q, v, and B are, but that still leaves m and R. It seems to me as if both quantities are dependent of each other and I must know one to get the other. Is there a way around this?
Thank you both!

Eureka! Two equations, two unknowns!
 
Must say I don't understand kine's eureka. Do you, Ivan ?
 

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