Particle's path in magnetic field

• ivanwho49
In summary: I had never heard before. So I just started to reason logically, not knowing a thing about how the tube was connected. I was quite pleased with myself, because I had no idea what I was talking about.
ivanwho49

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)

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?

Last edited:
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)

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?

1 person
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 ?

1. What is the direction of a particle's path in a magnetic field?

The direction of a particle's path in a magnetic field is determined by the right-hand rule. If the particle's velocity is perpendicular to the magnetic field, the path will be circular. If the velocity is not perpendicular, the path will be a helix.

2. How does the strength of the magnetic field affect a particle's path?

The strength of the magnetic field affects the radius of the particle's path. A stronger magnetic field will result in a smaller radius, while a weaker magnetic field will result in a larger radius.

3. Can a particle's path be affected by other forces in addition to the magnetic field?

Yes, a particle's path can be affected by other forces such as electric fields and gravitational fields. These additional forces will alter the shape and direction of the particle's path.

4. What is the relationship between the charge of a particle and its path in a magnetic field?

The charge of a particle plays a significant role in its path in a magnetic field. A positively charged particle will move in a different direction than a negatively charged particle, due to the direction of the magnetic force acting on each charge.

5. How does the speed of a particle affect its path in a magnetic field?

The speed of a particle does not affect the shape of its path in a magnetic field, but it does affect the size of the radius. A faster-moving particle will have a larger radius, while a slower-moving particle will have a smaller radius.

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