# Particle's path in magnetic field

1. Mar 4, 2014

### ivanwho49

1. The problem statement, all variables and given/known data

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)

2. Relevant equations

I'm not sure.

3. 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: Mar 4, 2014
2. Mar 4, 2014

### ivanwho49

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..?

3. Mar 4, 2014

### Staff: Mentor

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?

4. Mar 4, 2014

### kinematics

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?

5. Mar 4, 2014

### ivanwho49

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.

6. Mar 4, 2014

### ivanwho49

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!

7. Mar 5, 2014

### BvU

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.

8. Mar 5, 2014

### kinematics

Eureka! Two equations, two unknowns!

9. Mar 5, 2014

### BvU

Must say I don't understand kine's eureka. Do you, Ivan ?