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Homework Help: Particle's path in magnetic field

  1. Mar 4, 2014 #1
    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)
    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. jcsd
  3. Mar 4, 2014 #2
    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..?
  4. Mar 4, 2014 #3


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    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?
  5. Mar 4, 2014 #4
    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?
  6. Mar 4, 2014 #5
    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.
  7. Mar 4, 2014 #6
    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!
  8. Mar 5, 2014 #7


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    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.
  9. Mar 5, 2014 #8
    Eureka! Two equations, two unknowns!
  10. Mar 5, 2014 #9


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