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Momentum and Lorentz-force

  1. Feb 29, 2008 #1
    [SOLVED] Momentum and Lorentz-force

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

    I have two questions.

    1) Please take a look at page 22 in this document (look at the bottom):

    http://walet.phy.umist.ac.uk/P615/Notes.pdf

    It says that the momentum in GeV/c for a particle which moves in a magnetic field is given by:

    p = 0.3*B*r, where B is in Teslas and r is the radius in metres.

    I can derive p = qBr, but how does it become p = 0.3Br when p is measured in GeV/c?

    2) If I have a proton moving in a magnetic field, where the magnetic field is perpendicular to the velocity-vector of the proton and the proton turns right, does the magnetic field point into or out of the plane?

    3. The attempt at a solution
    1) Ok, I know that the net-force is mv^2/r = q(E + v x B) = qvB, so p = q*B*r. But this is in m/s. How do I convert to GeV/c to obtain the expression above?

    2) I tried using the right-hand-rule: The index finger in the direction of the velocity, the thumb in the direction of F (right, since it is turning this way) and then the middle-finger points upwards, which is the direction of B? Can you confirm this?

    Thanks in advance.

    Sincerely Niles.
     
    Last edited: Feb 29, 2008
  2. jcsd
  3. Feb 29, 2008 #2
    You've got the index and the middle fingers mixed up.
    1ev=1.6e(-19) joules.
     
  4. Feb 29, 2008 #3
    Thanks on the right-hand-rule.

    Can you show me how it is derived? I've spent several hours today trying to do it, but I keep messing it up.
     
  5. Feb 29, 2008 #4
    Ok, I can start myself.

    First I have p = qvB, where [p] = kg*m/s = J*s/m.

    I want this in GeV/c. We know that 1 eV = 1.602 E -19 J, so I multiply J*s/m with (1,602 E -19)^(-1) eV/J. This is where I am stuck
     
  6. Feb 29, 2008 #5

    kdv

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    First, your derivation of qvB is not correct since everything is relativistic here so p i snot mv but [itex] \gamma m v [/itex] and so on. But I think that your final result is correct.

    Now, the equation given in the text does not contain the charge so you know it can't be the complete equation. It seems as if they are assuming a charge equal to the elementary charge. This will give a factor of [tex]1.6 \times 10^{-19} [/tex] which will cancel out with the factor coming from the conversion of eV to Joules. There is a factor of 10^9 coming from the GeV. And finally, don't forget the factor of c, the speed of light (which why there will be a factor of 1/3 approx 0.3).
     
  7. Feb 29, 2008 #6

    kdv

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    My mistake..It won't be 1/3, it will simply be 3x10^8/10^9 = 0.3 .
     
  8. Feb 29, 2008 #7

    kdv

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    we can't answer your second question without knowing in what direction the proton i smoving!
     
  9. Mar 1, 2008 #8
  10. Mar 1, 2008 #9
    Ok, I know that the units for q*B*r = kg*m/s which equals 1.87*10^18 GeV/c. Where to go from here? I don't know what to do with the q in Coulombs.

    If we say that the charge q = 1.602*10^(-19) eV, then I get Br = 0.3 GeV/c - but that's wrong?
     
    Last edited: Mar 1, 2008
  11. Mar 1, 2008 #10
    The charge cannot be in ev. Thats a unit of energy.
     
  12. Mar 1, 2008 #11
    Then what steps am I missing to get 0.3Br?
     
  13. Mar 1, 2008 #12

    kdv

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    The formula is qBR which, as you pointed out, is in kg m/s which is the same as Joule/(m/s).


    Let's start by getting rid of the charge. Let's stay in SI units for now. Let's say we want a formula of the form K BR where K is some numerical constant such that when people plug in the value of B and R in Tesla and in meters, they get the correct momentum in SI units. Then obviously the formula is

    [tex] 1.6 \times 10^{-19} B R [/tex]
    will give the momentum in SI units if B and R are plugged in in tesla and meters.

    Now, this will give the momemtum in J/(m/s)

    Now let's say that you want people to plug in the values of B and R in meters and in Tesla but you want the answer to come out in GeV/(m/s). Since [itex]1\,GeV = 1.6 \times10^{-10} J[/itex], you must divide the equation by that factor.

    So the formula

    [tex] \frac{1.6 \times 10^{-19}}{1.6 \times 10^{-10}} BR = 10^{-9} BR [/tex] will give the momentum in GeV/(m/s) if B and R are given in tesla and in meters.

    Now you want the final result to be given in GeV/c. Since c is approximately 3x10^8 m/s, you must multiply the equation above by c to get the momentum in GeV/c, So finally, 0.3BR will give the momentum in GeV/c if B and R are given in Tesla and in meters.
     
  14. Mar 1, 2008 #13
    That figures, but only for a unit or test charge...
     
  15. Mar 1, 2008 #14

    kdv

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    Like I said in my post #5, the equation makes only sense if we assume that the particle has an elementary charge [itex] e [/itex]. For particle physics experiments, this makes sense since one would not be working with nuclei (other than a single proton) or atoms.
     
  16. Mar 2, 2008 #15
    Thanks - I see my method wasn't correct.
     
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