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Proton travelling through a magnetic field

  1. Jun 30, 2011 #1
    1. The problem statement, all variables and given/known data

    I am having trouble with a basic deflection problem - a single proton travelling through a magnetic field:

    v = Velocity of the proton = 6 000 000 meters per second (1/50th the speed of light, so no relativistic effects)

    B = magnetic field strength = 0.5 Tesla

    Charge of proton q = + 1.60217e-19 Coulombs

    Mass of proton m = 1.67262e-27 kg

    Direction = zero radians


    2. Relevant equations

    Force = q*v*B sin(theta)

    Radius=(Mass*sqr(Velocity))/Force

    Period:=(2*pi*Radius)/Velocity

    Force=Mass*Acceleration


    3. The attempt at a solution

    First I find the maximum force on the proton due to its charge and velocity through the field:

    Force = q*v*B sin(theta)

    In this case the proton is travelling perpendicular to the field, so sin(theta) = 1

    and the magnitude of the force is simply:

    Force = q*v*B

    = 4.8065e-13 Newtons

    Now that is fine, and if I want the radius of the circle described by the proton in the field I can use:

    Radius=(Mass*sqr(Velocity))/Force or r=mv/qB

    = 0.125 meters or 125mm

    Which is particularly satisfying, because it relates such high speed, low mass and charge into a realistic spatial dimension.

    I can also find the time that it takes to complete one revolution:

    Period:=(2*pi*Radius)/Velocity

    =131.19e-9 seconds or 131 nanoseconds.


    But what I would like to find is the instantaneous position and direction after a given time, say 1 nanosecond.

    I tried dividing the circumference by 131 nanoseconds, but I feel like this is cheating. :frown:

    I also tried to use

    Acceleration = Force/mass, but in this case the acceleration is only a change in direction, not in velocity. :confused:
     
  2. jcsd
  3. Jun 30, 2011 #2

    ehild

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    Homework Helper
    Gold Member

    It is no cheating if you calculate the angular position as
    alpha = 2pi*(t/T). And you know that the proton moves along a circle with constant speed, so the velocity vector is tangent to the circle. Knowing alpha, you also get the direction of the velocity.

    ehild
     
  4. Jun 30, 2011 #3
    Thanks for your reply. You describe angular position, but I require cartesian coordinates in order to simulate this:

    I calculate alpha = 2pi*(t/T), using t = 1e-9 (1 nanosecond), and T (Period of one revolution) = 131.19e-9 seconds

    This gives me an alpha of 0.0478 (is that in Radians?)

    To attempt to convert that into an instantaneous position (my original problem),

    I use Y = Sin(alpha)*Radius and X = Cos(alpha)*Radius

    So, in in one nanosecond the proton is now at:

    X = 0.1258 meters
    Y = 0.00599 meters

    Is that correct?
     
  5. Jun 30, 2011 #4

    ehild

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    it is correct, but I got x=0.1248 m.

    ehild
     
  6. Jun 30, 2011 #5
    Ah, I think I see the error with my calculation.

    I am now using X = Cos(alpha)*(Velocity*1e-9)

    That is, if velocity is distance per second, and I require distance per nanosecond, hence(Velocity*1e-9). This delivers x and y values that describe a circle with the correct radius!

    Thanks!
     
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