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Homework Help: The magnetic field of moving point charges

  1. Apr 9, 2010 #1
    1. The problem statement, all variables and given/known data

    A proton has a velocity of 1.0 * 10^2 m/s in the x direction + 2.0*10^2 m/s in the y direction and is located in the z=0 plane at x=3.0m, y=4.0m at some time t=T. Find the magnetic field in z=0 plane at x=2.0m, y=4.0m.

    2. Relevant equations

    magnetic field is given by: B= (mu/4pi)(qv*R/r^2)

    where mu= magnetic constant= 4pi*10^-7 N/A^2
    pi= pie= 3.14
    q= charge
    v= velocity
    R= unit vector that points to the field point P from the charge q

    3. The attempt at a solution

    I dont understand how to do this question, since we are not given charge! Also, we cant use the biot-savart law, dB= (mu/4pi)(I dl*R/r^2) since we do not have current given either (I).

    Please help, and in detail so I can understand, preferrably by tomorrow since its due then. Thank-you
  2. jcsd
  3. Apr 9, 2010 #2
    I think you are expected to develop the formula for the magnetic field due to a point charge in uniform motion.

    The elementary charge, (+e for a proton, -e for an electron) is a number you should remember, as it comes up often, and is the conversion factor from Joules to eV.

    [tex]e = 1.6\cdot 10^{-19} C[/tex]
  4. Apr 9, 2010 #3
    Yes, that makes sense! But how would I go about doing this, since the velocity is in both the x and y direction? I think I have to split it up into a parallel and perpendicular component. However, if I did that, what do I do for the R and r components? I am a bit lost with this one. would r= sqrt(2^2 + 2^2) ? and R= 2m in x dir - 2m in y dir,

    then R= R/r = (2m in x dir - 2m in y dir)/ (2sqrt(2))
    R= (sqrt(2) in x dir) - (sqrt(2) in y dir)

    therefore: B= 10^(-7)*(v)(R)/ 8^2

    I am still lost as to what to do with v and R in the last steps. Can you help with this too?
  5. Apr 9, 2010 #4
    sorry r^2 in the equation would be= (sqrt(8))^2 which equals 8
  6. Apr 9, 2010 #5
    The formula you posted for the magnetic field of a point charge is wrong in terms of direction, though it provides the correct magnitude (For low velocity charges, like the one in question)

    The one way I found to get the correct formula is to look at the charge in its rest frame, where it only produces a simple static electric field and no magnetic field, and to then transform that field into the frame moving with velocity -v relative to the rest frame.

    That will give you the EM field in the lab frame in which the particle is moving with velocity +v.

    The formulas are completely relativistic, so I'd like to know if you've learned the subject this way, or if you were simply given the formula you posted, since it seems pretty incorrect to me and going into the transform if you haven't studied it won't be too useful. The B field produced is perpendicular to the radius vector, not along it!
    Last edited: Apr 9, 2010
  7. Apr 9, 2010 #6
    No, i was just given this formula. However, in another question similar to this one except that the velocity is strictly horizontal, this formula is used to determine the magnetic field. I dont really understand your method though. Are you saying to use the eqn for the electric field?
  8. Apr 9, 2010 #7
    No no, nevermind. :)
    The simple formula is quite enough here, but it is missing a crucial component, and that is its direction.

    The direction of the magnetic field produced by a charge in uniform motion is [tex]\hat v \times \hat r[/tex] where [tex]\hat v[/tex] is the unit vector in the direction of motion of the particle producing the field, and [tex]\hat r[/tex] is the unit vector from the point charge to the point of interest.

    Finding the magnitude is straightforward. All you need is the distance, [tex]r[/tex] between the point charge and the point of interest at t=T. As for the direction, you need to understand what [tex]\hat v \times \hat r[/tex] is! Draw a 3d diagram first and look at what's perpendicular to what. Remember that the cross product of two vectors is perpendicular to both those vectors!

    An easier way to picture it is to look at a point charge moving in the +x direction only, and to then focus in on the yz plane, the B field should be contained completely in that plane, and be perpendicular to the [tex]r[/tex] vector.

    Now that you have that in hand, try and find the direction and magnitude of the field.
  9. Apr 9, 2010 #8
    Well first, I dont understand what to do with the "located in the z=0 plane at x=3 m, y= 4m" part of the question. Do I use that as the starting point, then point the magnetic field down 2 and to the side 2 since they say the magnetic field is located at x=2m, y=2m?
  10. Apr 9, 2010 #9
    You are given the position of the charge at t=T. You know that the magnetic field relative to the instantaneous position of the charge is given by:

    [tex]\vec B = \frac{\mu_0}{4\pi}\frac{qv}{r^2}(\hat v \times \hat r)[/tex]

    Find [tex]r[/tex] and [tex]\hat v \times \hat r[/tex] to find all you need to know about the field.
  11. Apr 9, 2010 #10
    But how do I find v*r?
    I handed in my assignment, however, I would still like to know how to do this. Thank-you, by the way, for your help. appreciate it!
  12. Apr 10, 2010 #11
    You know what [tex]\hat v[/tex] is, by definition it is [tex]\frac{\vec v}{|v|}[/tex], and [tex]\hat r = \frac{\vec r}{|r|}[/tex].

    Now all you need to do is take the cross product between the two:
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