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Magnetic Field of a Moving Point charge

  1. Mar 17, 2012 #1
    1. The problem statement, all variables and given/known data
    A point charge of q1=3.6nC is moving with speed 4.5 x 107m/s parallel to the y-axis along the line x=3m. The magnetic field produced by this charge at the origin when it is at the point x=3m, y=4m is approximately:



    2. Relevant equations
    vector B= (magnetic constant/4pi) (q(vector v X unit vector r)/r2)
    magnetic constant= 4pi x 10-7 (Tesla x meter)/Ampere

    3. The attempt at a solution
    The problem here is a curious cross product outcome.
    vector v= 4.5 x 107j = <0, 4.5 x 107, 0>
    vector r= <3-0, 4-0, 0-0>= <3,4,0>
    magnitude of r= sqrt(32+42)= 5
    unit vector r= vector r/ magnitude of vector r= <3,4,0>/5

    [itex]\vec{v}[/itex]X[itex]\hat{r}[/itex]= <0, 4.5 x 107, 0> X <3, 4, 0> =

    ((4.5 x 107(0) - 0(4))i - (0(0) - 0(3))j + (0(4) - 4.5 x 107(3))k)/5 =

    (-13.5 x 107j)/5= <0, 0, -13.5 x 107>/5

    Wouldn't that imply that the cross product vector is in the -Z direction? However, using right hand rules, shouldn't the cross product vector be in the positive Z direction?
     

    Attached Files:

    Last edited: Mar 17, 2012
  2. jcsd
  3. Mar 17, 2012 #2

    Redbelly98

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    Welcome to PF.

    In the Biot Savart Law, the direction of r is from the current element or charge to the point where B is being calculated. So r is <-3,-4,0> in this case.

    p.s. I presume the charge is moving in the +y direction?
     
  4. Mar 17, 2012 #3
    Ahhhhh. So I had it reversed.
    Yes, the charge is moving in the +y direction.

    So the cross product would now be: <0, 4.5 x 107, 0>m/s X <-3, -4, 0>/5 =
    (0--(3/5)(4.5 x 107)) k= +2.7 x 107m/s k

    Then vector B= (magnetic constant/ 4pi) (3.6 x 10-9C)(2.7 x 107m/s k) / (25m2)= 3.89 x 10-10 T k. Which is equal to 0.39nT k

    Thank you for your assistance :)
     
    Last edited: Mar 17, 2012
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