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Electron Moving in a Uniform Magnetic Field

  1. May 8, 2009 #1
    1. The problem statement, all variables and given/known data
    An electron that has a velocity with x component 2.3 x 106 m/s and y component 2.9 x 106 m/s moves through a uniform magnetic field with x component 0.040 T and y component -0.12 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.


    2. Relevant equations
    F = q vXB


    3. The attempt at a solution

    Since both magnetic field and velocit were given in component form I figured they could be combined into one (meaning one for velocity and one for magnetic field). After doing

    a = sqrt[ (ax)^2 + (ay)^2] for both V and B I got V =3.7 x10 ^6 m/s, angle = 5.158 degrees; B = 1.2649 x 10^-1, angle = -71.565.

    Then I added the two angles (I figured their sum is the total difference b/w the B and V vectors?) and plugged in:

    F = q v X B
    F= (1.6 x 10^-19)(3.7 x 10^6)(1.2649 x 10^-1)sin (-66.407)
    F = -6.862 x 10^-14 N

    For part b) it should be the same # as the charge of the proton and electron is the same (only sign is different and we use absolute values in that equation).

    I hope that my approach is ok, though, assuming my approach is ok I think I might have messed up the combining of the two angles. Any help/guidance would be greatly appreciated. Thank you in advance.

    P.S. I thought I should mention that no diagram is provided. Thanks again.
     
  2. jcsd
  3. May 8, 2009 #2

    dx

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    You got the angle wrong. Also, this is all much easier to do using the components. Both v and B are in the xy plane, so their cross product will have only a z-component. The z component of q(v x B) is just q(vxBy - vyBx).
     
  4. May 8, 2009 #3
    Hmm by using that, I get an answer of -6.27 x 10^-14, which is incorrect....Just out of curiosity, what would be the correct angle, had I not realized that components were easier? Thanks again, especially for that godly-quick response.
     
  5. May 8, 2009 #4

    dx

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    They asked for the magnitude, so you should remove the minus sign.
     
  6. May 8, 2009 #5

    dx

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    Assuming you calculated the angle of each of the vectors correctly, the angle between them would be 71.565 + 5.158 = 76.723. Just draw a picture and it should be obvious why.
     
  7. May 9, 2009 #6
    Duh! I can't believe I didn't see that! Also, the answer yielded with your method is correct. Thank you very much!
     
  8. May 9, 2009 #7

    dx

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    No problem.
     
  9. May 9, 2009 #8
    what about the component qE of the lorentz force?
     
  10. May 9, 2009 #9

    dx

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    E is zero here.
     
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