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Finding Magnetic Force

  1. Aug 6, 2015 #1
    1. The problem statement, all variables and given/known data
    An electron (charge = –1.6 x 10^–19 C) is moving at 3.0 x 10^5 m/s in the positive x direction. A magnetic field of 0.80 T is in the positive z direction. The magnetic force on the electron is:

    http://edugen.wileyplus.com/edugen/art2/common/pixel.gif

    a) 0 N

    b) 3.8 x 10^–14 N in the positive z direction

    c) 3.8 x 10^–14 N in the negative z direction

    d) 3.8 x 10^–14 N in the positive y direction

    e) 3.8 x 10^–14 N in the negative y direction

    2. Relevant equations
    F(b) =q(vxb)

    3. The attempt at a solution
    Force is 3.8x10^-14

    In terms of direction, I was told that it was the positive y direction, but isn't the charge negative so it should be the negative y direction?
     
  2. jcsd
  3. Aug 6, 2015 #2

    berkeman

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    Staff: Mentor

    I get it in the positive y direction, because X x Y = Z. Can you post a diagram?
     
  4. Aug 6, 2015 #3

    berkeman

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    Staff: Mentor

    BTW, why are all of the answers wrong?
     
  5. Aug 6, 2015 #4
    They are all wrong? I got the same force as the other force answers....

    I know X x Y = Z, but from my textbook "If q is negative, then the force and cross product have opposite signs and thus must be in opposite directions. "
    Shouldn't this then reverse it from positive y direction to negative y direction?
     
  6. Aug 6, 2015 #5

    TSny

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    But what is X x Z ?
     
  7. Aug 7, 2015 #6
    You mean qv x F? I am not sure I follow?
     
  8. Aug 7, 2015 #7
    F = –1.6 x 10^–19 (3.0 x 10^5 x0.8)
    F =-3.84 x 10^–14

    Admittedly I get a negative answer, but at least it matches numerically.
     
  9. Aug 7, 2015 #8
    I think you mean it's magnitude matches .

    The direction - assuming z - axis leaves the plane , how are you getting force in the -ve y direction ?
    Velocity vector's cross product with the magnetic field is in the -ve y direction . So , a -ve charge implies the opposite of this , i.e. , in the +ve y direction .

    Hope this helps .
     
  10. Aug 7, 2015 #9
    Again I would get the +ve y direction answer as you do, except the part in the textbook that confuses me is this: 'If q is negative, then the force and cross product have opposite signs and thus must be in opposite directions' q is negative isn't it? Thus the positive y becomes negative y. Oh boy how lost am I now...
     
  11. Aug 7, 2015 #10
    That's what I have said . v×B ( in vector form ) is towards -ve y direction , but electron charge is -ve , so force is in the +ve direction .
     
  12. Aug 7, 2015 #11
    Ah ok, I think I get that now.

    So am I correct in concluding firstly that F = 3.84 x 10^–14 or as berkeman hinted at, is that incorrect?
     
  13. Aug 7, 2015 #12
    Although you'll have to check the magnitude .
     
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