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Lorentz Law field equation applied in a field

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    An electron is travelling at 5.0 × 106 m s−1 directly into the page at the point P shown in Figure 3. P is 4.0 cm from O. Again, D = 10 cm and the current i in each wire has magnitude of 1.0 A with the directions that you found in part (c). Find the magnitude and direction of the force that the electron experiences.

    2. Relevant equations
    F = q(E(r) + v x B(r) (The Lorentz law field equation)
    E(r) = Q / 4πε r^2 (The electric field)
    B(r) = μ i / 2π r (The magnetic field)

    3. The attempt at a solution
    I have i = 1.0A, v = 5.0E6 ms^-1, q = -1.6E-19 C. I do have an answer (but I'm not sure of two things:

    (i) Should I be using the equation: F = q(E(r) + v x B(r)) ? or the more straight forward F = qvB(r)

    (ii) Should I be looking at the resultant force from both conductors left and right.

    Attached Files:

    Last edited: Feb 15, 2012
  2. jcsd
  3. Feb 15, 2012 #2
    It sounds like you are very close:

    The more straightforward F = qvB(r) is true when you don't have an electric field and the velocity is perpendicular to the magnetic field, however you must be careful in the direction as there are some minus signs in a cross product like that.

    Rather than splitting up the force from both conductors (which would be possible), you may find it easier to consider three components. For instance if you only consider forces forces along the axis joining the two wire, then because the electron has v=0 in this direction the magnetic fields do not contribute to the force, but you still need to consider the resulting electric field from the wires.

    This should flow easily enough especially if you write your force equation with vectors (just define axis if they have not done so for you)
  4. Feb 15, 2012 #3
    It looks like as the velocity points into the page (+z), the magnetic field points down (-y) then the force points right (+x) axis using the right hand rule... Perpendicular to both and because the electron is negatively charged.

    As for the magnitude I'm not really sure I understand what you mean by...the magnetic field do not contribute to the force... I have a magnitude of about 5 nN does his sound about right?
    Last edited: Feb 15, 2012
  5. Feb 17, 2012 #4
    It looks as though there is no electric field so F = qvB is going to be the way ahead.
  6. Feb 17, 2012 #5
    You may use that F=qVB when there is no electric field and the magnetic field is perpendicular to the direction of motion. To understand this consider that

    [itex] \overline{v} \times \overline{B}=|\overline{v}||\overline{B}| \sin( \theta) \hat{n}[/itex]

    Since the magnetic field will be a loops in the x-y plane (if the electron travels into the page which is the z direction). Therefore the angle between them [itex] \theta [/itex] will be 90° and you may simply use F=qvB (so long as E=0), then work out the direction of [itex] \hat{n}[/itex].

    However this is NOT the only force it will experience since the two wire's hold a current. There will be an electric field which the net force will be zero half way between them, however you are not half way in between so there will be an overall electric field that contributes to the force as well.
  7. Feb 17, 2012 #6
    I take it you are referring to Gauss Law? Not just E(r) = Q / 4πε r^2
  8. Feb 17, 2012 #7
  9. Feb 17, 2012 #8
    Thanks very much for the input! I've calculated both F = q(E(r) + v x B(r)) and also F =qvB and the results show virtually the same answer (a difference of 0.0000000034) proving that the electric field has a negligible effect on the outcome.
  10. Feb 17, 2012 #9
    Well done, that is fine remember to include the direction as well as the magnitude!

    All the best
  11. Feb 17, 2012 #10
    Thank you for your help!
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