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Question about forces due to magnetism

  1. Jun 12, 2015 #1

    DCN

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    The force on a particle moving through a magnetic field is given by:
    $$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    F} = q\,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} \times \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    B} $$
    Is the velocity ##\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} ## relative to the source of magnetic field or some third party observer?
     
  2. jcsd
  3. Jun 12, 2015 #2

    Dale

    Staff: Mentor

    The velocity is relative to any inertial reference frame. It is not necessarily relative to the source, particularly if the source is moving non inertially.
     
  4. Jun 13, 2015 #3

    DCN

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    If that's the case, then please explain this.

    Let's say that there is a test charge ##q## whose position from the origin is given by the radial vector ##\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    r} ## and there is a source charge ##Q## at the origin moving to the right along the positive ##x##-axis with a constant velocity ##{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over v} _s}## with respect to ##q##.

    Now, in the rest frame of ##q##, ##q## has a velocity of magnitude ##0##, so it will not feel any magnetic effects. The force it will feel will be solely from the electric field generated by ##Q##. The force in the rest frame of ##q## is then:
    $$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    F} = {{Qq} \over {4\pi {\varepsilon _0}}}{{\hat r} \over {{{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    r} } \right\|}^2}}}$$
    If we consider a difference reference frame in which we are moving to the right with a velocity of ##{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} _o}##, then ##q## will have a velocity of ## - {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} _o}## and ##Q## will have a velocity of ##{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} _s} - {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} _o}##. The magnetic field then produced by ##Q## is:
    $$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    B} = {{{\mu _0}Q} \over {4\pi }}{{\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} }_s} - {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} }_o}} \right) \times \hat r} \over {{{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    r} } \right\|}^2}}}$$
    The force on ##q## due to the magnetic field will then be:
    $$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    F} = - q\,{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} }_o} \times \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    B} $$
    $$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    F} = - {{{\mu _0}Qq} \over {4\pi {{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    r} } \right\|}^2}}}\,{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} }_o} \times \left( {\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} }_s} - {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} }_o}} \right) \times \hat r} \right)$$
    So the total force the particle will experience in the moving reference frame is:
    $${{Qq} \over {4\pi {\varepsilon _0}}}{{\hat r} \over {{{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    r} } \right\|}^2}}} - {{{\mu _0}Qq} \over {4\pi {{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    r} } \right\|}^2}}}\,{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} }_o} \times \left( {\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} }_s} - {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
    v} }_o}} \right) \times \hat r} \right)$$
    Which is clearly not equal to the force found in the stationary reference frame. Am I missing something here?
     
  5. Jun 13, 2015 #4

    Vanadium 50

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    Staff Emeritus
    Science Advisor
    Education Advisor

    Yes. The fields have to transform as well when you change frames.
     
  6. Jun 13, 2015 #5

    DCN

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    How does that work?
     
  7. Jun 13, 2015 #6

    Vanadium 50

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    Staff Emeritus
    Science Advisor
    Education Advisor

    I don't understand the question. There is a Lorentz transformation for electromagnetic fields. You change frames, you need to use it to determine how the fields transform.
     
  8. Jun 14, 2015 #7

    jtbell

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    Staff: Mentor

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