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B field question.

  1. Apr 3, 2012 #1
    If i have an electric dipole in a B field, it wont experience a torque right.
    Because the electric dipole is not moving. So it should just stay put.
    Just want to check my understanding.
  2. jcsd
  3. Apr 4, 2012 #2


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    As long as there is no electric field in the system of the dipole (and ignoring stuff like gravity and so on), it does not feel any force/torque.
  4. Apr 4, 2012 #3
    ok thanks for your answer, but if the dipole was moving then it would experience a torque.
  5. Apr 4, 2012 #4
    It depends on the way in which it moves. The dipole would only experience a torque on it if it were rotating with respect to the B field. If the dipole was moving without rotation, it would not experience a torque but would experience a force on its center of mass. If it was only rotating, it would feel a torque but not a force on its center of mass.
  6. Apr 5, 2012 #5


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    Are you sure about that? I got confused by trying to combine B,v,d in my mind, so I calculated it:

    Represent the dipole by a positive charge q at position [itex]a=(a_x,a_y,a_z)[/itex] and a negative charge -q at position -a.

    A velocity [itex]\vec{v}[/itex] of the dipole generates the total force [itex]\vec{F}=\vec{F_q}+\vec{F_{-q}}=q(\vec{v} \times \vec{B})+(-q)(\vec{v} \times \vec{B})=0[/itex].
    The torque based on this movement is [itex]\vec{M}=\vec{a} \times \vec{F_q} - \vec{a} \times \vec{F_{-q}}=2q \vec{a} \times(\vec{v} \times \vec{B})[/itex]
    Another way to see this is to transform the problem in the system of the dipole: The magnetic field gets an added electric component, which can generate a torque but not a net force.

    Now, let the dipole rotate with angular velocity [itex]\vec{\omega}[/itex]. The positive charge then moves with [itex]\vec{\omega} \times \vec{a}[/itex] and the negative charge with the negative value of that.

    Therefore, the total force is [itex]\vec{F}=\vec{F_q}+\vec{F_{-q}}=q\left((\vec{\omega} \times \vec{a}) \times \vec{B}\right)+(-q)\left(((\vec{\omega} \times -\vec{a}) \times \vec{B}\right)=2q \left((\vec{\omega} \times \vec{a}) \times \vec{B}\right)[/itex]
    The torque is [itex]\vec{M}=\vec{a} \times \vec{F_q} - \vec{a} \times \vec{F_{-q}}=0[/itex] by symmetry as [itex]\vec{F_q}=\vec{F_{-q}}[/itex].

    Replacing 2qa by the dipole moment d and allowing both movement and rotation at the same time finally gives the general formulas:

    [tex]\vec{F}=\left((\vec{\omega} \times \vec{d}) \times \vec{B}\right)[/tex]
    [tex]\vec{M}= \vec{d} \times (\vec{v} \times \vec{B})[/tex]

    I wonder how the solutions of these equations (together with J d/dt omega = M and m d/dt v = F) look like.
  7. Apr 5, 2012 #6
    You are correct. I somehow was thinking about a dipole with the same signed charge on both ends (that's last time I comment before my coffee).
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