1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: B field question.
  1. A question on a<b (Replies: 82)

  2. B-field phase question (Replies: 1)

  3. B field question? (Replies: 5)

Loading...