Electric Dipole in B Field: No Torque

[cragar]

If i have an electric dipole in a B field, it won't experience a torque right.
Because the electric dipole is not moving. So it should just stay put.
Just want to check my understanding.

 

[mfb]

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.

 

[cragar]

ok thanks for your answer, but if the dipole was moving then it would experience a torque.

 

[Jasso]

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.

 

[mfb]

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 a=(a_x,a_y,a_z) and a negative charge -q at position -a.

A velocity \vec{v} of the dipole generates the total force \vec{F}=\vec{F_q}+\vec{F_{-q}}=q(\vec{v} \times \vec{B})+(-q)(\vec{v} \times \vec{B})=0.
The torque based on this movement is \vec{M}=\vec{a} \times \vec{F_q} - \vec{a} \times \vec{F_{-q}}=2q \vec{a} \times(\vec{v} \times \vec{B})
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 \vec{\omega}. The positive charge then moves with \vec{\omega} \times \vec{a} and the negative charge with the negative value of that.

Therefore, the total force is \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)
The torque is \vec{M}=\vec{a} \times \vec{F_q} - \vec{a} \times \vec{F_{-q}}=0 by symmetry as \vec{F_q}=\vec{F_{-q}}.

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

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

I wonder how the solutions of these equations (together with J d/dt omega = M and m d/dt v = F) look like.

 

[Jasso]

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 a=(a_x,a_y,a_z) and a negative charge -q at position -a.

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).