- 2,802
- 605
In the wikipedia page for guiding center,the following lines are written about curvature drift of charged particles.
My problem is the part that tells curl of B is zero in a vacuum.
Although I know Maxwell equations permit such a situation(with [itex]\vec{B}=\vec{B}(\mbox{only space variables}) \ and \ \vec{E}=0[/itex]),I don't understand how that can happen.
Obviously a current has caused the existence of the magnetic field and so there should be a changing electric field.But there is no electric field and that seems strange to me.
I'll appreciate any ideas.
Thanks
The expression for the grad-B drift above can be rewritten for the case when [itex]\nabla B[/itex] is due to the curvature. This is most easily done by realizing that in a vacuum, Ampere's Law is [itex]\nabla\times\vec{B} = 0[/itex]. In cylindrical coordinates chosen such that the azimuthal direction is parallel to the magnetic field and the radial direction is parallel to the gradient of the field, this becomes
[itex] \nabla\times\vec{B} = \frac{1}{r} \frac{\partial}{\partial r} \left( r B_\theta \right) \hat{z} = 0 [/itex]
Since[itex]r B_\theta[/itex]is a constant, this implies that
[itex] \nabla B = - B \frac{\vec{R}_c}{R_c^2} [/itex]
and the grad-B drift velocity can be written
[itex] \vec{v}_{\nabla B} = -\frac{\epsilon_\perp}{q} \frac{\vec{B}\times \vec{R}_c}{R_c^2 B^2} [/itex]
My problem is the part that tells curl of B is zero in a vacuum.
Although I know Maxwell equations permit such a situation(with [itex]\vec{B}=\vec{B}(\mbox{only space variables}) \ and \ \vec{E}=0[/itex]),I don't understand how that can happen.
Obviously a current has caused the existence of the magnetic field and so there should be a changing electric field.But there is no electric field and that seems strange to me.
I'll appreciate any ideas.
Thanks