When a conductor moves through an electric field, charge carriers inside it experience the Lorentz force, which takes the form $\mathbf{F} = q \mathbf{v} \times \mathbf{B}$. As the positive and negative charges separate, they will also attract each other due to the electric force $q\mathbf{E}$. Since the forces are pointing in opposite directions, one could show that $\mathbf{E} = -\mathbf{v} \times \mathbf{B}$ (see image below). However, once everything is in equilibrium, the sum of the forces is zero, and so $q\mathbf{v} \times \mathbf{B} - q\mathbf{E} = 0$, which would imply that $\mathbf{E} = \mathbf{v} \times \mathbf{B}$. What is the correct relationship between the electric field and $\mathbf{v} \times \mathbf{B}$?

enter image description here

Image Source: IEEE Transactions on Magnetics, Vol. 38, No. 2, p. 1334


1 Answer 1


The correct relationship is $\mathbf{E} = -\mathbf{v} \times \mathbf{B}.$

The total force on a "proton" (a positive hole) in the rod is given by

$$ \mathbf{F} = \mathbf{F}_{\rm electric} + \mathbf{F}_{\rm magnetic} = q\mathbf{E} + q\mathbf{v}\times\mathbf{B}.$$

They are equilibrium, so the total force is zero, and thus

$$ q\mathbf{E} + q\mathbf{v}\times\mathbf{B} = 0 $$ $$ \mathbf{E} + \mathbf{v}\times\mathbf{B} = 0 $$ $$ \mathbf{E} = -\mathbf{v} \times \mathbf{B}. $$

  • $\begingroup$ To follow this up: Suppose the direction of the velocity in the image were flipped. Now the top is positive and the bottom is negative. If the polarization is related to the electric field by $\mathbf{P} = \epsilon_0 \mathbf{E}$ and surface charge density is given by $\sigma = \mathbf{P} \cdot \mathbf{n}$ where $\mathbf{n}$ is the surface normal. Thus the E field (and polarization) would point down. Thus $\mathbf{P} \cdot \mathbf{n}$ at the top of the conductor would be negative, even though the top of the conductor is positive. How is this explained? $\endgroup$ Commented May 11, 2018 at 2:29
  • $\begingroup$ @BillyKalfus Polarization density is only relevant in dielectrics - you are talking about a conductor. $\endgroup$
    – J. Murray
    Commented May 11, 2018 at 2:54
  • $\begingroup$ Hm, okay - then how could I find the surface charge densities at the top and bottom of the conductor? $\endgroup$ Commented May 11, 2018 at 2:55
  • $\begingroup$ @BillyKalfus If you can calculate the electric field immediately outside the surface of the conductor, then $\sigma = \epsilon_0 |\vec E|$ in accordance with Gauss' law. This is not a trivial task, though - surface charges will accumulate on the sides of the conductor, not just at the top and bottom, and the resulting electric field is quite complicated. $\endgroup$
    – J. Murray
    Commented May 11, 2018 at 3:05

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