I came across a problem which is as follows:

A bar of thickness $d$ is made to move with a velocity v perpendicular to a uniform magnetic field of induction $B$. Top and bottom faces of the bar are parallel to the lines of magnetic field.


(a) Find surface charge densities on the upper and the bottom faces of the bar, assuming material of the bar to be a perfect conductor.

(b) Find the surface charge densities on the upper and the bottom faces of the bar, assuming material of the bar to be a dielectric of dielectric constant $K$.

Part (a) is pretty simple. I just equated the force on the moving electron due to magnetic field and the electric field produced due to the charges induced on the plates as follows $$qvB=qE$$ where $E$ is the electric field due to the induced charges

as $E=\sigma/\epsilon$. I get the charge density on plates as $\epsilon Bv$.

My Doubts:

It is part (b) that I cannot understand. My doubts regarding it are :

Doubt 1) The dielectric constant gives the relation between the external electric field and the field induced due to the charges of the dielectric. But in this case there is no external field, so how to actually use the dielectric constant to find electric field in a dielectric .

Doubt 2) Also secondly I don’t understand how the force on an electron will be affected due to the material being a dielectric.I thought that the same amount of charge as in the case of conductor should be produced to equalise the force of magnetic field but that surely isn’t the case. So what actually is happening?


1 Answer 1


One way to look at it is to do a Lorentz boost with $-\vec v$, then:

$$\vec B'=\gamma \vec B$$ $$ \vec E'=\gamma(-\vec v \times \vec B)$$

so $E'=\gamma v B$, which differs from your (non-relativistic) results by:

$$\gamma=\frac 1{\sqrt{1-\frac{v^2}{c^2}}}$$

In this frame, the block is at rest in a magnetic field and a perpendicular electric field. For the conductor, that will give a surface charge density satisfying:



$$\sigma'=\epsilon_0E'=\epsilon_0\gamma v B$$

Now boost that back to the frame of the picture, where the surface charge is invariant, but the surface area contracts by the Lorentz factor:

$$\sigma=\gamma\sigma'=\epsilon_0\gamma^2 v B$$

The same reasoning can be applied to (b).


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