If you move your apparatus in that way, then what is a uniform magnetic field in the frame of reference of the copper plate becomes both a magnetic and an electric field in the "laboratory" frame of reference. e.g. See [here][1].

The observed electric field would be 
 $$\vec{E}' = \gamma \vec{v} \times \vec{B}\ ,$$
where $\gamma$ is the usual Lorentz factor.

The observed magnetic field in the laboratory frame would be
$$ \vec{B}' = \gamma \vec{B}\ . $$

In the rest frame of the plate, the net force on a charged particle is zero, since the particle would be at rest with respect to the uniform B-field and $\vec{E}=0$.

In the laboratory frame, the force is the sum of that due to $\vec{E}'$ and that due to the motion of the charged particles, moving at $-\vec{v}$ with respect to $\vec{B}'$. Thus
$$ \vec{F}' = q(\vec{E}' + (-\vec{v})\times \vec{B'}) = q( \gamma \vec{v} \times \vec{B} - \gamma \vec{v}\times \vec{B}) = 0\ . $$

There would be no separation of the charged particles in either frame of reference.

  [1]: https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#The_E_and_B_fields