 Solution

Add a uniform electric field satisfying $\mathbf{E}=-{\mathbf{v}}_0\times \mathbf{B}$. Then the Lorentz force $q(\mathbf{E}+{\mathbf{v}}_0\times \mathbf{B})$ vanishes for the desired velocity. With the axes used above, $\mathbf{E}=E\hat{\mathbf{y}}$, perpendicular to both ${\mathbf{v}}_0$ and $\mathbf{B}$, withThe crossed fields form a velocity selector. All particles must first have traversed the same electric potential difference from rest: that is what converts selection of speed into selection of charge-to-mass ratio. Equal charge-to-mass ratios yield equal speeds even if the separate masses and charges differ. For other ratios, the initial transverse force is $q(E-vB)$ and is nonzero.