Past exam of the physics 2 paper of the University of Oxford / 2021 resit / 3

Let $U>0$ be the drop in electric potential along the positive electric charge's path. By conservation of energy, $qU=m{v}^2/2$, so the nonrelativistic exit speed is $v=\sqrt{2qU/m}$. Apply a uniform magnetic field perpendicular to the incident velocity and to the desired orbital plane. The magnetic Lorentz force is perpendicular to the velocity, does no work, and supplies the centripetal force:For example, $\mathbf{v}=v\hat{\mathbf{x}}$ and $\mathbf{B}=B\hat{\mathbf{z}}$ give initial deflection along $-y$ for positive $q$. Reversing the magnetic field reverses the bending. A uniform electric field alone would give constant acceleration and a parabolic path, not uniform circular motion. The displayed result assumes $qU\ll m{c}^2$; the exact relativistic momentum relation would give $R=\sqrt{q^2{U}^2/c^2+2mqU}/(qB)$.

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.