Past exam of the physics 2 paper of the University of Oxford / 2013 / 7

The upper positive plate gives a downward field $E=V/a$, while $\mathbf{B}$ points into the page. For a rightward electron, the electric force is upward and the magnetic force downward. Its initial speed isThese are equal to the precision intended by the rounded plate spacing: the arrangement is a velocity selector. Accordingly,It is not appropriate to calculate a large electric deflection and ignore magnetic bending. For completeness, if all printed values are treated literally, let $v_d=E/B$, $u=v_0-v_d$, and ${\omega }_c=eB/m_e$. The exact nonrelativistic crossed-field motion iswith $v_x=v_d+u\cos {\omega }_{c}t$ and $v_y=-u\sin {\omega }_{c}t$. At the exit solve $x(t_e)=0.50$ m, then use $\theta =atan2(v_y,v_x)$. The small rounding mismatch produces nanometre-scale oscillations, with $|\theta |\le {\sin }^{-1}(|u|/v_d)\simeq 6.32\times 10^{-4}$ rad. Its precise oscillatory exit angle is not robust to the supplied rounding.

In uniform fields, a rigid neutral electric dipole with both charges moving at the selected common velocity has zero net translational force. To leading nonrelativistic order its torque is $\mathbf{\tau }=\mathbf{p}\times (\mathbf{E}+\mathbf{v}\times \mathbf{B})$, which also vanishes at the velocity-selector condition. Thus it continues undeflected and, initially nonrotating, does not start rotating in the ideal balanced fields. An unbalanced effective field could exert a torque; the dipole orientation is not otherwise specified.