= Solution
Choose the entry point as $(0,0)$, the initial <velocity> as $v\hat{\mathbf x}$, and the slit as $(d,a)$, with $a,d>0$. A <magnetic field> perpendicular to the page changes the direction of the <velocity>, while the <Lorentz force> does no <work>. For positive $q$, an upward initial <force> requires the <magnetic flux density> to point into the page. The resulting <circular motion> has centre $(0,r)$, since its initial tangent is horizontal.
The slit must lie on the <circle>, giving $d^2+(a-r)^2=r^2$ and hence $r=(d^2+a^2)/(2a)$. Equating the <centripetal force> to the magnetic part of the <Lorentz force>, $mv^2/r=|q|vB$, gives
$$
\boxed{|B|=\frac{2mva}{|q|(d^2+a^2)},\qquad \mathbf B=-\frac{2mva}{q(d^2+a^2)}\hat{\mathbf z}.}
$$
A negative <electric charge> requires the opposite direction of <magnetic flux density>. These expressions assume nonrelativistic <motion> and negligible energy loss through <electromagnetic radiation>.
For an opaque screen, reaching the slit must be the \i[first] encounter with $x=d$. The <velocity> there has $v_x/v=(d^2-a^2)/(d^2+a^2)$, so passage from left to right requires $a<d$, as in the drawn arrangement. At $a=d$ the <trajectory> merely touches the screen. If $a>d$, the same <circle> first meets the screen at height $d^2/a<a$, where the particle is blocked; its later passage through $(d,a)$ does not solve the opaque-screen problem. This qualification distinguishes a geometric <circle> through two points from an accessible slit <trajectory>.
Back to article page
