 Solution

The Lorentz force is $d\mathbf{p}/dt=q\mathbf{v}\times \mathbf{B}$. Because it is perpendicular to the velocity, it changes the direction of momentum without changing its magnitude. For a charged particle in a uniform magnetic field, the radius of curvature is $R=p/(|q|B)$. Over the short track, the bending angle is $\varphi \simeq L/R$. Equivalently, integrate the force for a time $L/v$ while neglecting the small rotation in its direction. The magnitude of the acquired momentum component perpendicular to the incident direction isIts direction follows $q\mathbf{v}\times \mathbf{B}$; a signed component can therefore be written $qBL$ after fixing an oriented bending axis. Chord length and arc length agree at this order in $L/R$.