A proton moving with a speed u in the x - y plane along the positive x axis enters at y = 0 in a region of uniform magnetic field B directed into the x - y plane as shown in figure . After sometime, the proton leaves the region with a speed v at co-ordinate y. Then:

 

$\overrightarrow{\mathrm{F}}=\mathrm{q}\left(\overrightarrow{\mathrm{u}}\times \overrightarrow{\mathrm{B}}\right)$. Since force F is perpendicular to u, it. does no work on the particle. Hence the speed of the proton remains unchanged, i.e. v = u. From Fleming's Left Hand rule, the force is directed upwards. Hence the proton, after completing a semicircle in the region of magnetic field leaves the magnetic region at positive y-coordinate.