A particle is projected into a plane perpendicular to a uniform...

When a charged particle is projected perpendicular to a uniform magnetic field, it moves in a circular path of radius $r = \dfrac{mv}{qB}$.

The area enclosed by this circular path is $A = \pi r^2 = \pi \left( \dfrac{mv}{qB} \right)^2 = \dfrac{\pi m^2 v^2}{q^2 B^2}$.

The kinetic energy of the particle is $K = \dfrac{1}{2} m v^2$, which gives $m^2 v^2 = 2mK$.

Substituting this in the expression for area, we get:

$A = \dfrac{2\pi m K}{q^2 B^2}$

Since $m$, $q$, and $B$ are constants, $A \propto K$.

Thus, the area enclosed by the path is directly proportional to the kinetic energy of the particle.