JEE Main 2025 Apr 7 Shift 1, Physics Q17: Motion of a Charged Particle in a Magnetic Field

Circular motion in a perpendicular magnetic field gives

$\frac{m{v}^2}{r}=qvB\Rightarrow mv=qBr\Rightarrow p=qBr$

Kinetic energy in terms of momentum:

$E=\frac{p^2}{2m}=\frac{q^2{B}^2{r}^2}{2m}$

Rearranging:

$r^2=\frac{2m}{q^2{B}^2}E\Rightarrow r=\sqrt{\frac{2m}{q^2{B}^2}}\sqrt{E}\propto \sqrt{E}$

So $r$ grows as $\sqrt{E}$: a concave-down curve rising and gradually flattening. This matches option (4).

$\therefore$ Option (4).