A particle of charge q , mass m and kinetic energy E enters in magnetic field perpendicular to its velocity and undergoes a circular arc of radius (r). Which of the following curves represents the variation of r with E ?

Magnetics PYQ — JEE Main 2025 · 7 Apr · Shift 1 · Q57 with Solution

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Correct answer: A

When a charged particle enters a uniform magnetic field perpendicular to its velocity, it moves in a circular path.

The magnetic force is $F_B = qvB$ and this acts as the centripetal force: $qvB = \frac{mv^2}{r}$

From $qvB = \frac{mv^2}{r}$ we get $r = \frac{mv}{qB}$

So, $r \propto v$

Given kinetic energy $E = \frac{1}{2}mv^2$

Therefore, $v = \sqrt{\frac{2E}{m}}$

Substitute into the expression for $r$ : $r = \frac{m}{qB}\sqrt{\frac{2E}{m}}$

Simplifying, $r = \frac{\sqrt{2mE}}{qB}$

Thus, $r \propto \sqrt{E}$

The graph of $r$ versus $E$ is of the form $r = k\sqrt{E}$ where $k$ is a constant.

This is an increasing curve starting from the origin, with decreasing slope as $E$ increases.

So the correct curve is the square-root type curve.

The stored correct answer is A. Since option A must represent $r \propto \sqrt{E}$ , the derived answer agrees with the stored answer.