JEE Physics 2025 Question with Solution

Given: A particle of charge $q$ and mass $m$ enters a magnetic field perpendicular to its velocity and moves in a circle of radius $r$.

Find: The graph of $r$ versus kinetic energy $E$.

For circular motion in a magnetic field,

$$r = \frac{mv}{qB}$$

Also, kinetic energy is

$$E = \frac{1}{2}mv^2$$

So,

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

Substituting in the expression for $r$,

$$r = \frac{m}{qB}\sqrt{\frac{2E}{m}} = \frac{\sqrt{2mE}}{qB}$$

Therefore,

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

Hence $r$ increases with $E$, but the rate of increase gradually decreases. So the required curve is option (4).

Therefore, the correct option is D.

Given: Magnetic force provides the centripetal force for the charged particle.

Find: The correct $r$-versus-$E$ graph.

Using magnetic force equals centripetal force,

$$\frac{mv^2}{r} = qvB$$

This gives

$$mv = qBr$$

Now,

$$E = \frac{1}{2}mv^2$$

Using $v = \frac{qBr}{m}$,

$$E = \frac{1}{2}m\left(\frac{qBr}{m}\right)^2$$ $$E = \frac{q^2B^2r^2}{2m}$$

So,

$$r^2 \propto E$$

which means

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

Thus the graph is an increasing curve concave downward, matching option (4).

Therefore, the correct option is D.