JEE Physics 2026 Question with Solution

Given: A current carrying solenoid is placed vertically. A particle of mass $m$ and charge $Q$ is released from rest and moves along the axis of the solenoid.

Find: The acceleration $a$ of the charged particle.

The magnetic force on a charged particle is

$$\vec{F}_m = Q(\vec{v} \times \vec{B})$$

Inside a long solenoid, the magnetic field $\vec{B}$ is along the axis of the solenoid. Since the solenoid is vertical, $\vec{B}$ is vertical.

Initially, the particle is released from rest, so $\vec{v} = 0$. Therefore,

$$\vec{F}_m = Q(0 \times \vec{B}) = 0$$

As the particle starts moving, its velocity remains along the axis of the solenoid, so $\vec{v}$ is parallel or anti-parallel to $\vec{B}$.

Hence, the magnitude of magnetic force is

$$F_m = |Q|vB\sin\theta$$

Here $\theta = 0^\circ$ or $180^\circ$, so

$$\sin\theta = 0$$

Therefore,

$$F_m = 0$$

So the only force acting on the particle is gravitational force:

$$F_g = mg$$

Using Newton's second law,

$$F_{\text{net}} = ma = mg$$

Thus,

$$a = g$$

Therefore, the acceleration of the charged particle is equal to $g$, so the correct option is D.