An ideal solenoid is kept with its axis vertical. Current I₀ is flowing in the solenoid. A charge Q is thrown downward inside the solenoid at acceleration of the charge particle is then

To determine the acceleration of the charged particle, let us analyze the forces acting on it when it is thrown downward inside the solenoid.

First, we consider the magnetic field inside an ideal solenoid. An ideal solenoid has a uniform magnetic field directed parallel to its central axis. Since the axis of the solenoid is vertical, the magnetic field $B$ inside the solenoid is also directed vertically (either vertically upward or vertically downward, depending on the direction of the current $I_0$).

Second, we analyze the magnetic force acting on the moving charged particle. The magnetic force $F_m$ on a charge $Q$ moving with velocity $v$ in a magnetic field $B$ is given by the Lorentz force formula:
$F_m=Q(v\times B)$
The magnitude of this force is:
$F_m=\left|Q\right|vB\sin \left(\theta \right)$
where $\theta$ is the angle between the velocity vector of the charge and the magnetic field vector.

In this scenario, the charge is thrown vertically downward inside the solenoid. Since the magnetic field lines are also vertical (parallel to the axis), the velocity vector $v$ is parallel or antiparallel to the magnetic field vector $B$. Therefore, the angle $\theta$ between them is either 0 degrees or 180 degrees.

Since $\sin \left(0\right)=0$ and $\sin \left(180\right)=0$, the magnetic force acting on the charge is:
$F_m=0$

Because the magnetic force is zero, the only force acting on the particle in the vertical direction is the gravitational force:
$F_g=mg$
where $m$ is the mass of the particle and $g$ is the acceleration due to gravity.

Using Newton's second law, the acceleration $a$ of the particle is:
$a=\frac{F_g}{m}=\frac{mg}{m}=g$

Thus, the acceleration of the charged particle is equal to the acceleration due to gravity, $g$.

Therefore, the correct option is a = g.