A

To derive the time period of the revolution of a charged particle moving in a magnetic field, we start by considering the forces acting on the charge.

Step 1: The magnetic force acting on a charged particle moving in a magnetic field is given by the equation:
$$ F = qvB \sin \theta $$
Since the motion is perpendicular to the magnetic field, \( \theta = 90^\circ \) and \( \sin \theta = 1 \), therefore:
$$ F = qvB $$

Step 2: For the charged particle to move in a circle, this magnetic force must provide the centripetal force required for circular motion. The centripetal force is given by:
$$ F_{c} = \frac{mv^2}{r} $$
where \( m \) is the mass of the particle, \( v \) is its velocity, and \( r \) is the radius of the circular path.

Step 3: Setting the magnetic force equal to the centripetal force gives us:
$$ qvB = \frac{mv^2}{r} $$
Rearranging this equation to find the radius \( r \) leads to:
$$ r = \frac{mv}{qB} $$

Step 4: The time period \( T \) of revolution is related to the circumference of the circle and the velocity of the particle:
$$ T = \frac{2\pi r}{v} $$
Substituting the expression for \( r \) into this equation, we get:
$$ T = \frac{2\pi \left(\frac{mv}{qB}\right)}{v} = \frac{2\pi m}{qB} $$

Step 5: Examining this final expression for the time period, we see that \( T \) depends on the mass of the particle \( m \), the charge of the particle \( q \), and the magnetic field strength \( B \). However, it is independent of the velocity \( v \).

Hence, the time period of revolution is independent of the magnetic field strength.
Therefore, the correct answer is A.