1. Magnetic force on a moving charge

   For a charged particle moving in a magnetic field, $\vec F = q(\vec v \times \vec B).$

   Hence, the magnetic force is always perpendicular to both $\vec v$ and $\vec B$.

2. Check the Reason (R)

   The statement says: Moving charged particle experiences magnetic force perpendicular to its direction of motion.

   This is true, because from $\vec F = q(\vec v \times \vec B),$ we get $\vec F \perp \vec v.$

   So, (R) is true.

3. Effect on speed and energy

   Work done by a force is related to the component of force along displacement. Since magnetic force is always perpendicular to velocity, it does no work: $P = \vec F \cdot \vec v = 0.$

   Therefore, the kinetic energy remains constant: $K = \frac{1}{2}mv^2 = \text{constant}.$

   Since kinetic energy remains constant, the speed $v$ also remains constant.

   So, (A) is true.

4. Does (R) correctly explain (A)?

   Yes. Because the magnetic force being perpendicular to motion implies zero work done, which means no change in kinetic energy and hence no change in speed.

   Therefore, (R) is the correct explanation of (A).

5. Evaluate the options

   • A: Both (A) and (R) are true and (R) is the correct explanation of (A). ✅
   • B: Incorrect
   • C: Incorrect
   • D: Incorrect

Final Answer: $\boxed{A}$