A particle of mass M and charge Q moving with velocity v describes a circular path of radius R when subjected to a uniform transverse magnetic field of induction B. The work done by the field when the particle completes one full circle is :

Options

A. `((mv^(2))/R)2piR`
B. Zero
C. `BQ2piR`
D. `BQv2piR`

Explanation

To solve the problem, we need to determine the work done by the magnetic field on a charged particle moving in a circular path. Here's the step-by-step solution:

### Step 1: Understand the Motion of the Charged Particle
A particle with mass \( M \) and charge \( Q \) is moving with a velocity \( v \) in a circular path of radius \( R \) due to a uniform magnetic field \( B \). The magnetic force acts as the centripetal force that keeps the particle in circular motion.

### Step 2: Identify the Forces Acting on the Particle
The magnetic force \( F \) acting on the charged particle can be expressed using the formula:
\[ F = QvB \sin(\theta) \]
where \( \theta \) is the angle between the velocity vector and the magnetic field. In this case, since the magnetic field is transverse (perpendicular) to the velocity of the particle, \( \theta = 90^\circ \) and thus \( \sin(90^\circ) = 1 \). Therefore, the magnetic force becomes:
\[ F = QvB \]

### Step 3: Determine the Direction of the Magnetic Force
The magnetic force acts perpendicular to the velocity of the particle, which means it changes the direction of the velocity but not its magnitude. This is crucial because it implies that the kinetic energy of the particle remains constant.

### Step 4: Calculate the Work Done by the Magnetic Field
Work done \( W \) by a force is given by the equation:
\[ W = F \cdot d \cdot \cos(\phi) \]
where \( \phi \) is the angle between the force and the displacement. Since the magnetic force is always perpendicular to the displacement of the particle (which is tangential to the circular path), \( \phi = 90^\circ \) and thus \( \cos(90^\circ) = 0 \).

Therefore, the work done by the magnetic field when the particle completes one full circle is:
\[ W = F \cdot d \cdot \cos(90^\circ) = 0 \]

### Conclusion
The work done by the magnetic field when the particle completes one full circle is:
\[ \boxed{0} \]