A particle of mass `M` and charge `Q` moving with velocity `vec(v)` describe 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. (a)`((mv^(2))/R)2piR`
B. (b)zero
C. (c)`BQ2piR`
D. (d)`BQv2R`

Explanation

To solve the problem, we need to analyze the motion of a charged particle in a magnetic field and determine the work done by the magnetic field when the particle completes one full circular path.

### Step-by-step Solution:

1. **Understand the Forces Acting on the Particle**:
The charged particle of mass \( M \) and charge \( Q \) is moving in a circular path due to the magnetic force acting on it. The magnetic force \( \vec{F} \) acting on the particle is given by the Lorentz force equation:
\[ \vec{F} = Q (\vec{v} \times \vec{B}) \]
where \( \vec{v} \) is the velocity of the particle and \( \vec{B} \) is the magnetic field.

2. **Direction of the Force**:
The magnetic force \( \vec{F} \) is always perpendicular to the velocity \( \vec{v} \) of the particle because it is a cross product. This means that the force does not do any work on the particle since work is defined as:
\[ W = \vec{F} \cdot \vec{d} \]
where \( \vec{d} \) is the displacement. If \( \vec{F} \) is perpendicular to \( \vec{d} \), then \( W = 0 \).

3. **Complete One Full Circle**:
As the particle moves in a circular path, it completes one full circle. During this motion, the magnetic force continuously changes the direction of the particle's velocity but does not change its speed. Since the force is always perpendicular to the displacement, the work done by the magnetic field over one complete revolution is:
\[ W = 0 \]

4. **Conclusion**:
Therefore, the work done by the magnetic field when the particle completes one full circle is:
\[ \text{Work done} = 0 \]

### Final Answer:
The work done by the field when the particle completes one full circle is \( 0 \).