1. Magnetic force on a moving charge

   A charge $Q$ moving with velocity $\vec v$ in a magnetic field $\vec B$ experiences the Lorentz force $\vec F = Q(\vec v \times \vec B).$

2. Direction of magnetic force

   Since the field is transverse, $\vec B \perp \vec v$ at every instant. Therefore, the magnetic force is always perpendicular to the velocity.

   So, $\vec F \cdot \vec v = 0.$

3. Work done by force

   The rate of doing work is $\frac{dW}{dt} = \vec F \cdot \vec v.$

   Since $\vec F \cdot \vec v = 0$, $\frac{dW}{dt} = 0.$

   Hence the magnetic field does no work on the particle at any instant.

4. Over one full circular revolution

   Because the magnetic force only changes the direction of velocity, not its magnitude, the kinetic energy remains constant. Therefore, the total work done over one complete circle is $W = 0.$

5. Option check

   • A: $\left(\frac{Mv^2}{R}\right)2\pi R$ This is force $\times$ distance, but this is not valid here because force is always perpendicular to displacement.
   • B: zero ✅
   • C: $BQ\,2\pi R$ ❌ dimensionally not work.
   • D: $BQv\,2\pi R$ ❌ though dimensions match work, magnetic force does no work.

Therefore, the correct answer is B: zero.