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Step-by-Step Solution
Step 1: Recognize the Nature of the Magnetic Force
When a charged particle of charge $Q$ and mass $M$ moves with velocity $\vec{v}$ in a uniform magnetic field of magnitude $B$, it experiences a force given by $ \vec{F} = Q \,\vec{v} \times \vec{B}$. This force always acts perpendicular to the velocity of the particle.
Step 2: Understand the Particle’s Circular Motion
The force acting on the charged particle provides the centripetal force that causes it to move in a circular path of radius $R$. Because the magnetic force is always directed towards the center of the circular path (perpendicular to the instantaneous velocity), the displacement at every point on the path is tangential, while the force is radial.
Step 3: Calculate the Work Done by the Magnetic Field
The infinitesimal work done by a force $ \vec{F} $ for an infinitesimal displacement $ d\vec{s} $ is given by
$ dW = \vec{F} \cdot d\vec{s} = F \, ds \,\cos \theta, $
where $ \theta $ is the angle between the force and displacement vectors. In uniform circular motion under a magnetic field, this angle $ \theta $ is $90^\circ$, because $\vec{F}$ is always perpendicular to $d\vec{s}$.
Thus, for each infinitesimal displacement, $ \cos 90^\circ = 0 $. Hence, $ dW = 0 $.
Step 4: Conclude the Net Work Done After One Complete Circle
Since the instantaneous work done $ dW $ at every point is zero, even after completing one full revolution (or any number of revolutions), the total work done by the magnetic field is
$ W_{\text{net}} = \int dW = 0. $
Final Answer
The net work done by the magnetic field when the particle completes one full circle is zero.
