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(C) The magnetic force $\vec{F}_m$ acting on a charged particle moving in a magnetic field is given by $\vec{F}_m = Q(\vec{v} 	imes \vec{B})$.Since t

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Zero

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(C) The magnetic force $\vec{F}_m$ acting on a charged particle moving in a magnetic field is given by $\vec{F}_m = Q(\vec{v} 	imes \vec{B})$.Since the force $\vec{F}_m$ is always perpendicular to the velocity vector $\vec{v}$,the force is always directed towards the center of the circular path (centripetal force).Work done $W$ is defined as the line integral of the force over the displacement: $W = \int \vec{F} \cdot d\vec{s}$.Since the force $\vec{F}_m$ is always perpendicular to the displacement $d\vec{s}$ at every point on the circular path,the dot product $\vec{F}_m \cdot d\vec{s} = F_m ds \cos(90^\circ) = 0$.Therefore,the total work done by the magnetic field on the particle in one full circle is $0$.

$A$ particle of mass $M$ and charge $Q$ moving with velocity $\vec{v}$ 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

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