Question

A particle of mass $m$ and charge $q$ is moving with speed $v$ describes a circular path of radius $r$ when subjected to a uniform transverse magnetic field $B$. The work done by the magnetic field when the particle completes one full circle is -

• A. $2\pi qvrB$
• B. $2\pi m{v}^2$
• C. $\pi m{v}^2$
• D. zero

Answer 1

The correct option is D zero

Magnetic force $\overrightarrow{F}$ on a moving charge $q$ with velocity $\overrightarrow{v}$ in a magnetic field $\overrightarrow{B}$ is given by$$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$$
Since, the magnetic force is $\perp$ to $\overrightarrow{v}$, so, force is always $\perp$ to the motion of charged particle $q$.

$\therefore$Work done,

$W=\overrightarrow{F}.d\overrightarrow{s}=Fds\cos {90}^{\circ }=0$

$[d\overrightarrow{s}=$ displacement$]$

$\therefore$ Work done by magnetic field on the charged particle is zero.

Hence, option $\left(d\right)$ is the correct answer.