Question

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

• A. $zero$
• B. $BQ2\pi R$
• C. $BQv\left(2\pi R\right)$
• D. $\left(\frac{M{v}^2}{R}\right)\left(2\pi R\right)$

Answer 1

Answer: (A)

$zero$

${\overrightarrow{F}}_{mag}=q(\overrightarrow{v}\times \overrightarrow{B})$
Work done on the particle $W=\int Pdt$
But, $P={\overrightarrow{F}}_{mag}.\overrightarrow{v}=q(\overrightarrow{v}\times \overrightarrow{B}).\overrightarrow{v}=0$ since $\overrightarrow{v}\times \overrightarrow{B}\perp \overrightarrow{v}$
$\therefore W=0$.
Also, one can look at it from displacement perspective. Since the particle comes back to the same position after one cycle, $\overrightarrow{ds}$ is zero.
$\Rightarrow W=0$