Question

A force $\overrightarrow{F}$, acting on an electric charge $q$, in presence of an electro-magnetic field, moves the charge parallel to the magnetic field with velocity $\overrightarrow{\nu }$. Then $\overrightarrow{F}$ is equal to (where $\overrightarrow{E}$ and $\overrightarrow{B}$ are electric field and magnetic field respectively) _________.

• A. $q\overrightarrow{E}$
• B. $q(\overrightarrow{\nu }\times \overrightarrow{E})$
• C. $q(\overrightarrow{\nu }\times \overrightarrow{B})$
• D. $q\overrightarrow{B}$

Answer 1

The correct option is C $q(\overrightarrow{\nu }\times \overrightarrow{B})$

The direction of the magnetic force on a moving charge is perpendicular to the plane formed by v and B and follows right hand rule. The force is given by

$F$ = q$(v\times B)$

= qvBsin $\theta$