Question

The force on a charged particle due to electric and magnetic fields is given by $\overrightarrow{F}=q\overrightarrow{E}+q\overrightarrow{\nu }\times \overrightarrow{B}$. Suppose $\overrightarrow{E}$ is along the X-axis and $\overrightarrow{B}$ along the Y-axis. In what direction and with what minimum speed ν should a positively charged particle be sent so that the net force on it is zero?

Answer 1

According to the problem, the net electric and magnetic forces on the particle should be zero.



i.e., $\overrightarrow{F}=q\overrightarrow{E}+q\left(\overrightarrow{\nu }\times \overrightarrow{B}\right)=0$
$\Rightarrow E=-\left(\overrightarrow{\nu }\times \overrightarrow{B}\right)$
So, the direction of $\overrightarrow{\nu }\times \overrightarrow{B}$ should be opposite to the direction of $\overrightarrow{E}$. Hence, $\overrightarrow{\nu }$ should be along the positive z-direction.

Again, E = νB sin θ
$\Rightarrow \mathrm{\nu }=\frac{E}{B}\sin \mathrm{\theta }$
For ν to be minimum, $\mathrm{\theta }=90°\mathrm{and},\mathrm{thus},{\nu }_{\min }=\frac{E}{B}$
So, the particle must be projected at a minimum speed of $\frac{E}{B}$ along the z-axis.