The force on a charged particle due to electric and magnetic fields is given by $\vec{F}=q \vec{E}+q \vec{v} \times \vec{B}$. Suppose $\vec{E}$ is along the X-axis and $\vec{B}$ along the $Y$-axis. In what direction and with what minimum speed v should a positively charged particle be sent so that the net force on it is zero?
$\mathrm{F}=\mathrm{q}(\mathrm{E}+\mathrm{v} \times \mathrm{B})$
Now, for net force to be 0 , we must have $E=-(v \times B)$
So, the direction of $E$ must be opposite to that of $(v \times B)$, so $v$ must be in $Z$ axis and its magnitude is $E /(B \sin \theta)$. For $v$ to be minimum. $\theta=90^{\circ}$, so $v=E / B$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.