A charged particle of charge 4 mC enters a uniform magnetic field of induction $\overline{\mathrm{B}}=3\overline{\mathrm{i}}+6\overline{\mathrm{j}}+6\overline{\mathrm{k}}$ tesla with a velocity $\overline{\mathrm{v}}=4\overline{\mathrm{i}}-\mathrm{x}\overline{\mathrm{j}}+\mathrm{y}\overline{\mathrm{k}}$. If the particles continues to move un-deviated, then the magnitude of velocity of the particle is

Given data

Magnetic field

$\overrightarrow{B}=3\overrightarrow{i}+6\overrightarrow{j}+ 6\overrightarrow{k}$

Velocity

 $\overrightarrow{v}=4\overrightarrow{i}-x\overrightarrow{j}+ y\overrightarrow{k}$.

Charge $q=4 mC$

Concepts used moving charges and magnetism

Explanation

According to force on a moving charge in a magnetic field

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$

$q=4\text{ mC}$

$=4\times {10}^{-3}\text{ C}$

If force is zero v must be parallel to B

$$\begin{gathered} \frac{6}{-x}=\frac{3}{4} \\ x=-8 \end{gathered}$$

$$\begin{gathered} \frac{6}{y}=\frac{3}{4} \\ y=8 \end{gathered}$$

Solving these equations simultaneously then $x=y=3$

 so the velocity of the particle is 

$\overrightarrow{v}=4\overrightarrow{i}-8\overrightarrow{j}+8\overrightarrow{k}\text{ m/s}$.

The magnitude of velocity

$\sqrt{4^2+(-8{)}^2+8^2}$

$=12 m/s$