A charged particle carrying charge

Question:

A charged particle carrying charge $1 \mu \mathrm{C}$ is moving with velocity $(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \mathrm{ms}^{-1}$. If an external magnetic field of $(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}) \times 10^{-3} \mathrm{~T}$ exists in the region where the particle is moving then the force on the particle is $\overrightarrow{\mathrm{F}} \times 10^{-9} \mathrm{~N}$. The vector $\vec{F}$ is :

1. $-0.30 \hat{\mathrm{i}}+0.32 \hat{\mathrm{j}}-0.09 \hat{\mathrm{k}}$

2. $-300 \hat{\mathrm{i}}+320 \hat{\mathrm{j}}-90 \hat{\mathrm{k}}$

3. $-30 \hat{\mathrm{i}}+32 \hat{\mathrm{j}}-9 \hat{\mathrm{k}}$

4. $-3.0 \hat{\mathrm{i}}+3.2 \hat{\mathrm{j}}-0.9 \hat{\mathrm{k}}$

Correct Option: , 3

Solution:

$\overrightarrow{\mathrm{F}}=9(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})$ (Force on charge particle moving

in magnetic field)

$\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}) \times 10^{-3}$

$=\left(\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 5 & 3 & -6\end{array}\right) \times 10^{-3}$

$=[\hat{\mathrm{i}}[-18-12]-\hat{\mathrm{j}}[-12-20]+\hat{\mathrm{k}}[6-15]] \times 10^{-3}$

$=[\hat{\mathrm{i}}[-30]+\hat{\mathrm{j}}[32]+\hat{\mathrm{k}}[-9]] \times 10^{-3}$

Force $=10^{-6}[-30 \hat{\mathrm{i}}+32 \hat{\mathrm{j}}-9 \hat{\mathrm{k}}] \times 10^{-3}$

$=10^{-9}[-30 \hat{\mathrm{i}}+32 \hat{\mathrm{j}}-9 \hat{\mathrm{k}}]$