Question

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

• A. $\left(-0.30\hat{i}+0.32\hat{j}-0.09\hat{k}\right)$
• B. $\left(-3.0\hat{i}+3.2\hat{j}-0.9\hat{k}\right)$
• C. $\left(-30\hat{i}+32\hat{j}-9\hat{k}\right)$
• D. $\left(-300\hat{i}+320\hat{j}-90\hat{k}\right)$

Answer 1

The correct option is C

$\left(-30\hat{i}+32\hat{j}-9\hat{k}\right)$

Step1: Given data.

Charge, $q=1\mu C$

Particle moving with velocity, $V=$$\left(2\hat{i}+3\hat{j}+4\hat{k}\right)$$m{s}^{-1}$

External magnetic filed, $B=$ $\left(5\hat{i}+3\hat{j}-6\hat{k}\right)$ $\times$${10}^{-3}T$

Magnetic force on particle, $F_m=$ $\overrightarrow{F}\times {10}^{-9}$$N$

Step2: Finding the vector $\overrightarrow{F}$.

We know that,

Magnetic Force on charge particle moving with velocity in magnetic field.

$F_m=q\left(\overrightarrow{V}\times \overrightarrow{B}\right)$ …..$\left(i\right)$

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

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

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

$\Rightarrow \overrightarrow{V}\times \overrightarrow{B}$ $=\left(-30\hat{i}+32\hat{j}-9\hat{k}\right)\times {10}^{-3}$ ……$\left(ii\right)$

Putting equation $\left(ii\right)$ in equation $\left(i\right)$ we get.

$\Rightarrow$ $F_m=q\times \left(-30i^{}+32j^{}-9k\right)\times {10}^{-3}$

$\Rightarrow$ $F_m=$${10}^{-6}\times \left(-30\hat{i}+32\hat{j}-9\hat{k}\right)\times {10}^{-3}$ $\left[\because q=1\mu C={10}^{-6}C\right]$

$\Rightarrow$ $\overrightarrow{F}\times {10}^{-9}$$={10}^{-9}\times \left(-30\hat{i}+32\hat{j}-9\hat{k}\right)$ $\left[\because Given:F_m=\overrightarrow{F}\times {10}^{-9}\right]$

$\Rightarrow$ $\overrightarrow{F}=$$\left(-30\hat{i}+32\hat{j}-9\hat{k}\right)$

Hence, Option C is correct option. The vector $\overrightarrow{\mathbf{F}}$ is $\left(-30\hat{i}+32\hat{j}-9\hat{k}\right)$.