Question

A particle of mass $1\text{mg}$ and having charge $1\mu \text{C}$ is moving in a magnetic field, $\overrightarrow{B}=(2\hat{i}+3\hat{j}+\hat{k})\text{T}$ with velocity $\overrightarrow{v}=(2\hat{i}+\hat{j}$$-\hat{k})$$\text{km/s}$. Find the magnitude of acceleration.

• A. $3\sqrt{3}{\text{m/s}}^2$
• B. $2\sqrt{3}{\text{m/s}}^2$
• C. $4\sqrt{3}{\text{m/s}}^2$
• D. $5\sqrt{3}{\text{m/s}}^2$

Answer 1

The correct option is C $4\sqrt{3}{\text{m/s}}^2$

Force on a charged particle moving in a magnetic field $\overrightarrow{B}$ with velocity $\overrightarrow{v}$ is given by$$\overrightarrow{F_m}=q(\overrightarrow{v}\times \overrightarrow{B})$$So, acceleration of the particle, $$\overrightarrow{a}=\frac{\overrightarrow{F_m}}{m}=\frac{q(\overrightarrow{v}\times \overrightarrow{B})}{m}$$Given:

$\overrightarrow{v}=(2\hat{i}+\hat{j}-\hat{k})\times {10}^3\text{m/s}$

$\overrightarrow{B}=(2\hat{i}+3\hat{j}+\hat{k})\text{T}$

$q={10}^{-6}\text{C}$

$m={10}^{-3}\text{kg}$

Now,
$\overrightarrow{v}\times \overrightarrow{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 2& 3& 1\end{array}\right|\times {10}^3$

$=\left[\hat{i}\right(1+3)-\hat{j}(2+2)+\hat{k}(6-2\left)\right]\times {10}^3$

$=[4\hat{i}-4\hat{j}+4\hat{k}]\times {10}^3$

$\therefore \overrightarrow{a}=\frac{q(\overrightarrow{v}\times \overrightarrow{B})}{m}=\frac{{10}^{-6}(4\hat{i}-4\hat{j}+4\hat{k})\times {10}^3}{{10}^{-3}}$

$\Rightarrow \overrightarrow{a}=(4\hat{i}-4\hat{j}+4\hat{k}){\text{m/s}}^2$

$\Rightarrow |\overrightarrow{a}|=\sqrt{4^2+(-4{)}^2+4^2}=4\sqrt{3}{\text{m/s}}^2$

Hence, option $\left(C\right)$ is the correct answer.