Question

A charged particle is projected in a magnetic field $\overrightarrow{B}=(3\hat{i}+4\hat{j})\times {10}^{-2}\text{T}$. The acceleration of the particle is found to be $\overrightarrow{a}=(x\hat{i}+2\hat{j})\text{m/s}$. The value of $x$ is :

• A. $\frac{4}{3}$
• B. $\frac{-4}{6}$
• C. $\frac{-8}{3}$
• D. $\frac{5}{3}$

Answer 1

The correct option is C $\frac{-8}{3}$

The magnetic force acting on a moving charge in a magnetic field is given by

$\overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B})$

Thus, $\overrightarrow{F}$ is always perpendicular to the $\overrightarrow{V}$ as well as magnetic field $\overrightarrow{B}$.

Hence, acceleration $\left(\overrightarrow{a}\right)$ of charge will also be perpendicular to the magnetic field vector $\left(\overrightarrow{B}\right)$ because $\overrightarrow{a}=\overrightarrow{F}/m$.

$\therefore \overrightarrow{a}\cdot \overrightarrow{B}=0$

$(x\hat{i}+2\hat{j})\cdot (3\hat{i}+4\hat{j})\times {10}^{-2}=0$

$(3x+8)\times {10}^{-2}=0$

$\therefore x=\frac{-8}{3}$

Hence, option $\left(c\right)$ is the correct choice.

Why this question ? Tip: Application of dot product is very useful when we are dealing with quantities that are mutually perpendicular. If $\overrightarrow{a}\perp \overrightarrow{b}$, then $\overrightarrow{a}\cdot \overrightarrow{b}=0$.