Question

In magnetic field $\overrightarrow{B}=(2\hat{i}+3\hat{j})\text{T}$ , velocity of a charged particle is given in table-1. Corresponding magnetic force is given in table-2. Match the two tables.
Table-1 Table-2

$\left(\text{A}\right)(4\hat{i}-3\hat{j})\text{m/s}$	$\left(\text{P}\right)$ zero.
$\left(\text{B}\right)(4\hat{i}-6\hat{j})\text{m/s}$	$\left(\text{Q}\right)$ Maximum.
$\left(\text{C}\right)\left(6\hat{k}\right)\text{m/s}$	$\left(\text{R}\right)$ May be along positive $z$- direction.
$\left(\text{S}\right)$ May be along negative $z$- direction.

• A. $\text{A}\to \text{P},\text{B}\to \text{Q},\text{C}\to \text{R}$
• B. $\text{A}\to \text{Q},\text{B}\to \text{R},\text{C}\to \text{P}$
• C. $\text{A}\to \text{R},\text{B}\to \text{P},\text{C}\to \text{Q}$
• D. $\text{A}\to \text{S},\text{B}\to \text{P},\text{C}\to \text{Q}$

Answer 1

The correct option is C $\text{A}\to \text{R},\text{B}\to \text{P},\text{C}\to \text{Q}$

Magnetic force acting on a particle moving in the presence of external uniform magnetic field is given by $${\overrightarrow{F}}_m=q(\overrightarrow{v}\times \overrightarrow{B})$$$\text{Case-I}:$ Velocity $\overrightarrow{v}=4\hat{i}-3\hat{j}$

Then , $\overrightarrow{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& (-3)& 0\\ 2& 3& 0\end{array}\right|=18\hat{k}\text{N}$

$\text{Case-II}:$ Velocity $\overrightarrow{v}=4\hat{i}-6\hat{j}$

Then , $\overrightarrow{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& (-6)& 0\\ 2& 3& 0\end{array}\right|=0\hat{k}\text{N}$

$\text{Case-III}:$ Velocity $\overrightarrow{v}=6\hat{k}$

Then , $\overrightarrow{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 0& 6\\ 2& 3& 0\end{array}\right|=12\hat{j}-18\hat{i}\text{N}\Rightarrow |F|\approx 21.63\text{N}$

Hence, option (c) is the correct answer.