Question

A magnetic field $(4\times {10}^{-3}\hat{k})\text{T}$ exerts a force $(4\hat{i}+3\hat{j})\times {10}^{-10}\text{N}$ on a particle having a charge ${10}^{-9}\text{C}$ and going in the $x-y$ plane. The velocity $\left(\text{in m/s}\right)$ of the particle is

• A. $100\hat{i}-25\hat{j}$
• B. $75\hat{i}+100\hat{j}$
• C. $100\hat{i}+75\hat{j}$
• D. $-75\hat{i}+100\hat{j}$

Answer 1

The correct option is D $-75\hat{i}+100\hat{j}$

Given:
$\overrightarrow{B}=(4\times {10}^{-3}\hat{k})\text{T};q={10}^{-9}\text{C}$
$\overrightarrow{F}=(4\hat{i}+3\hat{j})\times {10}^{-10}\text{N}$

Magnetic force on a moving charged particle in a magnetic field is given by

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

$(4\hat{i}+3\hat{j}){10}^{-10}={10}^{-9}[(V_x\hat{i}+V_y\hat{j}+V_z\hat{z})\times 4\times {10}^{-3}\hat{k}]$

$4\hat{i}+3\hat{j}=\left[V_x\right(-\hat{j})+V_y\hat{i})\times 4\times {10}^{-2}$

$400\hat{i}+300\hat{j}=(V_y\hat{i}-V_x\hat{j})\times 4$

$100\hat{i}+75\hat{j}=V_y\hat{i}-V_x\hat{j}$

$\therefore V_y=100;V_x=-75$

So, the velocity of the particle will be

$\overrightarrow{V}=V_x\hat{i}+V_y\hat{j}$

$\overrightarrow{V}=-75\hat{i}+100\hat{j}$

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