Question

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

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

Answer 1

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

Magnetic force on a charge particle is,

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

$(4\hat{i}+3\hat{j})\times {10}^{-10}={10}^{-9}[(v_x\hat{i}+v_y\hat{j})\times (4\times {10}^{-3}\hat{k})]$

$(4\hat{i}+3\hat{j})\times {10}^{-10}=4\times {10}^{-12}[(v_x\hat{i}+v_y\hat{j})\times \left(\hat{k}\right)]$

$(4\hat{i}+3\hat{j}){10}^2$ = $4v_x\widehat{(-j)}+4v_y\widehat{\left(i\right)}$

Comparing $\hat{i}$ and $\hat{j}$ we get,

$400=4v_y\Rightarrow v_y=100$

$300=-4v_x\Rightarrow v_x=-75$

$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.