Question

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

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

Answer 1

The correct option is B $-75\overrightarrow{i}+100\overrightarrow{j}$

Let the velocity of the particle be $\overrightarrow{V}=V_x\hat{i}+V_y\hat{j}+V_z\hat{k}$

Force acting on the particle $\overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B})$

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

Or $400\hat{i}+300\hat{j}=-4V_x\hat{j}+4V_y\hat{i}$

$⟹V_y=100$ and $V_x=-75$ and $V_z=0$

So, the velocity $\overrightarrow{V}=-75\hat{i}+100\hat{j}$

Option B is correct.