Question

A magnetic field of $\left(4.0\times {10}^{-3}\overrightarrow{k}\right)T$ exerts a force of $\left(4.0\overrightarrow{i}+3.0\overrightarrow{j}\right)\times {10}^{-10}N$ on a particle having a charge of $1.0\times {10}^{-9}C$ and going in the X-Y plane. Find the velocity of the particle.

Answer 1

$\begin{array}{c}F=q(v\times B)\\ clearly\\ \overrightarrow{v}=a\overrightarrow{i}+b\overrightarrow{j}\\ \Rightarrow \left(4.0\overrightarrow{i}+3.0\overrightarrow{j}\right)\times {10}^{-10}={10}^{-9}\left(\left(a\overrightarrow{i}+b\overrightarrow{j}\right)\times 4\times {10}^{-3}\right)\\ \Rightarrow \left(4.0\overrightarrow{i}+3.0\overrightarrow{j}\right)\times {10}^{-1}=\left(-4a\overrightarrow{j}+4b\overrightarrow{i}\right)\times {10}^{-3}\\ \Rightarrow a=-100,b=75\\ so,\overrightarrow{v}=75\overrightarrow{j}-100\overrightarrow{i}\end{array}$

So, velocity of the particle is $75\overrightarrow{j}-100\overrightarrow{i}$