Question

A electron experiences a force $(4.0\hat{i}+3.0\hat{j})\times {10}^{-13}N$ in a uniform magnetic field when its velocity is $2.5\hat{k}\times {10}^{7}m{s}^{-1}.$ The magnetic field vector $\overrightarrow{B}$ is

• A. $-0.075\hat{i}+0.1\hat{j}$
• B. $0.1\hat{i}+0.075\hat{j}$
• C. $-0.075\hat{i}-0.1\hat{j}+\hat{k}$
• D. $-0.075\hat{i}-0.1$

Answer 1

The correct option is A $-0.075\hat{i}+0.1\hat{j}$

We have,

$\overrightarrow{7}=(4\hat{i}+3\hat{j})\times {10}^{-13}0$

$\overrightarrow{v}=2.5k\times {10}^{7}m/s$

We know that $\overrightarrow{7}=-e\overrightarrow{v}\times \overrightarrow{B}$

Let $\overrightarrow{B}=x\hat{j}+y\hat{j}+z\hat{k}$, Then

$\overrightarrow{v}\times \overrightarrow{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 0& 2.5\times {10}^7\\ x& y& z\end{array}\right|=2.5\times {10}^7[x\hat{j}-y\hat{i}]$

SO, $\overrightarrow{7}=-16\times {10}^{-19}\times 2.5\times {10}^7[x\hat{j}-y\hat{i}]$

$\Rightarrow [4\hat{i}+3\hat{j}]\times {10}^{-13}=-4\times {10}^{12}[x\hat{j}-y\hat{i}]$

$\Rightarrow 4\hat{i}+3\hat{j}=40y\hat{i}-40x\hat{j}$

$y=\frac{4}{40},x=\frac{3}{40},z=0$

$\overrightarrow{B}=x\hat{i}+y\hat{j}+z\hat{k}=-0.075\hat{i}+0.1\hat{j}$

Option $A$ is correct