Question

In the product

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})=q\overrightarrow{v}\times (B\hat{i}+B\hat{j}+B_0\hat{k})$

For $q=1$ and $\overrightarrow{v}=2\hat{i}+4\hat{j}+6\hat{k}$,
$\overrightarrow{F}=4\hat{i}-20\hat{j}+12\hat{k}$

What will be the complete expression for $\overrightarrow{B}$?

• A. $6\hat{i}+6\hat{j}-8\hat{k}$
• B. $-8\hat{i}-8\hat{j}-6\hat{k}$
• C. $-6\hat{i}-6\hat{j}-8\hat{k}$
• D. $8\hat{i}+8\hat{j}-6\hat{k}$

Answer 1

The correct option is C $-6\hat{i}-6\hat{j}-8\hat{k}$

Given:

$q=1$
$\overrightarrow{v}=2\hat{i}+4\hat{j}+6\hat{k}$
$\overrightarrow{B}=B\hat{i}+B\hat{j}+B_0\hat{k}$
$\overrightarrow{F}=4\hat{i}-20\hat{j}+12\hat{k}$

Now,

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

$\Rightarrow 4\hat{i}-20\hat{j}+12\hat{k}=1\left[\right(2\hat{i}+4\hat{j}+6\hat{k})\times (B\hat{i}+B\hat{j}+B_0\hat{k}\left)\right]$

$\Rightarrow 4\hat{i}-20\hat{j}+12\hat{k}=(4B_0-6B)\hat{i}-(2B_0-6B)\hat{j}+(-2B)\hat{k}$

On comparing and solving, we get,

$B=-6$ and $B_0=-8$

Therefore,

$\overrightarrow{B}=B\hat{i}+B\hat{j}+B_0\hat{k}=-6\hat{i}-6\hat{j}-8\hat{k}$

Hence, option $\left(C\right)$ is the correct answer.