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In the product

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

$=q\overrightarrow{v}\times \left ( B\hat{i}+B\hat{j}+B_{0}\hat{k} \right )$

For $q=1 and \overrightarrow{v}=2\hat{i}+4\hat{j}+6\hat{k} and$

$\overrightarrow{F}=4\hat{i}-20\hat{j}+12\hat{k}$

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

Option: 1
$-8\hat{i}-8\hat{j}-6\hat{k}$

Option: 2
$-6\hat{i}-6\hat{j}-8\hat{k}$

Option: 3
$8\hat{i}+8\hat{j}-6\hat{k}$

Option: 4
$6\hat{i}+6\hat{j}-8\hat{k}$

Answers (1)

best_answer

$\bar{F}=q\left ( \bar{v}\times \bar{B} \right )=q\bar{v}\times \left ( B\hat{i}+B\hat{j}+B_{0}\hat{k} \right )$

$\left [ 4\hat{i}-20\hat{j}+12\hat{k} \right ]=\left [ 2\hat{i}+4\hat{j}+6\hat{k} \right ]\times \left [ B\hat{i}+B\hat{j}+B_{0}\hat{k} \right ]$

$=\hat{i}\left ( 4B_{0}-6B \right )-\hat{j}\left ( 2B_{0}-6B \right )$

$+\hat{k}\left ( -2B \right )$

$-2B=12$

$B=-6$

$4B_{0}-6B=4$

$2B_{0}-6B=20$

$2B_{0}=-16,B_{0}=-8$

$B=-6$

$\bar{B}=B\hat{i}+B\hat{j}+B_{0}\hat{K}$

$\bar{B}=-6\hat{i}-6\hat{j}-8\hat{k}$

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manish

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