Question

In the product
$\vec{F}=q(\vec{v}\times \vec{B})$
$=q\vec{v}\times \left(B\hat{i}+B\hat{j}+B_0\hat{k}\right)$
For $q=1$ and $\vec{v}=2\hat{i}+4\hat{j}+6\hat{k}$ and
$\vec{F}=4\hat{i}-20\hat{j}+12\hat{k}$
What will be the complete expression for $\vec{B}$ ?

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

Answer 1

Answer: (B)

$\vec{F}=q(\vec{V}\times \vec{B})$
$\vec{F}\cdot \vec{B}=0$
$\vec{V}\times \vec{B}=\hat{i}\left[4B_0-6B\right]+\hat{j}\left[6B-2B_0\right]+\hat{k}[2B-4B]$
$=\hat{i}\left[4B_0-6B\right]+\hat{j}\left(6B-2B_0\right)-\hat{k}(2B)$
$4B_0-6B=4\dots (1)$
$6B-2B_0=-20\dots (2)$
$-2B=12\dots (3)$
Solving (1), (2) and (3) we get
$B=-6$ and $B_0=-8$