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)

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