91Ó°ÊÓ

$$\begin{gathered} q=2 \\ \stackrel{→}{v}=2.0\stackrel{Áåœ}{i}+4.0\stackrel{Áåœ}{j}+6.0\stackrel{Áåœ}{k} \\ \stackrel{→}{F}=4.0\stackrel{Áåœ}{i}-20\stackrel{Áåœ}{j}+12\stackrel{Áåœ}{k} \end{gathered}$$

Using the formula for cross-product we can find the vector $\stackrel{\mathbf{→}}{\mathbf{B}}$from the given vectors$\stackrel{\mathbf{→}}{\mathbf{F}}$and$\stackrel{\mathbf{→}}{\mathbf{v}}$.

The vector product is written as,

$\left(\stackrel{→}{a}×\stackrel{â‡\P Ä}{b}\right)=\left(a_y{b}_z-b_y{a}_z\right)\stackrel{Áåœ}{i}+\left(a_z{b}_x-b_z{a}_x\right)\stackrel{Áåœ}{j}+\left(a_x{b}_y-b_y{a}_x\right)\stackrel{Áåœ}{k}$ (i)

Use equation (i) to write the cross product between $\stackrel{→}{v}$and $\stackrel{→}{B}$.

$$\begin{gathered} \stackrel{→}{F}=q\stackrel{→}{v}×\stackrel{→}{B} \\ {F}_x\stackrel{Áåœ}{i}+F_y\stackrel{Áåœ}{j}+F_z\stackrel{Áåœ}{k}=q\left(v_y{B}_z-v_z{B}_y\right)\stackrel{Áåœ}{i}+q\left(v_z{B}_x-v_x{B}_z\right)\stackrel{Áåœ}{j}+q\left(v_x{B}_y-v_y{B}_x\right)\stackrel{Áåœ}{k} \end{gathered}$$

Comparing the above equation with the given vector$\stackrel{→}{F}$, we get

$$\begin{gathered} 2\left(4B_z-6B_y\right)=4 \\ 2\left(6B_x-2B_z\right)=-2 \\ 2\left(2B_y-4B_x\right)=12 \end{gathered}$$

As $B_x=B_y$, above equation becomes,

$$\begin{gathered} 2\left(2B_y-4B_y\right)=12 \\ 4B_y-8B_y=12 \\ -4B_y=12 \\ {B}_y=-3 \end{gathered}$$

Inserting this value in above equation we get,

$$\begin{gathered} 2\left(4B_z-6\left(-3\right)\right)=4 \\ {B}_z=-4 \end{gathered}$$

Calculate the value of $B_x$from the above equations.

$$\begin{gathered} B_x=B_y \\ {B}_x=-3 \end{gathered}$$

Therefore, the vector$\stackrel{→}{\mathrm{B}}$ can be written as$-3.0\stackrel{Áåœ}{i}-3.0\stackrel{Áåœ}{j}-4.0\stackrel{Áåœ}{k}$