In the product
$\stackrel{}{}$ F → = q ( v → × B → ) = q v → × B i ^ + B j ^ + B 0 k ^
For q = 1 and $\overrightarrow{\mathrm{v}}=2\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+6\widehat{\mathrm{k}}\text{ and }\overrightarrow{\mathrm{F}}=4\widehat{\mathrm{i}}-20\widehat{\mathrm{j}}+12\hat{\mathrm{k}}$
What will be the complete expression for $\overrightarrow{\mathrm{B}}$ ?

The correct answer is:
$-6\hat{i}-6\hat{j}-8\hat{k}$

Step 1: Understand the given formula and values
The force experienced by a charge $q$ moving with velocity $\overrightarrow{v}$ in a magnetic field $\overrightarrow{B}$ is given by the cross product:
$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$
From the problem description and the visible image options, the given parameters are:
Charge: $q=1$
Velocity: $\overrightarrow{v}=2\hat{i}+4\hat{j}+6\hat{k}$
Magnetic field vector: $\overrightarrow{B}=B\hat{i}+B\hat{j}+B_0\hat{k}$
Force vector: $\overrightarrow{F}=4\hat{i}-20\hat{j}+12\hat{k}$

Step 2: Calculate the cross product of velocity and magnetic field
Since $q=1$, we have $\overrightarrow{F}=\overrightarrow{v}\times \overrightarrow{B}$. Let us compute the cross product using a determinant:
$\overrightarrow{v}\times \overrightarrow{B}=\left|\begin{array}{cc}\hat{i}& $ \hat{j}$\hat{k}\\ 2& 4& 6\\ B& B{B}_0\end{array}\right|$

Expanding this determinant along the first row gives:
$\overrightarrow{v}\times \overrightarrow{B}=\hat{i}(4B_0-6B)-\hat{j}(2B_0-6B)+\hat{k}(2B-4B)$

Simplifying the terms:
$\overrightarrow{v}\times \overrightarrow{B}=(4B_0-6B)\hat{i}-(2B_0-6B)\hat{j}-2B\hat{k}$

Step 3: Equate components to find B and B₀
Comparing the calculated force components with the given force $\overrightarrow{F}=4\hat{i}-20\hat{j}+12\hat{k}$:
Comparing the $\hat{k}$ component:
$-2B=12\Rightarrow B=-6$

Comparing the $\hat{i}$ component:
$4B_0-6B=4$
Substitute the value of $B=-6$:
$4B_0-6(-6)=4$
$4B_0+36=4$
$4B_0=-32\Rightarrow B_0=-8$

Step 4: Verify the remaining component
Let us check the $\hat{j}$ component to verify consistency:
$-(2B_0-6B)=-2B_0+6B$
Substitute $B_0=-8$ and $B=-6$:
$-2(-8)+6(-6)=16-36=-20$
This perfectly matches the given component of $-20\hat{j}$.

Step 5: Write the complete expression for B
Using the values found:
$\overrightarrow{B}=-6\hat{i}-6\hat{j}-8\hat{k}$