In the product
$\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}})=\mathrm{q}\overrightarrow{\mathrm{v}}\times \left(\mathrm{B}\widehat{\mathrm{i}}+\mathrm{B}\widehat{\mathrm{j}}+{\mathrm{B}}_0\hat{\mathrm{k}}\right)$
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 option is:
$-6\widehat{\mathrm{i}}-6\widehat{\mathrm{j}}-8\hat{\mathrm{k}}$

Step 1: Understand the Given Quantities
We are given the magnetic force relation:
$\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}})$
where:
• $q=1$
• $\overrightarrow{v}=2\widehat{i}+4\widehat{j}+6\widehat{k}$
• $\stackrel{��}{B}=B\widehat{i}+B\widehat{j}+B_0\widehat{k}$
• $\overrightarrow{F}=4\widehat{i}-20\widehat{j}+12\widehat{k}$

Step 2: Compute the Cross Product $\overrightarrow{v}\times \overrightarrow{B}$
Using the determinant form for the cross product of two vectors:
$\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 2& 4& 6\\ B& B& B_0\end{array}\right|$
Expanding this determinant, we get:
$\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}}=\widehat{\mathrm{i}}(4{\mathrm{B}}_0-6\mathrm{B})-\widehat{\mathrm{j}}(2{\mathrm{B}}_0-6\mathrm{B})+\widehat{\mathrm{k}}(2\mathrm{B}-4\mathrm{B})$
Simplifying the terms:
$\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}}=(4{\mathrm{B}}_0-6\mathrm{B})\widehat{\mathrm{i}}+(6\mathrm{B}-2{\mathrm{B}}_0)\widehat{\mathrm{j}}-2\mathrm{B}\widehat{\mathrm{k}}$

Step 3: Equate to the Force Vector
Since $q=1$, we have $\overrightarrow{F}=\overrightarrow{v}\times \overrightarrow{B}$:
$4\widehat{\mathrm{i}}-20\widehat{\mathrm{j}}+12\widehat{\mathrm{k}}=(4{\mathrm{B}}_0-6\mathrm{B})\widehat{\mathrm{i}}+(6\mathrm{B}-2{\mathrm{B}}_0)\widehat{\mathrm{j}}-2\mathrm{B}\widehat{\mathrm{k}}$

Step 4: Solve for the Coefficients B and B₀
Equating the $\widehat{k}$ components:
$-2\mathrm{B}=12\Rightarrow \mathrm{B}=-6$
Equating the $\widehat{i}$ components:
$4{\mathrm{B}}_0-6\mathrm{B}=4$
Substitute $B=-6$ into the equation:
$4{\mathrm{B}}_0-6(-6)=4$
$4{\mathrm{B}}_0+36=4$
$4{\mathrm{B}}_0=-32\Rightarrow {\mathrm{B}}_0=-8$
We can verify with the $\widehat{j}$ components:
$6\mathrm{B}-2{\mathrm{B}}_0=6(-6)-2(-8)=-36+16=-20$
This perfectly matches the given $\widehat{j}$ component of force, $-20$.

Step 5: Write the Vector Expression for B
Using the values $B=-6$ and $B_0=-8$, we substitute them back into:
$\overrightarrow{\mathrm{B}}=\mathrm{B}\widehat{\mathrm{i}}+\mathrm{B}\widehat{\mathrm{j}}+{\mathrm{B}}_0\hat{\mathrm{k}}$
This yields:
$\overrightarrow{\mathrm{B}}=-6\widehat{\mathrm{i}}-6\widehat{\mathrm{j}}-8\widehat{\mathrm{k}}$