In the below product, What will be the complete expression for B ?

To find the complete expression for the magnetic field vector $\overrightarrow{B}$, we start with the given relationship for magnetic force:
$\overrightarrow{F}=\; q\; (\overrightarrow{v}\times \overrightarrow{B})$

From the given information in the problem:
• Charge, $q\; =\; 1$
• Velocity, $\overrightarrow{v}=\; 2\hat{i}+\; 4\hat{j}+\; 6\hat{k}$
• Magnetic field, $\overrightarrow{B}=\; B\hat{i}+\; B\hat{j}+\; B$₀ k^
• Magnetic Force, $\overrightarrow{F}=\; 4\hat{i}-\; 20\hat{j}+\; 12\hat{k}$

Substitute $q\; =\; 1$ into the force equation:
$\overrightarrow{F}=\overrightarrow{v}\times \overrightarrow{B}$

Now, let's compute the cross product $\overrightarrow{v}\times \overrightarrow{B}$ using the determinant method:
$\overrightarrow{v}\times \overrightarrow{B}=\; det\; |\; [\hat{i},\hat{j},\hat{k}]\; ,\; [\; 2\; ,\; 4\; ,\; 6\; ]\; ,\; [\; B\; ,\; B\; ,\; B$₀ ] |
Expanding this determinant gives:
$\overrightarrow{v}\times \overrightarrow{B}=\; (4\; B$₀ - 6 B) i^ - (2 B₀ - 6 B) j^ + (2 B - 4 B) k^
Simplifying the terms:
$\overrightarrow{v}\times \overrightarrow{B}=\; (4\; B$₀ - 6 B) i^ + (6 B - 2 B₀) j^ - 2 B k^

We equate the components of the computed cross product to the components of the given force vector $\overrightarrow{F}=\; 4\hat{i}-\; 20\hat{j}+\; 12\hat{k}$:
1) For the $\hat{k}$-component:
$-2\; B\; =\; 12$
$B\; =\; -6$

2) For the $\hat{j}$-component:
$6\; B\; -\; 2\; B$₀ = -20
Substitute $B\; =\; -6$ into the equation:
$6(-6)\; -\; 2\; B$₀ = -20
$-36\; -\; 2\; B$₀ = -20
$-2\; B$₀ = 16
$B$₀ = -8

3) Let's verify with the $\hat{i}$-component:
$4\; B$₀ - 6 B = 4(-8) - 6(-6) = -32 + 36 = 4
This matches the $\hat{i}$-component of the given force, confirming that our values are correct.

Substituting $B\; =\; -6$ and $B$₀ = -8 back into the expression for $\overrightarrow{B}$:
$\overrightarrow{B}=\; -6\hat{i}-\; 6\hat{j}-\; 8\hat{k}$