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In the product vector F = q(vector v x B) = q vector v x (Bi + Bj + B0k)For q = 1 and

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In the product

\(\vec{F} = q(\vec{v} \times \vec{B})\)

= \(q\vec{v}\times (B\hat{i}+ B\hat{j}+B_0\hat{k})\)

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

(1) \(-8\hat{i}- 8\hat{j} - 6\hat{k})\)

(2) \(-6\hat{i}- 6\hat{j} - 8\hat{k})\)

(3) \(-8\hat{i}+ 8\hat{j} - 6\hat{k})\)

(4) \(6\hat{i}+6\hat{j} - 8\hat{k})\)

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Correct answer: (2) \(-6\hat{i}- 6\hat{j} - 8\hat{k})\)

Explanation:

\(\vec{F} = q(\vec{v} \times \vec{B})\)

\(4\hat{i}- 20\hat{j} +12\hat{k}\) = 1\(\begin{vmatrix}
i & j & \hat{k} \\[0.3em]
2 & 4 & 6 \\[0.3em]
B & B & B_0
\end{vmatrix}\)

Comparing

⇒ 4 = 4B0 - 6B

-20 = -2B0 + 6B

12 = 2B - 4B 

Solving B = -6

B0 = -8

∴ \(\vec{B} = -6\hat{i}- 6\hat{j} - 8\hat{k})\)

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