In the product overrightarrow{F} = q(vec{v} t | vedclass

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(B) Given the Lorentz force equation: $\overrightarrow{F} = q(\vec{v} 	imes \overrightarrow{B})$.Since $q = 1$,we have $\overrightarrow{F} = \vec{v}

Vedclass · Accepted Answer

$-6 \hat{i} - 6 \hat{j} - 8 \hat{k}$

Vedclass · Answer

(B) Given the Lorentz force equation: $\overrightarrow{F} = q(\vec{v} 	imes \overrightarrow{B})$.Since $q = 1$,we have $\overrightarrow{F} = \vec{v} 	imes \overrightarrow{B}$.Let $\overrightarrow{B} = B \hat{i} + B \hat{j} + B_0 \hat{k}$.The cross product is given by the determinant:$\vec{v} 	imes \overrightarrow{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 4 & 6 \ B & B & B_0 \end{vmatrix} = \hat{i}(4B_0 - 6B) - \hat{j}(2B_0 - 6B) + \hat{k}(2B - 4B) = 4 \hat{i} - 20 \hat{j} + 12 \hat{k}$.Comparing the components:$1$) $4B_0 - 6B = 4$$2$) $-(2B_0 - 6B) = -20 \implies 2B_0 - 6B = 20$$3$) $2B - 4B = 12 \implies -2B = 12 \implies B = -6$.Substitute $B = -6$ into equation $(2)$:$2B_0 - 6(-6) = 20 \implies 2B_0 + 36 = 20 \implies 2B_0 = -16 \implies B_0 = -8$.Thus,$\overrightarrow{B} = -6 \hat{i} - 6 \hat{j} - 8 \hat{k}$.

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