An electric charge +q moves with velocity ove | vedclass

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Question

(B) The total Lorentz force acting on the charge is given by $\overrightarrow{F} = q\overrightarrow{E} + q(\overrightarrow{V} 	imes \overrightarrow{B

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) The total Lorentz force acting on the charge is given by $\overrightarrow{F} = q\overrightarrow{E} + q(\overrightarrow{V} 	imes \overrightarrow{B})$.First,calculate the cross product $\overrightarrow{V} 	imes \overrightarrow{B}$:$\overrightarrow{V} 	imes \overrightarrow{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 1 \ 1 & 1 & -3 \end{vmatrix}$$= \hat{i}(-12 - 1) - \hat{j}(-9 - 1) + \hat{k}(3 - 4)$$= -13\hat{i} + 10\hat{j} - \hat{k}$.Now,substitute this into the force equation:$\overrightarrow{F} = q[(3\hat{i} + \hat{j} + 2\hat{k}) + (-13\hat{i} + 10\hat{j} - \hat{k})]$$\overrightarrow{F} = q[(3 - 13)\hat{i} + (1 + 10)\hat{j} + (2 - 1)\hat{k}]$$\overrightarrow{F} = q[-10\hat{i} + 11\hat{j} + \hat{k}]$.The $y$-component of the force is the coefficient of $\hat{j}$,which is $11q$.

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