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Question

(D) According to the Lorentz force law,the total force on a charge $q$ is $\vec F = q(\vec E + \vec v 	imes \vec B)$.Since the charge experiences no

Vedclass · Accepted Answer

$\vec E = - v_0 B_0(14\hat j + 7\hat k)$

Vedclass · Answer

(D) According to the Lorentz force law,the total force on a charge $q$ is $\vec F = q(\vec E + \vec v 	imes \vec B)$.Since the charge experiences no force,$\vec F = 0$,which implies $\vec E + \vec v 	imes \vec B = 0$,or $\vec E = -(\vec v 	imes \vec B)$.Given $\vec v = v_0(3\hat i - \hat j + 2\hat k)$ and $\vec B = B_0(\hat i + 2\hat j - 4\hat k)$,we calculate the cross product $\vec v 	imes \vec B$:$\vec v 	imes \vec B = v_0 B_0 \begin{vmatrix} \hat i & \hat j & \hat k \ 3 & -1 & 2 \ 1 & 2 & -4 \end{vmatrix}$$= v_0 B_0 [\hat i((-1)(-4) - (2)(2)) - \hat j((3)(-4) - (2)(1)) + \hat k((3)(2) - (-1)(1))]$$= v_0 B_0 [\hat i(4 - 4) - \hat j(-12 - 2) + \hat k(6 + 1)]$$= v_0 B_0 [0\hat i + 14\hat j + 7\hat k] = v_0 B_0(14\hat j + 7\hat k)$.Therefore,$\vec E = -(\vec v 	imes \vec B) = -v_0 B_0(14\hat j + 7\hat k)$.

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