An electron enters an electric field with int | vedclass

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(D) The total Lorentz force on the electron is given by $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B})$.Given: $\vec{E} = 3\hat{i} + 6\hat{j} + 2\hat{

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(D) The total Lorentz force on the electron is given by $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B})$.Given: $\vec{E} = 3\hat{i} + 6\hat{j} + 2\hat{k} 	ext{ V m}^{-1}$,$\vec{B} = 2\hat{i} + 3\hat{j} 	ext{ T}$,$\vec{v} = 2\hat{i} + 3\hat{j} 	ext{ m s}^{-1}$,and $q = -1.6 	imes 10^{-19} 	ext{ C}$.First,calculate the magnetic force $\vec{F}_m = q(\vec{v} 	imes \vec{B})$.Since $\vec{v} = 2\hat{i} + 3\hat{j}$ and $\vec{B} = 2\hat{i} + 3\hat{j}$,the vectors are parallel $(\vec{v} \parallel \vec{B})$.Therefore,$\vec{v} 	imes \vec{B} = 0$,which implies $\vec{F}_m = 0$.Now,calculate the electric force $\vec{F}_e = q\vec{E}$.$\vec{F}_e = (-1.6 	imes 10^{-19}) (3\hat{i} + 6\hat{j} + 2\hat{k}) = -4.8 	imes 10^{-19}\hat{i} - 9.6 	imes 10^{-19}\hat{j} - 3.2 	imes 10^{-19}\hat{k} 	ext{ N}$.The magnitude of the force is $|\vec{F}| = |\vec{F}_e| = 1.6 	imes 10^{-19} 	imes \sqrt{3^2 + 6^2 + 2^2} = 1.6 	imes 10^{-19} 	imes \sqrt{9 + 36 + 4} = 1.6 	imes 10^{-19} 	imes \sqrt{49} = 1.6 	imes 10^{-19} 	imes 7 = 1.12 	imes 10^{-18} 	ext{ N}$.Since $1.12 	imes 10^{-18} 	ext{ N}$ is not among the options,the correct choice is $D$.

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