An electron moves with speed 2 times 10^5 m/ | vedclass

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(C) Given: Velocity $\vec{v} = 2 	imes 10^5 \hat{i} \ m/s$,Magnetic field $\vec{B} = (\hat{i} + 4\hat{j} - 3\hat{k}) \ T$,and charge $q = -1.6 	imes

Vedclass · Accepted Answer

$1.6 	imes 10^{-13}$

Vedclass · Answer

(C) Given: Velocity $\vec{v} = 2 	imes 10^5 \hat{i} \ m/s$,Magnetic field $\vec{B} = (\hat{i} + 4\hat{j} - 3\hat{k}) \ T$,and charge $q = -1.6 	imes 10^{-19} \ C$.Using the Lorentz force formula: $\vec{F} = q(\vec{v} 	imes \vec{B})$.$\vec{F} = -1.6 	imes 10^{-19} 	imes [2 	imes 10^5 \hat{i} 	imes (\hat{i} + 4\hat{j} - 3\hat{k})]$.$\vec{F} = -3.2 	imes 10^{-14} [(\hat{i} 	imes \hat{i}) + 4(\hat{i} 	imes \hat{j}) - 3(\hat{i} 	imes \hat{k})]$.Since $\hat{i} 	imes \hat{i} = 0$,$\hat{i} 	imes \hat{j} = \hat{k}$,and $\hat{i} 	imes \hat{k} = -\hat{j}$:$\vec{F} = -3.2 	imes 10^{-14} [0 + 4\hat{k} - 3(-\hat{j})] = -3.2 	imes 10^{-14} [4\hat{k} + 3\hat{j}]$.The magnitude of the force is $|\vec{F}| = 3.2 	imes 10^{-14} \sqrt{4^2 + 3^2}$.$|\vec{F}| = 3.2 	imes 10^{-14} 	imes \sqrt{16 + 9} = 3.2 	imes 10^{-14} 	imes 5$.$|\vec{F}| = 1.6 	imes 10^{-13} \ N$.

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