A particle of mass 1 times 10^{-26} ,kg and c | vedclass

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(A) Given: Mass $m = 1 	imes 10^{-26} \,kg$, Charge $q = 1.6 	imes 10^{-19} \,C$, Velocity $\vec{v} = 1.28 	imes 10^6 \hat{i} \,ms^{-1}$.Electric f

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along the positive $X$-axis

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(A) Given: Mass $m = 1 	imes 10^{-26} \,kg$, Charge $q = 1.6 	imes 10^{-19} \,C$, Velocity $\vec{v} = 1.28 	imes 10^6 \hat{i} \,ms^{-1}$.Electric field $\vec{E} = -102.4 	imes 10^3 \hat{k} \,NC^{-1}$.Magnetic field $\vec{B} = 8 	imes 10^{-2} \hat{j} \,Wbm^{-2}$.The Lorentz force on the particle is $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B})$.Calculate the magnetic force: $\vec{v} 	imes \vec{B} = (1.28 	imes 10^6 \hat{i}) 	imes (8 	imes 10^{-2} \hat{j}) = (1.28 	imes 8 	imes 10^4) (\hat{i} 	imes \hat{j}) = 10.24 	imes 10^4 \hat{k} = 1.024 	imes 10^5 \hat{k} \,Vm^{-1}$.Since $102.4 	imes 10^3 = 1.024 	imes 10^5$, we have $\vec{E} = -1.024 	imes 10^5 \hat{k} \,NC^{-1}$.Thus, $\vec{F} = q(-1.024 	imes 10^5 \hat{k} + 1.024 	imes 10^5 \hat{k}) = 0$.Since the net Lorentz force is zero, the particle will remain undeflected and continue moving along the positive $X$-axis.

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