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 given by $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B})$.First,calculate the magnetic force component: $\vec{v} 	imes \vec{B} = (1.28 	imes 10^6 \ \hat{i}) 	imes (8 	imes 10^{-2} \ \hat{j}) = (1.28 	imes 8 	imes 10^4) \ (\hat{i} 	imes \hat{j}) = 10.24 	imes 10^4 \ \hat{k} = 1.024 	imes 10^5 \ \hat{k} \ Vm^{-1}$.Comparing the magnitudes: $|E| = 102.4 	imes 10^3 = 1.024 	imes 10^5 \ Vm^{-1}$.Since $\vec{E} = -1.024 	imes 10^5 \ \hat{k} \ Vm^{-1}$ and $\vec{v} 	imes \vec{B} = 1.024 	imes 10^5 \ \hat{k} \ Vm^{-1}$,the net force $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B}) = q(-1.024 	imes 10^5 \ \hat{k} + 1.024 	imes 10^5 \ \hat{k}) = 0$.Since the net force is zero,the particle will continue to move in its original direction,which is along the positive $X$-axis.

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