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(D) For the electron to move undeflected along the $x$-axis,the net Lorentz force must be zero: $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B}) = 0$.Gi

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$5 	imes 10^{-3} \, T$,along $-z$ direction

Vedclass · Answer

(D) For the electron to move undeflected along the $x$-axis,the net Lorentz force must be zero: $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B}) = 0$.Given: $\vec{v} = 6 	imes 10^{6} \hat{i} \, m/s$ and $\vec{E} = 300 \, V/cm = 3 	imes 10^{4} \, V/m$ along $+y$ direction,so $\vec{E} = 3 	imes 10^{4} \hat{j} \, V/m$.The electric force on the electron is $\vec{F}_e = q\vec{E} = (-e)(3 	imes 10^{4} \hat{j}) = -3 	imes 10^{4} e \hat{j}$.To counteract this,the magnetic force $\vec{F}_m = q(\vec{v} 	imes \vec{B})$ must be in the $+y$ direction.Since $q = -e$,we need $(-e)(v\hat{i} 	imes \vec{B}) = +3 	imes 10^{4} e \hat{j}$.This implies $\vec{v} 	imes \vec{B}$ must be in the $-y$ direction. Since $\hat{i} 	imes (-\hat{k}) = \hat{j}$ and $\hat{i} 	imes \hat{k} = -\hat{j}$,we find that $\vec{B}$ must be along the $-z$ direction.Using the magnitude condition $E = vB$,we get $B = \frac{E}{v} = \frac{3 	imes 10^{4}}{6 	imes 10^{6}} = 0.5 	imes 10^{-2} = 5 	imes 10^{-3} \, T$.Thus,the magnetic field is $5 	imes 10^{-3} \, T$ along the $-z$ direction.

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