An electron (mass 9 times 10^{-31} kg and ch | vedclass

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(B) For the electron to move undeflected,the net Lorentz force must be zero: $\vec{F}_e + \vec{F}_m = 0$,which implies $\vec{F}_e = -\vec{F}_m$.Since

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$\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 	imes 10^2 \ V \ m^{-1}$

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(B) For the electron to move undeflected,the net Lorentz force must be zero: $\vec{F}_e + \vec{F}_m = 0$,which implies $\vec{F}_e = -\vec{F}_m$.Since $\vec{F}_m = q(\vec{v} 	imes \vec{B})$,the magnitude is $F_m = qvB \sin(90^\circ) = qvB$.The electric force is $F_e = qE$.Equating the magnitudes: $qE = qvB \implies E = vB$.Given $v = c/100 = (3 	imes 10^8) / 100 = 3 	imes 10^6 \ ms^{-1}$.Given $B = 9 	imes 10^{-4} \ T$.Calculating $E$: $E = (3 	imes 10^6) 	imes (9 	imes 10^{-4}) = 27 	imes 10^2 \ V \ m^{-1}$.For the forces to cancel,$\vec{E}$ must be perpendicular to both $\vec{v}$ and $\vec{B}$. Since $\vec{v}$ is perpendicular to $\vec{B}$,$\vec{E}$ must be perpendicular to $\vec{B}$.

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