A velocity selector consists of an electric f | vedclass

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Question

(A) For a particle to pass undeflected in a velocity selector,the electric force must balance the magnetic force: $qE = qvB$,which implies $E = vB$.Fi

Vedclass · Accepted Answer

$192 \, kVm^{-1}$

Vedclass · Answer

(A) For a particle to pass undeflected in a velocity selector,the electric force must balance the magnetic force: $qE = qvB$,which implies $E = vB$.First,calculate the velocity $v$ of the electron using its kinetic energy $K = 728 \, eV$:$K = \frac{1}{2} mv^2$$728 	imes 1.6 	imes 10^{-19} \, J = \frac{1}{2} 	imes 9.1 	imes 10^{-31} \, kg 	imes v^2$$v^2 = \frac{2 	imes 728 	imes 1.6 	imes 10^{-19}}{9.1 	imes 10^{-31}}$$v^2 = 256 	imes 10^{12} \, m^2/s^2$$v = 16 	imes 10^6 \, m/s$Now,calculate the electric field $E$:$E = vB = (16 	imes 10^6 \, m/s) 	imes (12 	imes 10^{-3} \, T)$$E = 192 	imes 10^3 \, V/m = 192 \, kV/m$.

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