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A velocity selector consists of electric field $\overrightarrow{ E }= E \hat{ k }$ and magnetic field $\overrightarrow{ B }= B \hat{ j }$ with $B =12 mT$.
The value $E$ required for an electron of energy $728 eV$ moving along the positive $x$-axis to pass undeflected is:
(Given, , ass of electron $=9.1 \times 10^{-31} kg$ )
$192\, k\,Vm ^{-1}$
$192\, m\, Vm ^{-1}$
$9600\, k\,Vm ^{-1}$
$16 \,k\,Vm ^{-1}$
Solution
$\overrightarrow{ E }= E \hat{ k } B =12 mT$
$\overrightarrow{ B }= B \hat{ j } \text { Energy }=728\,eV$
Energy $=\frac{1}{2} mv ^{2}$
$728\,eV =\frac{1}{2} \times 9.1 \times 10^{-31} \times v ^{2}$
$728 \times 1.6 \times 10^{-19}=\frac{1}{2} \times 9.1 \times 10^{-31} \times v ^{2}$
$v =16 \times 10^{6} m / s$
$E=v B$
$E =16 \times 10^{6} \times 12 \times 10^{-3}$
$E =192 \times 10^{3}\,V / m$
