Question

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, mass of electron $=9.1 \times 10^{31} kg$ )

• A. $192\, kVm ^{-1}$
• B. $192 \, m \, Vm ^{-1}$
• C. $9600\, kVm ^{-1}$
• D. $16 \, kVm ^{-1}$

Answer 1

Answer: (A)

$ \overrightarrow{ E }= E \hat{ k } B =12 mT$
$ \overrightarrow{ B }= B \hat{ j } \text { Energy }=728 eV $
$\text { 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 = vB $
$ E =16 \times 10^6 \times 12 \times 10^{-3} $
$ E =192 \times 10^3 V / m$