Question

A velocity selector consists of electric field $\overrightarrow{E}=E\hat{k}$ and magnetic field $\overrightarrow{B}=B\hat{j}$ with $B=12mT$. The value $E$ required for an electron of energy $728eV$ moving along the positive $x$-axis to pass undeflected is :
(Given, mass of electron $=9.1\times 10^{31}kg$ )

• A. $192kV{m}^{-1}$
• B. $192mV{m}^{-1}$
• C. $9600kV{m}^{-1}$
• D. $16kV{m}^{-1}$

Answer 1

Answer: (A)

$\overrightarrow{E}=E\hat{k}B=12mT$
$\overrightarrow{B}=B\hat{j}\text{ Energy }=728eV$
$\text{ Energy }=\frac{1}{2}m{v}^2$
$728eV=\frac{1}{2}\times 9.1\times 10^{-31}\times v^2$
$728\times 1.6\times 10^{-19}-\frac{1}{2}\times 9.1\times 10^{-31}\times v^2$
$v=16\times 10^{6}m/s$
$E=vB$
$E=16\times 10^6\times 12\times 10^{-3}$
$E=192\times 10^{3}V/m$