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A velocity selector consists of electric field $\mathrm{\overrightarrow{E}=E \: \hat{k}}$ and magnetic field $\mathrm{\overrightarrow{B}=B \: \hat{j}}$ with $\mathrm{B=12\: \mathrm{mT}}$ . The value of $\mathrm{E}$ required for an electron of energy $\mathrm{ 728 \: \mathrm{eV}}$ moving along the positive $\mathrm{ x-axis}$ to pass undeflected is :

$\mathrm{ \text { (Given, mass of electron }=9.1 \times 10^{-31} \mathrm{~kg} \text { ) }}$
Option: 1
$\mathrm{192\: \mathrm{kVm}^{-1}}$

Option: 2
$\mathrm{192\: \mathrm{mVm}^{-1}}$

Option: 3
$\mathrm{9600\: \mathrm{kVm}^{-1}}$

Option: 4
$\mathrm{16 \: \mathrm{kVm}^{-1}}$

Answers (1)

best_answer

$\mathrm{\bar{E}=E(\hat{k})}$

$\mathrm{\bar{B}=B(\hat{j})}$

$\mathrm{B=12\: mT}$

$\mathrm{ K E=\frac{1}{2} m v^2=728 \mathrm{ev} \rightarrow(1 )}$
For velocity selector,

$\mathrm{\bar{F}_E=-\bar{F}_m}$

$\mathrm{q \bar{E}=-q(\bar{V} \times \bar{B})}$

$\mathrm{E=e(V B) \sin 90^{\circ}}$

$\mathrm{(\because q=-e)}$

$\mathrm{v=\frac{E}{B} \rightarrow (2)}$

$\mathrm{E=V B}$

$\mathrm{E=\left(\frac{728 \times 1.6 \times 10^{-19} \times 2}{9.1 \times 10^{-31}}\right)^{1 / 2} \times 12 \times 10^{-3}}$

$\mathrm{E=192 \: \mathrm{kVm}^{-1}}$

Hence (1) is correct option

Posted by

Devendra Khairwa

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