A velocity selector consists of electric field $\vec{E}=E \,\hat{k}$ and magnetic field $\vec{B}=B \,\hat{j}$ with $B=12 \,m T$. The value of $E$ required for an electron of energy $728 \,\mathrm{e} V$ moving along the positive $x$-axis to pass undeflected is :
(Given, mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ )
Solution
<p>$v = {E \over B}$ and $K = {1 \over 2}m{v^2}$</p>
<p>$\Rightarrow \sqrt {{{2K} \over m}} \times B = E$</p>
<p>$$ \Rightarrow E = \sqrt {{{2 \times 728 \times 1.6 \times {{10}^{ - 19}}} \over {9.1 \times {{10}^{ - 31}}}}} \times 12 \times {10^{ - 3}}$$</p>
<p>$= 192000$ V/m</p>
About this question
Subject: Physics · Chapter: Electromagnetic Waves · Topic: Maxwell's Equations
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