JEE MAIN - Physics (2022 - 26th July Evening Shift - No. 16)
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 $$=9.1 \times 10^{-31} \mathrm{~kg}$$ )
$$192 \,\mathrm{kVm}^{-1}$$
$$192 \,\mathrm{mVm}^{-1}$$
$$9600 \,\mathrm{kVm}^{-1}$$
$$16 \,\mathrm{kVm}^{-1}$$
Explanation
$$v = {E \over B}$$ and $$K = {1 \over 2}m{v^2}$$
$$ \Rightarrow \sqrt {{{2K} \over m}} \times B = E$$
$$ \Rightarrow E = \sqrt {{{2 \times 728 \times 1.6 \times {{10}^{ - 19}}} \over {9.1 \times {{10}^{ - 31}}}}} \times 12 \times {10^{ - 3}}$$
$$ = 192000$$ V/m
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