A velocity selector consists of electric field $\vec{E} = E \hat{k}$ and magnetic field $\vec{B} = B \hat{j}$ 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 = 9.1×10^–31 kg)

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192 kVm^–1

16 kVm^–1

9600 kVm^–1

192 m Vm^–1

answer is A.

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Detailed Solution

$\begin{array}{l} \vec{E} = E \hat{k} B = 12 mT \\ \vec{B} = B \hat{j} Energy = 728 eV \\ Energy = \frac{1}{2} {mv}^{2} \\ 728 eV = \frac{1}{2} \times 9 .1 \times 10^{- 31} \times v^{2} \\ 728 \times 1 .6 \times 10^{- 19} = \frac{1}{2} \times 9 .1 \times 10^{- 31} \times v^{2} \\ v = 16 \times 10^{6} m / s \\ E = vB \\ E = 16 \times 10^{6} \times 12 \times 10^{- 3} \\ E = 192 \times 10^{3} V / m \end{array}$