Question

A velocity selector consists of electric field $\overrightarrow E=E\hat K$ and magnetic field $\overrightarrow B=B\hat j$ with $B = 12 \;mT$. The value of 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)

• A. $192\; kVm^{–1}$
• B. $192\; mVm^{–1}$
• C. $9600 \;kVm^{–1}$
• D. $16 \;kVm^{–1}$

Answer 1

Answer: (A)

$192\; kVm^{–1}$

Answer 2

$v = \frac{E}{B}$

thereafter,$K = \frac{1}{2}mv^2$

$⇒ \sqrt\frac{2K}{{m}} \times B = E$

$⇒ E = \sqrt{ \frac{2 \times 728 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} \times 12 \times 10^{-3}$

= $192000 \;V/m$