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

The Correct Option is A

\(v = \frac{E}{B}\)

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

\(⇒ \sqrt\frac{2K}{{m}} \times B = E\)

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

= \(192000 \;V/m\)