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)
- \(192\; kVm^{–1}\)
- \(192\; mVm^{–1}\)
- \(9600 \;kVm^{–1}\)
- \(16 \;kVm^{–1}\)
The Correct Option is A
Solution and Explanation
\(v = \frac{E}{B}\)
thereafter,\( K = \frac{1}{2}mv^2\)
\(⇒ \sqrt\frac{2K}{{m}} \times B = E\)
\(⇒ E = \sqrt{ \frac{2 \times 728 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} \times 12 \times 10^{-3}\)
= \(192000 \;V/m\)
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).