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)

Question

A thin half ring of radius $35 \text{ cm}$ is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at centre of the half ring is $100 \text{ V/m}$, then the value of $Q$ is ______ $\text{nC}$.
($\epsilon_o = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$ and $\pi = 3.14$)

Answer 1

To determine the value of the electric field $ E $ required for the electron to pass undeflected through the velocity selector, we need to set the forces on the electron due to the electric and magnetic fields equal. The velocity selector works by ensuring that the electric force $ F_e = eE $ and the magnetic force $ F_m = evB $ cancel each other out for a charged particle like an electron.

Given: $ B = 12 \; \text{mT} = 12 \times 10^{-3} \; \text{T} $.

Energy of the electron: $ 728 \; \text{eV} $.

First, convert the electron's energy from electron volts to joules: $ 1 \; \text{eV} = 1.6 \times 10^{-19} \; \text{J} $ thus, $ 728 \times 1.6 \times 10^{-19} \; \text{J} $.

Using the kinetic energy formula, we can solve for the velocity $ v $ of the electron: $ \frac{1}{2} mv^2 = 728 \times 1.6 \times 10^{-19} $.

Mass of the electron $ m = 9.1 \times 10^{-31} \; \text{kg} $. Solving for $ v $: \[ v = \sqrt{\frac{2 \times 728 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} \]

Calculate the velocity $ v $ (use a calculator for precise values), and then equate the electric and magnetic forces: $ eE = evB $.

Simplifying gives: $ E = vB $.

Substitute the values of $ v $ and $ B $ calculated and given earlier to solve for $ E $.

Performing the calculation, we find: \[ E = \left(\sqrt{\frac{2 \times 728 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} \right) \times (12 \times 10^{-3}) \]

This results in: $ E = 192 \; \text{kV/m} $ which matches the given correct answer option.

Thus, the correct answer is: $ 192 \; \text{kV/m} $.

Answer 2