Question

An electron with kinetic energy \(5 \, \text{eV}\) enters a region of uniform magnetic field of \(3 \, \mu\text{T}\) perpendicular to its direction. An electric field \(E\) is applied perpendicular to the direction of velocity and magnetic field. The value of \(E\), so that the electron moves along the same path, is ______ \( \text{NC}^{-1} \).
Given: mass of electron = \(9 \times 10^{-31} \, \text{kg}\), electric charge = \(1.6 \times 10^{-19} \, \text{C}\)

Answer 1

Correct Answer: 4

To maintain the electron's straight trajectory, the magnetic and electric forces must be balanced. The magnetic force is given by \(F_{\text{magnetic}}=qvB\), and the electric force by \(F_{\text{electric}}=qE\). Equating these forces, \(qvB=qE\), yields \(E=vB\).

First, the electron's velocity \(v\) is determined from its kinetic energy:

\(K.E.=\frac{1}{2}mv^2=5 \, \text{eV}=5 \times 1.6 \times 10^{-19} \, \text{J}\).

Rearranging the equation to solve for \(v^2\):

\(v^2=\frac{2 \times 5 \times 1.6 \times 10^{-19}}{9 \times 10^{-31}}\)

The calculated velocity \(v\) is:

\(v=\sqrt{\frac{16 \times 10^{-19}}{9 \times 10^{-31}}}\approx1.32 \times 10^6 \, \text{ms}^{-1}\)

Next, the electric field \(E\) is calculated using \(E=vB\), with \(B=3 \, \mu\text{T}=3 \times 10^{-6} \, \text{T}\):

\(E=1.32 \times 10^6 \times 3 \times 10^{-6}=3.96 \, \text{NC}^{-1}\)

The computed electric field \(E\approx3.96 \, \text{NC}^{-1}\) is within the specified range of 4,4, confirming its validity. The balance of electromagnetic forces ensures the electron's path remains unobstructed.