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

For the given condition of moving undeflected, the net force should be zero:
\[qE = qvB \implies E = vB\]
The velocity $v$ can be expressed in terms of kinetic energy:
\[v = \sqrt{\frac{2KE}{m}}\]
Substituting this into the expression for $E$:
\[E = \sqrt{\frac{2KE}{m}} \cdot B\]
Substituting the given values:
\[E = \sqrt{\frac{2 \cdot 5 \cdot 1.6 \times 10^{-19}}{9 \times 10^{-31}}} \cdot 3 \times 10^{-6}\]
Calculating:
\[E = \sqrt{\frac{16 \times 10^{-19}}{9 \times 10^{-31}}} \cdot 3 \times 10^{-6}\]
\[E = \sqrt{\frac{1.6 \times 10^{12}}{9}} \cdot 3 \times 10^{-6}\]
\[E = \sqrt{1.78 \times 10^{12}} \cdot 3 \times 10^{-6}\]
\[E = 4 \, \text{N/C}\]
Final Answer: $E = 4 \, \text{N/C}$.