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

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: $E = 4 \, \text{N/C}$.

Answer 2

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: $E = 4 \, \text{N/C}$.