An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of 10⁵ m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

Correct Answer: 8 x 10⁻²⁰ N

To find the magnitude of the force experienced by the electron, we can break the solution down into two main steps: finding the magnetic field produced by the current-carrying conductor, and then calculating the magnetic force on the moving electron.

Step 1: Calculate the magnetic field (B) produced by the conductor
The magnetic field at a perpendicular distance $r$ from an infinitely long straight conductor carrying current $I$ is given by Ampere's Law:
$B=\frac{{\mu }_{0}I}{2\pi r}$
Given data:
�� Current, $I=5\text{A}$
• Distance, $r=20\text{cm}=0.2\text{m}$
• Permeability of free space, ${\mu }_0=4\pi \times {10}^{-7}\text{T}·\text{m/A}$

Substituting these values into the formula:
$B=\frac{(4\pi \times {10}^{-7})\times 5}{2\pi \times 0.2}$
$B=\frac{2\times {10}^{-7}\times 5}{0.2}$
$B=\frac{{10}^{-6}}{0.2}=5\times {10}^{-6}\text{T}$

Step 2: Calculate the magnetic force (F) on the electron
The force experienced by a charge $q$ moving with velocity $v$ in a magnetic field $B$ is given by the Lorentz force formula:
$F=qvBsin\theta$
Where:
• Charge of an electron, $q=1.6\times {10}^{-19}\text{C}$
• Speed of the electron, $v={10}^5\text{m/s}$
• Since the electron is moving parallel to the conductor, its velocity vector is perpendicular to the concentric circular magnetic field lines around the conductor. Thus, the angle between the velocity vector and the magnetic field vector is $\theta =90°$ (and $sin90°=1$).

Substituting the values into the force equation:
$F=(1.6\times {10}^{-19})\times {10}^5\times (5\times {10}^{-6})\times 1$
$F=1.6\times 5\times {10}^{-19+5-6}$
$F=8\times {10}^{-20}\text{N}$

Therefore, the magnitude of the force experienced by the electron is $8\times {10}^{-20}\text{N}$.