Question

An infinitely long straight conductor carries a current of $5\, A$ as shown. An electron is moving with a speed of $10^{5} m / s$ 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.

• A. $4 \times 10^{-20} N$
• B. $8 \pi \times 10^{-20} N$
• C. $4 \pi \times 10^{-20} N$
• D. $8 \times 10^{-20} N$

Answer 1

The Correct Option is D

To determine the magnitude of the force experienced by an electron moving parallel to an infinitely long straight conductor carrying a current, we can use the following steps: 

1. Calculate the magnetic field generated by the conductor at the location of the electron.
2. Use the formula for the magnetic force on a moving charge to find the force on the electron.

Step 1: Calculate the Magnetic Field

The magnetic field \( B \) at a distance \( r \) from an infinitely long straight conductor carrying a current \( I \) is given by Ampère's Law:

\(B = \frac{\mu_0 I}{2 \pi r}\)

Where:

• \( \mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A \) (permeability of free space)
• \( I = 5 \, A \) (current through the conductor)
• \( r = 20 \, cm = 0.2 \, m \) (perpendicular distance from the conductor)

\(B = \frac{4 \pi \times 10^{-7} \cdot 5}{2 \pi \cdot 0.2}\)

\(B = \frac{20 \times 10^{-7}}{0.4}\)

\(B = 5 \times 10^{-6} \, T\)

Step 2: Calculate the Magnetic Force

The force \( F \) on a charge \( q \) moving with velocity \( v \) in a magnetic field \( B \) is given by:

\(F = qvB \sin \theta\)

For this case, the electron is moving parallel to the conductor, so the angle \( \theta \) between the direction of velocity and magnetic field is \( 90^\circ \), thus \( \sin 90^\circ = 1 \).

• Charge of an electron \( q = -1.6 \times 10^{-19} \, C \)
• Velocity \( v = 10^5 \, m/s \)
• Magnetic field \( B = 5 \times 10^{-6} \, T \)

\(F = 1.6 \times 10^{-19} \cdot 10^5 \cdot 5 \times 10^{-6}\)

\(F = 8 \times 10^{-20} \, N\)

Thus, the magnitude of the force experienced by the electron is \( 8 \times 10^{-20} \, N \). This matches the correct answer option: \( 8 \times 10^{-20} \, N \).