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 force experienced by an electron moving near an infinitely long straight conductor carrying a current, we use the magnetic force formula combined with the Biot-Savart law to find the magnetic field created by the current in the wire.

Step 1: Determine the magnetic field produced by the conductor 

The magnetic field \(B\) at a distance \(r\) from an infinitely long straight wire carrying current \(I\) is given by the Biot-Savart law:

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

where \(\mu_0\) is the permeability of free space, \(4\pi \times 10^{-7} \, T \cdot m/A\), \(I = 5 \, A\), and \(r = 20 \, cm = 0.2 \, m\).

Substituting the values, we get:

\(B = \frac{4\pi \times 10^{-7} \times 5}{2 \pi \times 0.2} = \frac{20 \times 10^{-7}}{0.4} = 5 \times 10^{-6} \, T\)

Step 2: Calculate the magnetic force on the moving electron

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

\(F = qvB\)

Here, \(q = -1.6 \times 10^{-19} \, C\) (charge of an electron), \(v = 10^5 \, m/s\), and \(B = 5 \times 10^{-6} \, T\).

Calculating the magnitude of the force:

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

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

Since the force is a vector, we take the magnitude, resulting in \(8 \times 10^{-20} \, N\).

This corresponds to the correct answer choice of \(8 \times 10^{-20} \, N\).