Question

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

• A. 8×10^-20N
• B. 4×10^-20N
• C. 8π×10^-20N
• D. 4π×10^-20N

Answer 1

The Correct Option is A

To solve this problem, we need to calculate the magnetic force experienced by the electron due to the current in the conductor. The force on a charge moving in a magnetic field is given by \(F = q(v \times B)\), where:

• \(q\) is the charge of the electron \((-1.6 \times 10^{-19} \, \text{C})\)
• \(v\) is the velocity of the electron \((10^{5} \, \text{m/s})\)
• \(B\) is the magnetic field created by the current-carrying conductor

The magnetic field \((B)\) at a distance \(r\) from a long straight conductor is given by Ampère's law:

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

where:

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

Substituting these values into the formula for \(B\):

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

Now, calculate the force using the formula \(F = q(v \times B)\):

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

\(F = -1.6 \times 5 \times 10^{-20}\)

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

The magnitude of the force is \(8 \times 10^{-20} \, \text{N}\).

Therefore, the correct answer is:

8×10^-20N