Question

A long straight wire carries a current of 10 A. A proton moves parallel to the wire at a distance of 0.05 m with a velocity of \( 2 \times 10^5 \, \text{m/s} \) in the same direction as the current. Find the magnitude of the magnetic force acting on the proton. (Given: Charge of proton \( q = 1.6 \times 10^{-19} \, \text{C} \), permeability of free space \( \mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A} \)).

• A. \( 2.56 \times 10^{-19} \, \text{N} \)
• B. \( 1.28 \times 10^{-19} \, \text{N} \)
• C. \( 5.12 \times 10^{-19} \, \text{N} \)
• D. \( 3.84 \times 10^{-19} \, \text{N} \)

Hint:

When calculating the magnetic force on a charged particle, ensure the angle between the velocity and the magnetic field is correctly identified. For a current-carrying wire, the magnetic field forms concentric circles, so the field is perpendicular to the direction of the current and the particle’s velocity if it moves parallel to the wire.

Answer 1

The Correct Option is B

The magnetic force on a charged particle moving in a magnetic field is given by the Lorentz force formula: \[ F = q v B \sin\theta \] Where: - \( q = 1.6 \times 10^{-19} \, \text{C} \) (charge of the proton), - \( v = 2 \times 10^5 \, \text{m/s} \) (velocity of the proton), - \( B \) is the magnetic field due to the current-carrying wire, - \( \theta \) is the angle between the velocity of the proton and the magnetic field. Since the proton moves parallel to the wire and the current, the magnetic field produced by the wire is perpendicular to the velocity of the proton. Thus, \( \theta = 90^\circ \), and \( \sin\theta = 1 \). First, calculate the magnetic field \( B \) at a distance \( r = 0.05 \, \text{m} \) from the wire using the formula for the magnetic field due to a long straight current-carrying wire: \[ B = \frac{\mu_0 I}{2 \pi r} \] Where: - \( \mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A} \), - \( I = 10 \, \text{A} \), - \( r = 0.05 \, \text{m} \). Substitute the values: \[ B = \frac{(4\pi \times 10^{-7}) \times 10}{2 \pi \times 0.05} \] \[ B = \frac{4 \times 10^{-6}}{0.1} = 4 \times 10^{-5} \, \text{T} \] Now, calculate the magnetic force: \[ F = q v B \sin\theta = (1.6 \times 10^{-19}) \times (2 \times 10^5) \times (4 \times 10^{-5}) \times 1 \] \[ F = 1.6 \times 2 \times 4 \times 10^{-19 + 5 - 5} \] \[ F = 12.8 \times 10^{-20} = 1.28 \times 10^{-19} \, \text{N} \] Thus, the magnitude of the magnetic force acting on the proton is \( 1.28 \times 10^{-19} \, \text{N} \).