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 in a magnetic field is determined by the Lorentz force formula: \[ F = q v B \sin\theta \] where \( q = 1.6 \times 10^{-19} \, \text{C} \) (proton charge), \( v = 2 \times 10^5 \, \text{m/s} \) (proton velocity), \( B \) is the magnetic field from the current-carrying wire, and \( \theta \) is the angle between the proton's velocity and the magnetic field. As the proton moves parallel to the wire and current, the wire's magnetic field is perpendicular to the proton's velocity, meaning \( \theta = 90^\circ \) and \( \sin\theta = 1 \). Initially, calculate the magnetic field \( B \) at a distance \( r = 0.05 \, \text{m} \) from the wire using the formula for a long straight current-carrying wire's magnetic field: \[ B = \frac{\mu_0 I}{2 \pi r} \] with \( \mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A} \), \( I = 10 \, \text{A} \), and \( r = 0.05 \, \text{m} \). Substituting the values: \[ B = \frac{(4\pi \times 10^{-7}) \times 10}{2 \pi \times 0.05} \] \[ B = \frac{4 \times 10^{-6}}{0.1} = 4 \times 10^{-5} \, \text{T} \] Subsequently, calculate the magnetic force: \[ F = q v B \sin\theta = (1.6 \times 1.0^{-19}) \times (2 \times 10^5) \times (4 \times 10^{-5}) \times 1 \] \[ F = 1.6 \times 2 \times 4 \times 10^{-19 + 5 - 5} \] \[ F = 12.8 \times 10^{-20} = 1.28 \times 10^{-19} \, \text{N} \] The magnitude of the magnetic force on the proton is \( 1.28 \times 10^{-19} \, \text{N} \).