Question

An electric current \( I \) enters and leaves a uniform circular wire of radius \( r \) through diametrically opposite points. A charged particle \( q \) moves along the axis of the circular wire and passes through its center at speed \( v \). The magnetic force on the particle when it passes through the center has a magnitude of:

• A. \( \frac{qv \mu_0 I}{2 \pi r} \)
• B. \( \frac{qv \mu_0 I}{\pi r} \)
• C. \( \frac{qv \mu_0 I}{r} \)
• D. 0

Hint:

The magnetic field at the center of a current-carrying circular loop is zero, resulting in zero magnetic force on a charged particle moving through the center of the loop.

Answer 1

The Correct Option is D

The magnetic force on a charged particle moving through a magnetic field is given by the formula: \[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \] where \( q \) is the charge of the particle, \( \mathbf{v} \) is the velocity vector, and \( \mathbf{B} \) is the magnetic field. In this problem, the magnetic field is produced by the current \( I \) flowing through a circular loop. According to the Biot-Savart law, the magnetic field on the axis of a current-carrying circular loop at a point along its axis (at the center of the loop) is zero. At the center of the loop, where the charged particle \( q \) passes through, the net magnetic field due to the current is zero. This is because the magnetic fields produced by each segment of the wire cancel each other out at the center. Therefore, at the center of the loop, there is no magnetic field. Since the magnetic field is zero at the center of the loop, the magnetic force on the particle, which depends on both the charge \( q \), velocity \( v \), and the magnetic field \( B \), is also zero. Thus, the magnitude of the magnetic force on the particle is \( \boxed{0} \).