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 with speed \( v \). The magnetic force on the particle when it passes through the center has a magnitude

• 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:

For particles moving along the axis of a current-carrying circular loop, the magnetic force at the center of the loop is zero.

Answer 1

The Correct Option is D

The magnetic field at the center of a uniformly charged circular loop with current \( I \) is given by: \[ B = \frac{\mu_0 I}{2r} \] The force on a charged particle moving with speed \( v \) in a magnetic field is given by: \[ F = qvB \sin\theta \] where \( \theta = 90^\circ \) because the particle moves along the axis of the circular loop, making the angle between the velocity and the magnetic field direction \( 90^\circ \). So, the force becomes: \[ F = qvB = qv \times \frac{\mu_0 I}{2r} \] At the center of the loop, the magnetic force is zero because the magnetic field at the center of the loop is directed along the axis and the velocity is also along this axis. Hence, the magnetic force is zero. Thus, the correct answer is (D).