An electric current I enters and leaves a uniform circular wire of radius a through diametrically opposite points. A charged paricle q moving along the axis of the circular wire passes through its centre at speed v. The magnetic force acting on the particle when it passes through the centre has a magnitude

Options

A. `qv(mu_(0)i)/(2a)`
B. `qv(mu_(0)i)/(2pia)`
C. `qv(mu_(0)i)/(a)`
D. zero

Answer

To solve the problem, we need to analyze the situation involving the circular wire carrying current and the charged particle moving through its center.

### Step-by-Step Solution:

1. **Understanding the Setup**: We have a circular wire of radius \( a \) with a current \( I \) flowing through it. The current enters and exits the wire at diametrically opposite points.

2. **Magnetic Field at the Center**: The circular wire creates a magnetic field around it due to the current. However, since the current is entering and leaving at opposite points, the contributions to the magnetic field at the center from both halves of the wire will cancel each other out. Therefore, the net magnetic field \( B \) at the center of the wire is zero:
\[ B = 0 \]

3. **Charged Particle Motion**: A charged particle with charge \( q \) is moving along the axis of the circular wire and passes through its center with speed \( v \).

4. **Magnetic Force on the Charged Particle**: The magnetic force \( F \) acting on a charged particle moving in a magnetic field is given by the equation:
\[ F = q \vec{v} \times \vec{B} \]
where \( \vec{v} \) is the velocity of the charged particle and \( \vec{B} \) is the magnetic field.

5. **Calculating the Force**: Since we have already established that the magnetic field \( B \) at the center is zero, we can substitute this into the equation for magnetic force:
\[ F = q v \times 0 = 0 \]

6. **Conclusion**: Therefore, the magnetic force acting on the charged particle when it passes through the center of the circular wire is zero.

### Final Answer:
The magnitude of the magnetic force acting on the particle when it passes through the center is \( 0 \).