A conducting circular loop of radius r carries a constant current i.It is placed in a uniform magnetic field B such that B is perpendicular to the plane of the loop. The magnetic force acting on the loop is

Options

A. `i r B `
B. `2pi r i B `
C. zero
D. `pi i B`

Explanation

To solve the problem of finding the magnetic force acting on a conducting circular loop carrying a constant current in a uniform magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a circular loop of radius \( r \) carrying a constant current \( i \).
- The magnetic field \( \mathbf{B} \) is uniform and perpendicular to the plane of the loop.

2. **Identify the Direction of Current and Magnetic Field**:
- Assume the current \( i \) flows in a clockwise direction when viewed from above.
- The magnetic field \( \mathbf{B} \) is directed downwards, perpendicular to the plane of the loop.

3. **Use the Formula for Magnetic Force**:
- The magnetic force \( \mathbf{F} \) on a current-carrying conductor in a magnetic field is given by:
\[ \mathbf{F} = i \int (\mathbf{dL} \times \mathbf{B}) \]
- Here, \( \mathbf{dL} \) is an infinitesimal length vector along the current direction.

4. **Evaluate the Force on Each Segment**:
- The loop can be considered as composed of many small segments \( \mathbf{dL} \).
- For each segment, the direction of \( \mathbf{dL} \) is tangential to the loop, and the magnetic field \( \mathbf{B} \) is constant and directed downwards.

5. **Direction of the Force**:
- According to the right-hand rule, for a segment of current \( \mathbf{dL} \) flowing in a clockwise direction and a downward magnetic field \( \mathbf{B} \), the force \( \mathbf{F} \) on each segment will point outward from the center of the loop.
- This means that for diametrically opposite segments of the loop, the forces will be equal in magnitude but opposite in direction.

6. **Net Force Calculation**:
- Since the loop is symmetrical, the forces acting on opposite segments will cancel each other out.
- Therefore, the net magnetic force \( \mathbf{F}_{\text{net}} \) acting on the entire loop is:
\[ \mathbf{F}_{\text{net}} = 0 \]

### Conclusion:
The magnetic force acting on the conducting circular loop is zero.

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