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 loop. What is the magnetic force acting on the loop?

Options

A. `2piRIB`
B. `2piRl^(2)B^(3)`
C. `piR^(2)IB`
D. zero

Answer

To solve the problem of finding the magnetic force acting on a conducting circular loop of radius \( r \) carrying a constant current \( i \) in a uniform magnetic field \( B \) that is perpendicular to the plane of the loop, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a circular loop with radius \( r \).
- The loop carries a current \( i \).
- A uniform magnetic field \( B \) is applied perpendicular to the plane of the loop.

2. **Magnetic Force on a Current-Carrying Conductor**:
- The magnetic force \( \mathbf{F} \) on a current-carrying conductor in a magnetic field is given by the equation:
\[ \mathbf{F} = I \mathbf{L} \times \mathbf{B} \]
- Here, \( I \) is the current, \( \mathbf{L} \) is the length vector of the conductor, and \( \mathbf{B} \) is the magnetic field vector.

3. **Applying the Formula to the Loop**:
- For a circular loop, we can think of the loop as being made up of many infinitesimally small segments, each carrying a current \( i \).
- However, since the magnetic field \( \mathbf{B} \) is uniform and perpendicular to the plane of the loop, the direction of the magnetic force on each segment will be in the plane of the loop.

4. **Net Force Calculation**:
- The key point here is that the forces on opposite segments of the loop will be equal in magnitude but opposite in direction. Therefore, they will cancel each other out.
- As a result, the net magnetic force acting on the entire loop will be zero.

5. **Conclusion**:
- Thus, the magnetic force acting on the loop is:
\[ \mathbf{F} = 0 \]

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

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