A non planar closed loop of arbitrary shape carryign a current I is placed in uniform magnetic field. The force acting on the loop

Options

A. is zero only for one orientation of loop in magnetic field
B. is zero for two symmetricaly located positions of loop in magnetic field
C. is zero for all orientations
D. is never zero

Explanation

To solve the problem of finding the force acting on a non-planar closed loop of arbitrary shape carrying a current \( I \) placed in a uniform magnetic field \( \mathbf{B} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Force on a Current-Carrying Wire**:
The force \( \mathbf{F} \) on a segment of wire carrying current \( I \) in a magnetic field \( \mathbf{B} \) is given by the formula:
\[ \mathbf{F} = I \mathbf{L} \times \mathbf{B} \]
where \( \mathbf{L} \) is the length vector of the wire segment in the direction of the current.

2. **Dividing the Loop into Segments**:
Consider the closed loop as being composed of multiple segments. For simplicity, we can analyze two segments of the loop that are opposite each other. Let’s denote the upper segment as \( L_1 \) and the lower segment as \( L_2 \).

3. **Analyzing the Forces on Each Segment**:
- For the upper segment \( L_1 \), the force acting on it due to the magnetic field is:
\[ \mathbf{F}_1 = I \mathbf{L}_1 \times \mathbf{B} \]
- For the lower segment \( L_2 \), the force acting on it is:
\[ \mathbf{F}_2 = I \mathbf{L}_2 \times \mathbf{B} \]
Since \( L_2 \) is in the opposite direction to \( L_1 \), we can express \( \mathbf{L}_2 \) as \( -\mathbf{L}_1 \).

4. **Calculating the Net Force**:
Now, substituting \( \mathbf{L}_2 = -\mathbf{L}_1 \) into the equation for \( \mathbf{F}_2 \):
\[ \mathbf{F}_2 = I (-\mathbf{L}_1) \times \mathbf{B} = -I \mathbf{L}_1 \times \mathbf{B} \]
Thus, we have:
\[ \mathbf{F}_1 = I \mathbf{L}_1 \times \mathbf{B} \]
\[ \mathbf{F}_2 = -I \mathbf{L}_1 \times \mathbf{B} \]

5. **Adding the Forces**:
The net force \( \mathbf{F}_{\text{net}} \) acting on the loop is the sum of the forces on the segments:
\[ \mathbf{F}_{\text{net}} = \mathbf{F}_1 + \mathbf{F}_2 = I \mathbf{L}_1 \times \mathbf{B} - I \mathbf{L}_1 \times \mathbf{B} = 0 \]

6. **Conclusion**:
Since the forces on opposite segments of the loop cancel each other out, the total force acting on the non-planar closed loop in a uniform magnetic field is:
\[ \mathbf{F}_{\text{net}} = 0 \]