A closed loop a current `I` lies in the xz-plane. The loop will experience a force if it is placed in a region occupied by uniform magnetic field along

Options

A. x-axis
B. y-axis
C. z-axis
D. none of these

Explanation

To solve the problem, we need to analyze the situation of a closed loop carrying a current `I` in the xz-plane when placed in a uniform magnetic field. Here is the step-by-step solution:

### Step 1: Understand the Configuration
The closed loop lies in the xz-plane, meaning that the current flows in a plane defined by the x and z axes. We need to determine how this loop interacts with a uniform magnetic field.

**Hint:** Visualize the loop in the xz-plane and identify the direction of the current.

### Step 2: Identify the Magnetic Field Direction
Let’s assume the uniform magnetic field is directed along the y-axis (B vector). This is a common scenario in such problems, but the direction can be specified differently.

**Hint:** Remember that the magnetic field can be oriented in any direction, but for simplicity, we often assume it is along one of the coordinate axes.

### Step 3: Divide the Loop into Two Parts
To analyze the forces acting on the loop, we can divide the loop into two segments: the upper segment (above the x-axis) and the lower segment (below the x-axis).

**Hint:** Consider how the loop can be split into two parts for easier analysis of forces.

### Step 4: Apply the Magnetic Force Formula
The magnetic force on a segment of the current-carrying loop can be calculated using the formula:
\[ \vec{F} = I \vec{L} \times \vec{B} \]
where \( \vec{L} \) is the length vector of the segment and \( \vec{B} \) is the magnetic field vector.

**Hint:** Remember that the cross product will give a force that is perpendicular to both the current direction and the magnetic field direction.

### Step 5: Calculate the Forces on Each Segment
For the upper segment of the loop, the force will be directed in one direction, and for the lower segment, the force will be directed in the opposite direction. Since the segments are equal in length and carry the same current, the magnitudes of the forces will be equal but opposite.

**Hint:** Use the right-hand rule to determine the direction of the force on each segment.

### Step 6: Analyze the Net Force on the Loop
Since the forces on the upper and lower segments are equal in magnitude and opposite in direction, they will cancel each other out. Therefore, the net force on the closed loop will be zero.

**Hint:** Consider how the forces interact with each other and what the net effect is.

### Conclusion
The closed loop carrying current `I` in the xz-plane will experience no net force when placed in a uniform magnetic field along the y-axis. Thus, the answer to the question is that the loop does not experience a force.

**Final Answer:** D) None of these.