A current carrying loop lies on a smooth horizontal plane. Then,

Options

A. it is possible to establish a uniform magnetic field in the region so that the loop starts rotating about its own axis.
B. it is possible to establish a uniform magnetic field in the region so that the loop will tip over over about any of the point.
C. it is not possible that loop will tip over about any of the point whatever be the direction of established magnetic field (uniform).
D. both (a) and (b) are correct.

Explanation

To solve the problem regarding the current-carrying loop on a smooth horizontal plane, we will analyze the effects of a magnetic field on the loop and determine the correct option based on the behavior of the loop under different magnetic field orientations.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a current-carrying loop lying on a smooth horizontal plane. The loop can be represented in the XY-plane, with the Z-axis being vertical.

2. **Magnetic Moment**:
- The magnetic moment \( \mathbf{M} \) of the loop is given by the formula:
\[ \mathbf{M} = n \cdot I \cdot \mathbf{A} \]
where \( n \) is the number of turns (which is 1 for a single loop), \( I \) is the current, and \( \mathbf{A} \) is the area vector of the loop, which is perpendicular to the plane of the loop.

3. **Torque on the Loop**:
- The torque \( \mathbf{\tau} \) acting on the loop due to an external magnetic field \( \mathbf{B} \) is given by:
\[ \mathbf{\tau} = \mathbf{M} \times \mathbf{B} \]
- The torque will cause the loop to rotate if it is not aligned with the magnetic field.

4. **Case 1: Magnetic Field in the X-direction**:
- If the magnetic field \( \mathbf{B} \) is in the X-direction, the angle between \( \mathbf{M} \) (which is in the Z-direction) and \( \mathbf{B} \) is 90 degrees. Thus, torque exists:
\[ \tau = M \cdot B \cdot \sin(90^\circ) = M \cdot B \]
- The torque will act in the plane of the loop, but it will not cause the loop to rotate about its own axis.

5. **Case 2: Magnetic Field in the Y-direction**:
- If \( \mathbf{B} \) is in the Y-direction, \( \mathbf{M} \) and \( \mathbf{B} \) are parallel. Therefore, the torque is zero:
\[ \tau = M \cdot B \cdot \sin(0^\circ) = 0 \]
- The loop will not rotate.

6. **Case 3: Magnetic Field in the Z-direction**:
- If \( \mathbf{B} \) is in the Z-direction, \( \mathbf{M} \) and \( \mathbf{B} \) are perpendicular. Thus, torque exists:
\[ \tau = M \cdot B \cdot \sin(90^\circ) = M \cdot B \]
- The torque will act in the X-direction, which can cause the loop to tip over but not rotate about its own axis.

7. **Conclusion**:
- In all cases analyzed, the loop does not rotate about its own axis. However, it can tip over when the magnetic field is applied in the Z-direction. Therefore, the correct option is that it is possible to establish a uniform magnetic field in a region so that the loop will tip over about any point.

### Final Answer:
The correct option is that it is possible to establish a uniform magnetic field in a region so that the loop will tip over about any point.