A current carrying circular loop of radius R is placed in the x-y plane with centre at the origin. Half of the loop with `xgt0` is now bent so that it now lies in the y-z plane.

Options

A. The magnitude of magnetic moment now diminishes.
B. The magnetic moment does not change.
C. The magnitude of `vec(B)` at `(0,0,z),zgtgtR` increases.
D. The magnitude of `vec(B)` at `(0,0,z), zgtgtR` is unchanged.

Explanation

To solve the problem, we need to analyze the situation of a current-carrying circular loop that is modified by bending half of it. We will calculate the magnetic moment before and after the modification.

### Step-by-Step Solution:

1. **Initial Setup**:
- We have a circular loop of radius \( R \) lying in the x-y plane with its center at the origin. The current flowing through the loop is \( I \).

2. **Magnetic Moment of the Circular Loop in the x-y Plane**:
- The magnetic moment \( m_1 \) of a circular loop is given by the formula:
\[ m_1 = I \cdot A \]
- The area \( A \) of the circular loop is:
\[ A = \pi R^2 \]
- Therefore, the magnetic moment in the x-y plane is:
\[ m_1 = I \cdot \pi R^2 \]

3. **Modification of the Loop**:
- Half of the loop (the part where \( x > 0 \)) is bent to lie in the y-z plane. This means we now have a semicircular loop in the y-z plane.

4. **Magnetic Moment of the Semicircular Loop in the y-z Plane**:
- The area of the semicircular loop is half of the full circular area:
\[ A = \frac{1}{2} \cdot \pi R^2 \]
- The magnetic moment \( m_2 \) for the semicircular loop is:
\[ m_2 = I \cdot \frac{1}{2} \cdot \pi R^2 \]
- Thus, we have:
\[ m_2 = \frac{I \cdot \pi R^2}{2} \]

5. **Direction of Magnetic Moments**:
- The magnetic moment \( m_1 \) (in the x-y plane) is directed along the z-axis (perpendicular to the plane).
- The magnetic moment \( m_2 \) (in the y-z plane) is directed along the x-axis.

6. **Calculating the Total Magnetic Moment**:
- The total magnetic moment \( M \) can be found using the Pythagorean theorem since the two moments are perpendicular to each other:
\[ M = \sqrt{m_1^2 + m_2^2} \]
- Substituting the values:
\[ M = \sqrt{(I \cdot \pi R^2)^2 + \left(\frac{I \cdot \pi R^2}{2}\right)^2} \]
- Simplifying this:
\[ M = \sqrt{(I^2 \cdot \pi^2 R^4) + \left(\frac{I^2 \cdot \pi^2 R^4}{4}\right)} \]
\[ M = \sqrt{I^2 \cdot \pi^2 R^4 \left(1 + \frac{1}{4}\right)} = \sqrt{I^2 \cdot \pi^2 R^4 \cdot \frac{5}{4}} \]
\[ M = I \cdot \pi R^2 \cdot \sqrt{\frac{5}{4}} = I \cdot \pi R^2 \cdot \frac{\sqrt{5}}{2} \]

7. **Conclusion**:
- The total magnetic moment \( M \) is less than the initial magnetic moment \( m_1 \). Therefore, we can conclude that the magnetic moment diminishes after bending half of the loop.