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 B at (0,0z), `z gt R` increases.
D. The magnitude of B at (0,0,z), `z gt gt R` is unchanges.

Explanation

To solve the problem, we need to analyze the magnetic moment and magnetic field of the current-carrying circular loop before and after it is bent. Here’s a step-by-step solution:

### Step 1: Determine the initial magnetic moment of the loop
- The initial magnetic moment \( \vec{M_0} \) of a circular loop is given by the formula:
\[ \vec{M_0} = I \cdot A \]
where \( I \) is the current and \( A \) is the area of the loop. The area \( A \) of a circular loop with radius \( R \) is:
\[ A = \pi R^2 \]
Therefore, the initial magnetic moment is:
\[ \vec{M_0} = I \cdot \pi R^2 \hat{k} \]

### Step 2: Analyze the bending of the loop
- The loop is bent such that the part of the loop where \( x > 0 \) is now in the \( y-z \) plane. The remaining part of the loop (where \( x < 0 \)) remains in the \( x-y \) plane.

### Step 3: Calculate the new magnetic moment after bending
- The new configuration consists of two parts:
1. The semicircular part in the \( y-z \) plane.
2. The remaining semicircular part in the \( x-y \) plane.

- The magnetic moment for each semicircular part is:
- For the semicircular part in the \( y-z \) plane:
\[ \vec{M_1} = I \cdot \left(\frac{\pi R^2}{2}\right) (-\hat{i}) \]
- For the semicircular part in the \( x-y \) plane:
\[ \vec{M_2} = I \cdot \left(\frac{\pi R^2}{2}\right) \hat{k} \]

### Step 4: Combine the magnetic moments
- The total magnetic moment \( \vec{M} \) after bending is:
\[ \vec{M} = \vec{M_1} + \vec{M_2} = -I \cdot \frac{\pi R^2}{2} \hat{i} + I \cdot \frac{\pi R^2}{2} \hat{k} \]

### Step 5: Calculate the magnitude of the new magnetic moment
- The magnitude of the new magnetic moment is given by:
\[ |\vec{M}| = \sqrt{\left(-I \cdot \frac{\pi R^2}{2}\right)^2 + \left(I \cdot \frac{\pi R^2}{2}\right)^2} \]
\[ = \sqrt{2 \left(I \cdot \frac{\pi R^2}{2}\right)^2} = I \cdot \frac{\pi R^2}{2} \sqrt{2} \]

### Step 6: Compare the new magnetic moment with the initial magnetic moment
- The initial magnetic moment was \( I \cdot \pi R^2 \), and the new magnetic moment is:
\[ |\vec{M}| = \frac{I \cdot \pi R^2}{\sqrt{2}} \]
- This shows that the new magnetic moment is less than the initial magnetic moment:
\[ |\vec{M}| < |\vec{M_0}| \]

### Step 7: Calculate the magnetic field at point \( (0, 0, z) \)
- The magnetic field \( \vec{B_0} \) at a point along the axis of the original loop is given by:
\[ B_0 = \frac{\mu_0}{4\pi} \cdot \frac{2 I \pi R^2}{(z^2 + R^2)^{3/2}} \hat{k} \]

- After bending, the magnetic field \( \vec{B} \) at the same point is:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{I \pi R^2}{(z^2 + R^2)^{3/2}} (-\hat{i}) \]

### Step 8: Compare the magnetic fields
- The magnitude of the magnetic field also decreases after bending:
\[ |B| < |B_0| \]

### Conclusion
- The new magnetic moment decreases compared to the original magnetic moment.
- The magnetic field at the point \( (0, 0, z) \) also decreases.