A steady current `i` flows in a small square lopp of wire of side `L` in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let `vecmu_(1)` and `vecmu_(2)` respectively denote the magnetic moments due to the current loop before and after folding. Then

Options

A. `vecmu_(2)=0`
B. `vecmu_(1)` and `vecmu_(2)`are in the same direction
C. `|vecmu_(1)|/|vecmu_(2)|=sqrt2`
D. `|vecmu_(1)|/|vecmu_(2)|=1/sqrt2`

Answer

To solve the problem, we need to find the magnetic moments \(\vec{\mu_1}\) and \(\vec{\mu_2}\) of the square loop before and after folding, and then determine the ratio \(\frac{\mu_1}{\mu_2}\).

### Step-by-Step Solution:

1. **Calculate \(\mu_1\) (Magnetic Moment Before Folding)**:
- The magnetic moment \(\mu\) of a current-carrying loop is given by the formula:
\[ \mu = I \cdot A \]
- For a square loop of side length \(L\), the area \(A\) is:
\[ A = L^2 \]
- Therefore, the magnetic moment before folding is:
\[ \mu_1 = I \cdot L^2 \]

2. **Understand the Folding Process**:
- When the loop is folded about its middle, half of the loop will lie in a vertical plane and the other half will remain in the horizontal plane.
- This creates two segments of the loop: one in the vertical plane and one in the horizontal plane.

3. **Calculate \(\mu_2\) (Magnetic Moment After Folding)**:
- After folding, the loop can be considered as two perpendicular segments:
- The vertical segment contributes to one magnetic moment \(\mu_{v}\).
- The horizontal segment contributes to another magnetic moment \(\mu_{h}\).
- Each segment has an area of \(\frac{L^2}{2}\) (since half of the area of the original loop is now in each plane).
- Thus, the magnetic moments for each segment are:
\[ \mu_{v} = I \cdot \frac{L^2}{2} \quad \text{(vertical segment)} \]
\[ \mu_{h} = I \cdot \frac{L^2}{2} \quad \text{(horizontal segment)} \]
- Since these two magnetic moments are perpendicular, the resultant magnetic moment \(\mu_2\) is given by:
\[ \mu_2 = \sqrt{\mu_{v}^2 + \mu_{h}^2} = \sqrt{\left(I \cdot \frac{L^2}{2}\right)^2 + \left(I \cdot \frac{L^2}{2}\right)^2} \]
\[ \mu_2 = \sqrt{2 \left(I \cdot \frac{L^2}{2}\right)^2} = I \cdot \frac{L^2}{\sqrt{2}} \]

4. **Find the Ratio \(\frac{\mu_1}{\mu_2}\)**:
- Now, we can find the ratio of the magnetic moments:
\[ \frac{\mu_1}{\mu_2} = \frac{I \cdot L^2}{I \cdot \frac{L^2}{\sqrt{2}}} = \frac{L^2}{\frac{L^2}{\sqrt{2}}} = \sqrt{2} \]

### Final Result:
Thus, the ratio of the magnetic moments before and after folding is:
\[ \frac{\mu_1}{\mu_2} = \sqrt{2} \]