Two wires of same length and carrying same current are in shape square and a circle. Ratio of their magnetic moments is:

Options

A. `4:pi`
B. `pi:2`
C. `2:pi`
D. `pi:4`

Explanation

To find the ratio of the magnetic moments of two wires shaped as a square and a circle, we will follow these steps:

### Step 1: Understand the problem
We have two wires of the same length and carrying the same current. One wire is shaped as a square, and the other as a circle. We need to find the ratio of their magnetic moments.

### Step 2: Define the magnetic moment
The magnetic moment (M) of a current-carrying loop is given by the formula:
\[ M = n \cdot I \cdot A \]
where:
- \( n \) = number of turns (which is 1 for a single loop),
- \( I \) = current,
- \( A \) = area enclosed by the loop.

Since both wires carry the same current and have one turn, we can simplify the formula to:
\[ M = I \cdot A \]

### Step 3: Calculate the area of the square
Let the side length of the square be \( A \). The perimeter of the square is:
\[ P_s = 4A \]
Since the lengths of both wires are equal, we will express the length of the circular wire in terms of \( A \).

### Step 4: Calculate the area of the circle
Let the radius of the circle be \( R \). The perimeter (circumference) of the circle is:
\[ P_c = 2\pi R \]

### Step 5: Set the perimeters equal
Since the lengths of the wires are the same:
\[ 4A = 2\pi R \]
From this, we can solve for \( R \):
\[ R = \frac{4A}{2\pi} = \frac{2A}{\pi} \]

### Step 6: Calculate the areas
Now we can calculate the areas:
- Area of the square:
\[ A_s = A^2 \]

- Area of the circle:
\[ A_c = \pi R^2 = \pi \left(\frac{2A}{\pi}\right)^2 = \pi \cdot \frac{4A^2}{\pi^2} = \frac{4A^2}{\pi} \]

### Step 7: Calculate the magnetic moments
Now we can find the magnetic moments:
- Magnetic moment of the square:
\[ M_s = I \cdot A_s = I \cdot A^2 \]

- Magnetic moment of the circle:
\[ M_c = I \cdot A_c = I \cdot \frac{4A^2}{\pi} \]

### Step 8: Find the ratio of magnetic moments
Now we can find the ratio of the magnetic moments:
\[ \frac{M_s}{M_c} = \frac{I \cdot A^2}{I \cdot \frac{4A^2}{\pi}} = \frac{A^2}{\frac{4A^2}{\pi}} = \frac{\pi}{4} \]

### Final Answer
Thus, the ratio of the magnetic moments of the square wire to the circular wire is:
\[ \frac{M_s}{M_c} = \frac{\pi}{4} \]