A thin circular wire carrying a current I has a magnetic moment M. The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment

Options

A. M
B. `(4)/(pi^(2))M`
C. `(4)/(pi)M`
D. `(pi)/(4)M`

Answer

To solve the problem, we need to determine the magnetic moment of a square wire carrying the same current as a circular wire. Let's go through the steps systematically.

### Step 1: Understand the Magnetic Moment of a Circular Wire
The magnetic moment \( M \) of a circular wire carrying current \( I \) is given by the formula:
\[ M = I \cdot A \]
where \( A \) is the area enclosed by the loop. For a circular wire of radius \( r \), the area \( A \) is:
\[ A = \pi r^2 \]
Thus, the magnetic moment for the circular wire can be expressed as:
\[ M = I \cdot \pi r^2 \]

### Step 2: Relate the Circumference of the Circle to the Side of the Square
When the shape of the wire is changed from a circle to a square, the length of the wire remains the same. The circumference of the circle is:
\[ C = 2\pi r \]
For a square with side length \( a \), the perimeter is:
\[ P = 4a \]
Setting these equal gives:
\[ 2\pi r = 4a \implies a = \frac{\pi r}{2} \]

### Step 3: Calculate the Area of the Square
The area \( A' \) of the square is given by:
\[ A' = a^2 = \left(\frac{\pi r}{2}\right)^2 = \frac{\pi^2 r^2}{4} \]

### Step 4: Calculate the Magnetic Moment of the Square Wire
The magnetic moment \( M' \) of the square wire carrying the same current \( I \) is:
\[ M' = I \cdot A' = I \cdot \frac{\pi^2 r^2}{4} \]

### Step 5: Relate the Magnetic Moments of the Circle and Square
Now, we can find the ratio of the magnetic moment of the square wire to that of the circular wire:
\[ \frac{M'}{M} = \frac{I \cdot \frac{\pi^2 r^2}{4}}{I \cdot \pi r^2} = \frac{\frac{\pi^2 r^2}{4}}{\pi r^2} = \frac{\pi}{4} \]
Thus, we can express \( M' \) in terms of \( M \):
\[ M' = \frac{\pi}{4} M \]

### Final Answer
The magnetic moment of the square wire is:
\[ M' = \frac{\pi}{4} M \]