Question

A thin circular wire carrying current \(I\) has magnetic moment \(M\). The shape of the wire is changed to a square and it carries the same current. It will have magnetic moment \(M_1\). The ratio \(M\) to \(M_1\) is

• A. \(\frac{2}{π}\)
• B. \(\frac{4}{π}\)
• C. \(\frac{π}{4}\)
• D. \(\frac{π}{2}\)

Hint:

Same wire length, so compare the areas of a circle and a square.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The magnetic moment of a current loop is \(M=IA\). The same wire is reshaped, so its length \(L\) is the same and the current is the same. Only the area changes.

Step 2: Circle
\[ L=2\pi r\Rightarrow r=\frac{L}{2\pi},\qquad A_c=\pi r^2=\frac{L^2}{4\pi} \]

Step 3: Square
\[ L=4s\Rightarrow s=\frac L4,\qquad A_s=\frac{L^2}{16} \]

Step 4: Ratio
\[ \frac{M}{M_1}=\frac{A_c}{A_s}=\frac{L^2/4\pi}{L^2/16}=\frac{4}{\pi} \]
This is option (B). The circle encloses the larger area for a given perimeter, so the ratio is greater than 1.

Final Answer:
For the same wire length the circle has area L squared over 4 pi and the square L squared over 16, giving 4/pi, option (B). \[ \boxed{\frac{4}{\pi}} \]