Question

Two wires of the same length and material are used to form a square loop and a circular loop respectively. If the same current is passed through both loops, then the ratio of the magnetic moment of the square loop to that of the circular loop is

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

Hint:

The magnetic moment depends on the area of the loop. When comparing the magnetic moments of different loops with the same wire length, calculate the area for both and then find the ratio.

Answer 1

The Correct Option is C

Step 1: Magnetic moment of a loop.
The magnetic moment \( M \) of a current-carrying loop is given by: \[ M = I \times A \] where \( I \) is the current passing through the loop, and \( A \) is the area of the loop. Step 2: Calculate the area of both loops.
For a square loop with side \( a \), the area is: \[ A_{\text{square}} = a^2 \] For a circular loop with radius \( r \), the area is: \[ A_{\text{circle}} = \pi r^2 \] Step 3: Relationship between the lengths of the loops.
The length of the square loop is \( 4a \), and the length of the circular loop is \( 2\pi r \). Since the total length of the wire is the same for both loops, we have: \[ 4a = 2\pi r \quad \Rightarrow \quad a = \frac{\pi r}{2} \] Step 4: Calculate the ratio of magnetic moments.
The magnetic moment of the square loop is: \[ M_{\text{square}} = I \times a^2 = I \times \left( \frac{\pi r}{2} \right)^2 = \frac{\pi^2 r^2 I}{4} \] The magnetic moment of the circular loop is: \[ M_{\text{circle}} = I \times \pi r^2 = I \times \pi r^2 \] Thus, the ratio of magnetic moments is: \[ \frac{M_{\text{square}}}{M_{\text{circle}}} = \frac{\frac{\pi^2 r^2 I}{4}}{\pi r^2 I} = \frac{\pi}{4} \] Step 5: Conclusion.
Thus, the ratio of the magnetic moment of the square loop to that of the circular loop is \( \frac{\pi}{4} \), which corresponds to option (C).