Question:
Two wires of same lengths are shaped into a square and a circle. If they carry same current, then the ratio of their magnetic moments is
Two wires of same lengths are shaped into a square and a circle. If they carry same current, then the ratio of their magnetic moments is
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For the same perimeter, a circle always encloses a larger area than any polygon, so its magnetic moment is larger.
Updated On: Apr 20, 2026
- \(2:\pi\)
- \(\pi:2\)
- \(\pi:4\)
- \(4:\pi\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Magnetic moment: \(M = IA\). Same wire length \(\Rightarrow\) different areas for square and circle.
Step 2: Detailed Explanation:
Let total wire length \(= L\). For square: \(a = \frac{L}{4}\), \(A_s = a^2 = \frac{L^2}{16}\). For circle: \(2\pi r = L \Rightarrow r = \frac{L}{2\pi}\), \(A_c = \pi r^2 = \frac{L^2}{4\pi}\). Ratio of magnetic moments: \(\frac{M_s}{M_c} = \frac{A_s}{A_c} = \frac{L^2/16}{L^2/(4\pi)} = \frac{4\pi}{16} = \frac{\pi}{4}\).
Step 3: Final Answer:
\[ \boxed{\pi : 4} \]
Magnetic moment: \(M = IA\). Same wire length \(\Rightarrow\) different areas for square and circle.
Step 2: Detailed Explanation:
Let total wire length \(= L\). For square: \(a = \frac{L}{4}\), \(A_s = a^2 = \frac{L^2}{16}\). For circle: \(2\pi r = L \Rightarrow r = \frac{L}{2\pi}\), \(A_c = \pi r^2 = \frac{L^2}{4\pi}\). Ratio of magnetic moments: \(\frac{M_s}{M_c} = \frac{A_s}{A_c} = \frac{L^2/16}{L^2/(4\pi)} = \frac{4\pi}{16} = \frac{\pi}{4}\).
Step 3: Final Answer:
\[ \boxed{\pi : 4} \]
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