Q141 / 150 · MHT CET 2026 (20 April Shift 2) · Physics
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$\dfrac{2}{\pi}$
- Correct answerB$\dfrac{4}{\pi}$
- C$\dfrac{\pi}{4}$
- D$\dfrac{\pi}{2}$
Solution
Let the length of the wire be $L$.
For the circular wire of radius $r$, the circumference is $L = 2\pi r \Rightarrow r = \dfrac{L}{2\pi}$.
The magnetic moment of the circular loop is $M = I A_c = I (\pi r^2) = I \pi \left(\dfrac{L}{2\pi}\right)^2 = \dfrac{I L^2}{4\pi}$.
For the square wire of side $a$, the perimeter is $L = 4a \Rightarrow a = \dfrac{L}{4}$.
The magnetic moment of the square loop is $M_1 = I A_s = I (a^2) = I \left(\dfrac{L}{4}\right)^2 = \dfrac{I L^2}{16}$.
The ratio of the magnetic moments is $\dfrac{M}{M_1} = \dfrac{\dfrac{I L^2}{4\pi}}{\dfrac{I L^2}{16}} = \dfrac{16}{4\pi} = \dfrac{4}{\pi}$.
Answer: $\dfrac{4}{\pi}$
For the circular wire of radius $r$, the circumference is $L = 2\pi r \Rightarrow r = \dfrac{L}{2\pi}$.
The magnetic moment of the circular loop is $M = I A_c = I (\pi r^2) = I \pi \left(\dfrac{L}{2\pi}\right)^2 = \dfrac{I L^2}{4\pi}$.
For the square wire of side $a$, the perimeter is $L = 4a \Rightarrow a = \dfrac{L}{4}$.
The magnetic moment of the square loop is $M_1 = I A_s = I (a^2) = I \left(\dfrac{L}{4}\right)^2 = \dfrac{I L^2}{16}$.
The ratio of the magnetic moments is $\dfrac{M}{M_1} = \dfrac{\dfrac{I L^2}{4\pi}}{\dfrac{I L^2}{16}} = \dfrac{16}{4\pi} = \dfrac{4}{\pi}$.
Answer: $\dfrac{4}{\pi}$
Hence, the correct answer is option: 'B'.