A wire of length \(L\) carries a current \(I\). It is first bent into a circular loop of one turn. It is then bent into a square loop of one turn. The ratio of the magnetic moment of the circular loop to that of the square loop is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(4:\pi\)
The magnetic moment of a loop is \( m=IA \). The current is the same in both loops, so the ratio of the moments is the ratio of the areas.
For the circle, \( 2\pi r=L \), so \( r=L/2\pi \). For the square, each side is \( L/4 \).
\( \begin{aligned} A_c &= \pi\left(\frac{L}{2\pi}\right)^{2} \\ &= \frac{L^{2}}{4\pi} \end{aligned} \)
\( \begin{aligned} A_s &= \left(\frac{L}{4}\right)^{2} \\ &= \frac{L^{2}}{16} \end{aligned} \)
\( \begin{aligned} \frac{m_c}{m_s} &= \frac{A_c}{A_s} \\ &= \frac{L^{2}/4\pi}{L^{2}/16} \\ &= \frac{4}{\pi} \end{aligned} \)
NCERT (2026-27):
Torque on a rectangular current loop in a uniform magnetic field – “… We define the magnetic moment of the current loop as,”
\( m=IA \)\( \qquad (4.22) \)