A circular loop of radius \( r \) and a square loop of side \( a \) carry equal currents. If the magnetic moments of the two loops have the same magnitude, the ratio \( a/r \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \sqrt{\pi} \)
The magnetic moment of a plane loop of area A carrying a current I is \( m=IA \). It depends only on the current and the area enclosed, whatever the shape of the loop. With equal currents, equal moments need equal areas.
For the circular and the square loop:
\( m_{\mathrm{circle}}=I\pi r^{2} \)
\( m_{\mathrm{square}}=Ia^{2} \)
Equating the two:
\( \begin{aligned} Ia^{2} &= I\pi r^{2} \\ a^{2} &= \pi r^{2} \end{aligned} \)
\( \displaystyle \frac{a}{r}=\sqrt{\pi} \)
Therefore, the side of the square must be \( \sqrt{\pi} \) times the radius of the circle.