A current carrying circular loop of radius R is placed in the xy-plane with a centre at the origin and magnetic moment M. Half of the loop with x > 0 is now bent so that it now lies in the yz-plane. New magnetic moment will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → M/√2
The magnetic moment of a current loop is \( \vec{m} = I\vec{A} \), perpendicular to the plane of the loop. For the full loop in the xy-plane,
\( \displaystyle M = I\pi R^{2} \quad \text{(along the z-axis)} \)
After bending, the wire forms two semicircles in perpendicular planes. Imagine a pair of equal and opposite currents along the common diameter (on the y-axis); they add nothing to the circuit, but they close each semicircle into its own loop of area \( \pi R^{2}/2 \) carrying current I.
\( \text{Half in the xy-plane}: M/2 \text{ along z} \)
\( \text{Half in the yz-plane}: M/2 \text{ along x} \)
The two moments are perpendicular, so they add as vectors:
\( \displaystyle M_{\mathrm{new}} = \sqrt{\left(\frac{M}{2}\right)^{2} + \left(\frac{M}{2}\right)^{2}} = \frac{M}{\sqrt{2}} \)