A wire carrying current I is bent in the shape ABCDEFA as shown, where rectangle ABCDA and ADEFA are perpendicular to each other. If the sides of the rectangles are of lengths a and b, then the magnitude and direction of magnetic moment of the loop ABCDEFA is

Solution

Physics sir · 3 steps

Add a fake wire along AD in both directions, and the bent loop splits into two flat rectangles.

1. Split the loop. Imagine current I flowing along AD and an equal current flowing back along DA. They cancel, so nothing changes. The wire is now two closed rectangles: ABCDA in the xy-plane and ADEFA in the xz-plane, each of area ab.

   Loop	Path	Right-hand rule gives	Moment
   ABCDA	+y, then -x, then -y, then +x	anticlockwise seen from above, so +k̂	abI k̂
   ADEFA	-x, then +z, then +x, then -z	(-î)×k̂ = ĵ, so +ĵ	abI ĵ

2. Add the vectors. Magnetic moment is a vector, so the two moments add. They are equal in size and perpendicular to each other.

   M⃗ = abI (ĵ + k̂), |M⃗| = √((abI)² + (abI)²) = √2 abI

3. So. The moment has magnitude √2 abI and points along ĵ/√2 + k̂/√2, which is option (c).

The idea to keep. For a bent, non-planar loop, add equal and opposite currents along a shared edge to make flat loops. Find each loop's IA⃗ separately, then add them as vectors.

Answer from the NTA final answer key. Solution by PrepWiser.