# Magnetic moment of the loop

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A current carrying wire is fixed on a cube of side l as shown in the figure find the magnetic moment of the loop ?
I Thought of doing it by $M = N I A$
but I couldn't find area vector .

There are 2 methods of calculating the magnetic moment - see wikipedia article on Magnetic Moment. The cube has side $a$ and the loop carries current $I$.

1. Addition of vector loop areas

Split the loop into 2 planar loops : ACGEA and ABCA. The area of ABCA is $\frac12 a^2$. The unit vector of its area is $j$. The area of loop ACGEA is $\sqrt2 a^2$. Its unit vector of area is $\frac{1}{\sqrt2}(i+k)$. The total vector area is
$A=\frac12 a^2 j + \sqrt2 a^2 \frac{1}{\sqrt2}( i+k)=a^2 (i+\frac12 j+k)$
so the magnetic moment is
$m=IA=Ia^2(i+\frac12 j+k)$.

2. Addition of moments of current elements

The moment is the sum of $\frac12 I r \times dl$ for each current element $dl$ in the loop. Shift the origin to corner A, then the moments of current elements EA, AB are zero. The moments of BC, CG, GE about A are respectively
$ai \times -ak=a^2j$
$(ai-ak) \times aj=a^2(k+i)$
$aj \times (-ai-ak)=a^2(k+i)$.
The magnetic moment of the loop is again
$m=\frac12 I (a^2j+a^2(k+i)+a^2(k+i))=Ia^2(i+\frac12 j+k)$.

answered Apr 4, 2017 by (28,806 points)
selected Apr 5, 2017