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 .

2 votes

Best answer

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

$m=\frac12 I (a^2j+a^2(k+i)+a^2(k+i))=Ia^2(i+\frac12 j+k)$.

...