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Magnetic moment of a current loop.

by center o bass
Tags: current, loop, magnetic, moment
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center o bass
May2-12, 01:22 PM
P: 490
Hello, calculating the magnetic moment of a current loop is trivial, but I want to do it with the general formula

[tex]\vec m = \frac{1}2 \int \vec r \times \vec J(\vec r) d^3\vec r[/tex]

The only thing which is stopping me is to find a good argument on why

[tex]\frac{1}{2}\int \vec r d\vec r = \vec A[/tex] where [tex]\vec A[/tex] is the area vector of the loop. Is there a formal way of proving this or any intuitive diagrams one can draw to show that it must be true.
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May2-12, 03:27 PM
P: 617
Start with:

[tex]\vec m = \frac{1}2 \int \vec r \times \vec J(\vec r) d^3\vec r[/tex]

If the current is confined to a single planar wire:

[tex]\vec J(\vec r) = I δ(z) δ(\vec r-\vec r' )[/tex]

which leads to:
[tex]\vec m = \frac{I}2 \int \vec r' \times \vec d r'[/tex]
[tex]\vec m = \hat{z}\frac{I}2 \int r' sin γ d r'[/tex]

Now a little bit of trigonometry reveals that sin γ d r' is just the incremental arc length ds of the arc perpendicular to the radial vector:
[tex]\vec m = \hat{z} I \int \frac{1}2 r' ds[/tex]
The incremental wedge swept out by each incremental arc is essentially just a triangle with area of one half the base length times height. Here r' is the length of the triangular wedge, and ds is the height of the wedge, so the integrand is just the area of the wedge:

[tex]\vec m =\hat{z} I \int d a[/tex]
[tex]\vec m = I \vec A[/tex]

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