# Magnetic moment of a current loop.

1. May 2, 2012

### center o bass

Hello, calculating the magnetic moment of a current loop is trivial, but I want to do it with the general formula

$$\vec m = \frac{1}2 \int \vec r \times \vec J(\vec r) d^3\vec r$$

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

$$\frac{1}{2}\int \vec r d\vec r = \vec A$$ where $$\vec A$$ 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.

2. May 2, 2012

### chrisbaird

$$\vec m = \frac{1}2 \int \vec r \times \vec J(\vec r) d^3\vec r$$

If the current is confined to a single planar wire:

$$\vec J(\vec r) = I δ(z) δ(\vec r-\vec r' )$$

$$\vec m = \frac{I}2 \int \vec r' \times \vec d r'$$
$$\vec m = \hat{z}\frac{I}2 \int r' sin γ d r'$$
$$\vec m = \hat{z} I \int \frac{1}2 r' ds$$
$$\vec m =\hat{z} I \int d a$$
$$\vec m = I \vec A$$