Now, that we have built the two sides of the equation let's review and put everything together to find the magnetic field outside of a long straight wire. This model of a long straight wire is usually pretty good for many situations where you want to determine the magnetic field near the wire. When you start to get farther from it, the ends of the wire become problematic and we might have to develop a new model.

Consider a long straight wire with a uniform current $I_{tot}$.

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As we have argued, the magnetic field curls around the wire. So long as the wire has a uniform steady current, that magnitude of that magnetic field will constant at any fixed distance from the wire. This lead us to derive the left-hand side of Ampere's Law for an Amperian loop of radius $R$ centered on the wire,

$$\oint \mathbf{B} \cdot d\mathbf{l} = B \oint dl = B 2\pi R.$$

For the purposes of this page, we will assume that $R$ is greater than the radius of the wire.

Given that our Amperian loop is larger than the radius of the wire, the total enclosed current is just $I_{tot}$. So the left hand side of Ampere's Law is simply,

$$mu_0 I_{enc} = \mu_0 I_{tot}.$$

Combining the two sides of Ampere's law, we can find the magnitude of the magnetic field produced by the wire at a distance $R$ from the center of the wire.

$$\oint \mathbf{B} \cdot d\mathbf{l} = mu_0 I_{enc}$$ $$B \oint dl = \mu_0 I_{tot}$$ $$B 2\pi R = \mu_0 I_{tot}$$ $$ B =\mu_0\dfrac{I_{tot}}{2\pi R}$$

This is exactly the result we obtained with Biot-Savart for a very long wire, but by doing a much more complicated integral.

What happens inside the wire? That is the subject of the example below.

* Magnetic field inside a wire