Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 184_notes:amp_law [2017/10/05 00:46] – [Outside a long straight wire] caballero | 184_notes:amp_law [2020/08/24 13:29] (current) – dmcpadden | ||
|---|---|---|---|
| Line 1: | Line 1: | ||
| + | Section 21.6 in Matter and Interactions (4th edition) | ||
| + | |||
| + | / | ||
| + | |||
| ===== Putting Ampere' | ===== Putting Ampere' | ||
| - | Now, that we have built the two sides of the equation | + | Now, that we have built the two sides of Ampere' |
| - | ==== Outside a long straight wire ==== | + | ==== Magnetic Field Outside a Long Straight Wire ==== |
| - | Consider | + | The last step here is to piece together the left and right sides of Ampere' |
| - | FIXME add figure | + | $$\oint \vec{B} \bullet d\vec{l} = \mu_0 I_{enc}$$ |
| - | === Left-hand side === | + | [{{ 184_notes: |
| - | 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' | + | For the left hand side, we 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 be constant at any fixed distance from the wire. This lead us to derive the left-hand side of Ampere' |
| - | $$\oint \mathbf{B} \cdot d\mathbf{l} = B \oint dl = B 2\pi R.$$ | + | $$\oint \vec{B} \bullet |
| - | For the purposes of this page, we will assume | + | where we said that $R$ is greater than the radius of the wire. |
| - | === Right-hand side === | + | For the right hand side then, our Amperian loop is larger than the radius of the wire, so the total enclosed current is just $I_{tot}$. |
| - | + | ||
| - | Given that our Amperian loop is larger than the radius of the wire, the total enclosed current is just $I_{tot}$. | + | |
| $$\mu_0 I_{enc} = \mu_0 I_{tot}.$$ | $$\mu_0 I_{enc} = \mu_0 I_{tot}.$$ | ||
| - | |||
| - | === Put it together === | ||
| Combining the two sides of Ampere' | Combining the two sides of Ampere' | ||
| - | $$\oint \mathbf{B} \cdot d\mathbf{l} = mu_0 I_{enc}$$ | + | $$\oint \vec{B} \bullet |
| $$B \oint dl = \mu_0 I_{tot}$$ | $$B \oint dl = \mu_0 I_{tot}$$ | ||
| $$B 2\pi R = \mu_0 I_{tot}$$ | $$B 2\pi R = \mu_0 I_{tot}$$ | ||
| $$ B =\mu_0\dfrac{I_{tot}}{2\pi R}$$ | $$ B =\mu_0\dfrac{I_{tot}}{2\pi R}$$ | ||
| - | This is exactly the result we obtained with Biot-Savart for a very long wire, but by doing a much more complicated integral. | + | This is [[184_notes: |
| - | + | ||
| - | What is the direction of this field? It circulates around the as dictated by the right-hand rule. | + | |
| - | + | ||
| - | What happens inside the wire? That is the subject of the example below. | + | |
| - | + | ||
| - | === Example === | + | |
| - | + | ||
| - | * Magnetic field inside a wire | + | |
| + | The final step here is to ask what is the direction of this B-field? With Ampere' | ||
| + | ==== Ampere' | ||
| + | There are a few cases (besides outside a long wire) where Ampere' | ||
| + | - Figure out and draw the general shape of the magnetic field. | ||
| + | - Choose an Amperian loop that a) goes through your observation point, b) follows the magnetic field (to simplify the dot product) and c) has a constant magnetic field along the length of the loop (to pull the B out of the integral). This lets you simplify the left side of the integral. | ||
| + | - Find the current enclosed by the loop (maybe using current density if you need a fraction of the total current). | ||
| + | - Solve for the magnitude of the magnetic field | ||
| + | - Double check the direction of the magnetic field using the right hand rule. | ||