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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 | ||
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+ | Section 21.6 in Matter and Interactions (4th edition) | ||
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+ | / | ||
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===== 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 | + | ∮→B∙d→l=μ0Ienc |
- | === 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' |
- | $$\oint \mathbf{B} \cdot d\mathbf{l} = B \oint dl = B 2\pi R.$$ | + | $$\oint \vec{B} \bullet |
- | For the purposes of this page, we will assume | + | where we said that R is greater than the radius of the wire. |
- | === Right-hand side === | + | For the right hand side then, our Amperian loop is larger than the radius of the wire, so the total enclosed current is just Itot. |
- | + | ||
- | Given that our Amperian loop is larger than the radius of the wire, the total enclosed current is just Itot. | + | |
μ0Ienc=μ0Itot. | μ0Ienc=μ0Itot. | ||
- | |||
- | === Put it together === | ||
Combining the two sides of Ampere' | Combining the two sides of Ampere' | ||
- | $$\oint \mathbf{B} \cdot d\mathbf{l} = mu_0 I_{enc}$$ | + | $$\oint \vec{B} \bullet |
B∮dl=μ0Itot | B∮dl=μ0Itot | ||
B2πR=μ0Itot | B2πR=μ0Itot | ||
B=μ0Itot2πR | B=μ0Itot2πR | ||
- | 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. |