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184_notes:amp_law [2017/10/27 01:21] – dmcpadden | 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' | ||
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==== Magnetic Field Outside a Long Straight Wire ==== | ==== Magnetic Field Outside a Long Straight Wire ==== | ||
- | Finally, let' | + | The last step here is to piece together the left and right sides of Ampere' |
- | $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$$ | + | $$\oint \vec{B} \bullet |
- | {{ 184_notes: | + | [{{ 184_notes: |
For the left hand side, we argued the magnetic field curls around the wire. So long as the wire has a uniform steady current, that magnitude of that magnetic field will be constant at any fixed distance from the wire. This lead us to derive the left-hand side of Ampere' | For the left hand side, we argued the magnetic field curls around the wire. So long as the wire has a uniform steady current, that magnitude of that magnetic field will be constant at any fixed distance from the wire. This lead us to derive the left-hand side of Ampere' | ||
- | $$\oint \vec{B} \cdot d\vec{l} = B \oint dl = B 2\pi R$$ | + | $$\oint \vec{B} \bullet |
where we said that R is greater than the radius of the wire. | where we said that R is greater than the radius of the wire. | ||
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Combining the two sides of Ampere' | Combining the two sides of Ampere' | ||
- | $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$$ | + | $$\oint \vec{B} \bullet |
B∮dl=μ0Itot | B∮dl=μ0Itot | ||
B2πR=μ0Itot | B2πR=μ0Itot | ||
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This is [[184_notes: | This is [[184_notes: | ||
- | Something that is important | + | The final step here is to ask what is the direction of this B-field? With Ampere' |
==== Ampere' | ==== Ampere' | ||
There are a few cases (besides outside a long wire) where Ampere' | There are a few cases (besides outside a long wire) where Ampere' | ||
- | - Figure out the general shape of the magnetic field. | + | - Figure out and draw the general shape of the magnetic field. |
- | - Choose an Amperian loop that a) follows the magnetic field (to simplify the dot product) and that b) 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. | + | - 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). | - 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 and double | + | - Solve for the magnitude of the magnetic field |
- | + | - Double | |
- | === Example === | + |