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184_notes:i_thru [2017/10/26 23:45] – dmcpadden | 184_notes:i_thru [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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+ | [[184_notes: | ||
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===== Current through a loop ===== | ===== Current through a loop ===== | ||
- | Now that we have the left bit of the equation, the next step is to talk about the right side of Ampere' | + | Now that we have the left side of the equation, the next step is to talk about the right side of Ampere' |
{{youtube> | {{youtube> | ||
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The point of Ampere' | The point of Ampere' | ||
- | $$\oint \vec{B}\cdot d \vec{l} = \mu_0 I_{enc}$$ | + | $$\oint \vec{B} \bullet |
The right hand side describes the amount of current enclosed by the Amperian loop - that is, how much current runs through the inside of the loop. The figure below describes relationship between the loop and the enclosed current. | The right hand side describes the amount of current enclosed by the Amperian loop - that is, how much current runs through the inside of the loop. The figure below describes relationship between the loop and the enclosed current. | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
- | For the purposes of these notes, let's assume | + | For the purposes of these notes, let's assume |
- | == Enclosing all the current == | + | === Enclosing all the current |
If we are looking for the magnetic field outside of the wire, we will choose to draw the loop so that it's radius is bigger than the radius of wire. In this case, the total current that runs through the wire will pass through the Amperian loop, meaning the loop will enclose all the current. This is the simplest of cases where Ienc=Itot. | If we are looking for the magnetic field outside of the wire, we will choose to draw the loop so that it's radius is bigger than the radius of wire. In this case, the total current that runs through the wire will pass through the Amperian loop, meaning the loop will enclose all the current. This is the simplest of cases where Ienc=Itot. | ||
- | == Enclosing some of the current == | + | === Enclosing some of the current === |
- | + | ||
- | When we want to find the magnetic field inside the wire, then we need pick the radius of the loop to be smaller than the radius of the wire. In this case, //some of the current will pass through the loop but not all of the current//. We need to be able to determine what fraction of the total current will pass through our loop. | + | |
+ | When we want to find the magnetic field inside the wire, then we need pick the radius of the loop to be smaller than the radius of the wire. In this case, **some of the current will pass through the loop but not all of the current**. We need to be able to determine what fraction of the total current will pass through our loop. | ||
- | To find Iencl, you will need to know the current | + | [{{ 184_notes: |
- | $$I_{enc} = J A_{enc} | + | Just like how we used [[184_notes: |
+ | $$J=\frac{I_{tot}}{A_{tot}}$$ | ||
+ | where $I_{tot}isthetotalcurrentgoingthroughthewireandA_{tot}$ | ||
- | {{ 184_notes:week10_5.png? | + | Once we know the current density, we can use that to find the total enclosed current ($I_{enc}$). If we multiply the current density by the enclosed area (the area of the Amperian loop you chose), then that will give use the fraction of the current that passes through the loop. Mathematically, |
- | where both of these areas are cross-sectional areas. | + | $$I_{enc} = J A_{enc} = \dfrac{I_{tot}}{A_{tot}} A_{enc}$$ |
- | ==== Extra Words ==== | + | [{{ 184_notes:week10_5.png?450|Current density when the amperian loop encloses an area smaller than the area of the wire}}] |
- | {{184_notes: | + | |
- | As we discussed earlier, our canonical example with be the long straight | + | where again Atot is the total cross-sectional area of the wire, Aenc |
- | In this case, we will consider that the wire is a bit thicker than the average wire, so that it has a current density J=I/A. We will also //__assume that the current density is uniform__// in this class (upper-division courses may address non-uniform current densities). That is, at every point in the wire the same amount of charge per unit time per unit area exists. This will help us understand the power of Ampere' |