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184_notes:examples:week10_current_segment [2017/11/03 17:08] – [Solution] tallpaul | 184_notes:examples:week10_current_segment [2021/07/07 18:01] (current) – schram45 | ||
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=====Magnetic Field from a Current Segment===== | =====Magnetic Field from a Current Segment===== | ||
The notes outline how to find the [[184_notes: | The notes outline how to find the [[184_notes: | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
===Facts=== | ===Facts=== | ||
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===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * The current is steady, and the wire segment is uniform. | + | * The current is steady: This means the current is not changing with time or space through our segment |
+ | * The segment is straight and uniform: This will simplify down our model to where the segment only extends in a single direction, simplifying the separation vector. Also, assuming | ||
===Representations=== | ===Representations=== | ||
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Below, we show a diagram with a lot of pieces of the Biot-Savart Law unpacked. We show an example d→l, | Below, we show a diagram with a lot of pieces of the Biot-Savart Law unpacked. We show an example d→l, | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
For now, we write d→l=⟨dx,−dy,0⟩ | For now, we write d→l=⟨dx,−dy,0⟩ | ||
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Notice that we can rewrite y as y=−x−L. This equation comes from the equation for a straight line, y=mx+b, where the slope of the line (or wire in this case) is m=−1 and the y-intercept of the wire is at b=−L. An alternate solution to this example could also be to rotate the coordinate system so that the x or y axis lines up with wire. If finding y is troublesome, | Notice that we can rewrite y as y=−x−L. This equation comes from the equation for a straight line, y=mx+b, where the slope of the line (or wire in this case) is m=−1 and the y-intercept of the wire is at b=−L. An alternate solution to this example could also be to rotate the coordinate system so that the x or y axis lines up with wire. If finding y is troublesome, | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
We can use geometric arguments to say that dy=dx. See the diagram above for an insight into this geometric argument. We can now plug in to express d→l and →r in terms of x and dx: | We can use geometric arguments to say that dy=dx. See the diagram above for an insight into this geometric argument. We can now plug in to express d→l and →r in terms of x and dx: | ||
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&= \frac{\mu_0}{2 \pi}\frac{I}{L}\hat{z} | &= \frac{\mu_0}{2 \pi}\frac{I}{L}\hat{z} | ||
\end{align*} | \end{align*} | ||
+ | |||
+ | You can try to do this by adjusting your x-y coordinate system as well (this is in the example video), and you will get the exact same solution. This is a great way to get some practice solving these problems and it gives you other solutions to check your answer with. |