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184_notes:examples:week10_current_segment [2017/11/02 21:55] – dmcpadden | 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 $\text{d}\vec{l}$, | Below, we show a diagram with a lot of pieces of the Biot-Savart Law unpacked. We show an example $\text{d}\vec{l}$, | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
- | For now, we write $$\text{d}\vec{l} = \langle \text{d}x, \text{d}y, 0 \rangle$$ | + | For now, we write $$\text{d}\vec{l} = \langle \text{d}x, |
- | and $$\vec{r} = \vec{r}_{obs} - \vec{r}_{source} = \langle 0,0,0 \rangle - \langle x, y, 0 \rangle = \langle -x, -y, 0 \rangle$$ | + | |
- | Notice that we can rewrite $y$ as $y=-x-L$. This equation comes from the equation for a straight line, $y=mx+b$, where the slope of the line (or wire in this case) is $m=-1$ and the y-intercept of the wire is at $b=-L$. An alternate solution to this example could also be to rotate the coordinate system so that the x or y axis lines up with wire. If finding $y$ is troublesome, | + | |
- | If take the derivative of the line equation | + | We write the $y$-component with a negative sign so that $\text{d}y$ can be positive. For the separation vector, we write |
+ | $$\vec{r} = \vec{r}_{obs} - \vec{r}_{source} = \langle 0,0,0 \rangle - \langle x, y, 0 \rangle | ||
+ | |||
+ | 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: | ||
+ | |||
+ | We can use geometric arguments to say that $\text{d}y=\text{d}x$. See the diagram above for an insight into this geometric argument. We can now plug in to express $\text{d}\vec{l}$ and $\vec{r}$ in terms of $x$ and $\text{d}x$: | ||
$$\text{d}\vec{l} = \langle \text{d}x, -\text{d}x, 0 \rangle$$ | $$\text{d}\vec{l} = \langle \text{d}x, -\text{d}x, 0 \rangle$$ | ||
$$\vec{r} = \langle -x, L+x, 0 \rangle$$ | $$\vec{r} = \langle -x, L+x, 0 \rangle$$ | ||
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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. |