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184_notes:examples:week10_current_ring [2017/10/31 22:19] – [Magnetic Field from a Ring of Charge] tallpaul | 184_notes:examples:week10_current_ring [2021/07/07 17:35] – schram45 | ||
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- | =====Magnetic Field from a Ring of Current===== | + | =====Challenge Example: |
Suppose you have a circular ring, in which which there is a current $I$. The radius of the ring is $R$. The current produces a magnetic field. What is the magnetic field at the center of the ring? | Suppose you have a circular ring, in which which there is a current $I$. The radius of the ring is $R$. The current produces a magnetic field. What is the magnetic field at the center of the ring? | ||
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===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * If we orient the ring in the $xy$-plane and look down, the current flows in the counterclockwise direction. | + | * If we orient the ring in the $xy$-plane and look down, the current flows in the counterclockwise direction: In order to start this problem we need to assign a direction for the current to flow in our loop. Without this assumption there would be no way to assign a dl vector. |
- | * The current is steady. | + | * The current is steady: This means the current is not changing with time or space through our ring and is just a constant. |
- | * There are no other contributions to the magnetic field. | + | * There are no other contributions to the magnetic field: Any outside currents or moving charges could create a magnetic field that could effect our solution. |
===Representations=== | ===Representations=== | ||
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{{ 184_notes: | {{ 184_notes: | ||
- | Let's start breaking down some of the components of the Biot-Savart Law we listed in our representations. We can say for now that $\text{d}\vec{l}$ is directed in the $\hat{\phi}$ direction. | + | Let's start breaking down some of the components of the Biot-Savart Law we listed in our representations. We can say for now that $\text{d}\vec{l}$ is directed in the $\hat{\phi}$ direction. |
+ | $$\text{d}\vec{l} = R\text{d}\phi\hat{\phi}$$ | ||
+ | We also represent the separation vector using a cylindrical unit vector, too: | ||
+ | $$\vec{r} = \vec{r}_{\text{obs}} - \vec{r}_{\text{source}} = 0 - R\hat{s} = -R\hat{s}$$ | ||
- | ------------- | + | Now, we combine the two vectors in their cross product: |
+ | $$\text{d}\vec{l} \times \vec{r} = (R\text{d}\phi\hat{\phi}) \times (-R\hat{s}) = R^2 \text{d}\phi (-\hat{\phi} \times \hat{s}) = R^2 \text{d}\phi \hat{z}$$ | ||
+ | Notice that even though the direction of $\hat{\phi}$ and $\hat{s}$ depend on the angle $\phi$ at which the vectors exist, their cross product, $\hat{z}$, does not depend at all on $\phi$. This will greatly simplify our integration later. | ||
+ | |||
+ | In order to set up the integration, | ||
- | For now, we write $$\text{d}\vec{l} = \langle \text{d}x, \text{d}y, 0 \rangle$$ | ||
- | and $$\vec{r} = \vec{r}_{obs} - \vec{r}_{source} = 0 - \langle x, y, 0 \rangle = \langle -x, -y, 0 \rangle$$ | ||
- | Notice that we can rewrite $y$ as $y=-L-x$. This is a little tricky to arrive at, but is necessary to figure out unless you rotate your coordinate axes, which would be an alternative solution to this example. If finding $y$ is troublesome, | ||
- | $$\text{d}\vec{l} = \langle \text{d}x, -\text{d}x, 0 \rangle$$ | ||
- | $$\vec{r} = \langle -x, L+x, 0 \rangle$$ | ||
- | Now, a couple other quantities that we see will be useful: | ||
- | $$\text{d}\vec{l} \times \vec{r} = \langle 0, 0, \text{d}x(L+x) - (-\text{d}x)(-x) \rangle = \langle 0, 0, L\text{d}x \rangle = L\text{d}x \hat{z}$$ | ||
- | $$r^3 = (x^2 + (L+x)^2)^{3/ | ||
- | The last thing we need is the bounds on our integral. Our variable of integration is $x$, since we chose to express everything in terms of $x$ and $\text{d}x$. Our segment begins at $x=-L$, and ends at $x=0$, so these will be the limits on our integral. Below, we write the integral all set up, and then we evaluate using some assistance some [[https:// | ||
\begin{align*} | \begin{align*} | ||
\vec{B} &= \int \frac{\mu_0}{4 \pi}\frac{I \cdot d\vec{l}\times \vec{r}}{r^3} \\ | \vec{B} &= \int \frac{\mu_0}{4 \pi}\frac{I \cdot d\vec{l}\times \vec{r}}{r^3} \\ | ||
- | &= \int_{-L}^0 \frac{\mu_0}{4 \pi}\frac{IL\text{d}x}{(x^2 + (L+x)^2)^{3/2}}\hat{z} \\ | + | &= \int_0^{2\pi} \frac{\mu_0}{4 \pi}\frac{I \cdot R^2 \text{d}\phi \hat{z}}{R^3} \\ |
- | &= \frac{\mu_0}{2 \pi}\frac{I}{L}\hat{z} | + | &= \frac{\mu_0}{4 \pi}\frac{I \hat{z}}{R} \int_0^{2\pi} |
+ | & | ||
\end{align*} | \end{align*} | ||
+ | |||
+ | We show the visual result below. | ||
+ | |||
+ | {{ 184_notes: |