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184_notes:examples:week10_current_ring [2017/10/31 22:56] – tallpaul | 184_notes:examples:week10_current_ring [2021/07/07 17:42] – 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. The length of our $\text{d}\vec{l}$ is $R\text{d}\phi$ which comes from the [[https:// | + | 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. |
$$\text{d}\vec{l} = R\text{d}\phi\hat{\phi}$$ | $$\text{d}\vec{l} = R\text{d}\phi\hat{\phi}$$ | ||
+ | |||
+ | <WRAP TIP> | ||
+ | ===Assumption=== | ||
+ | We assumed the current was flowing counterclockwise when the loop was viewed from the top. This means our current flows in the same direction as $\phi$. If we had assumed a clockwise current flow then this value would just need a negative in front of it. | ||
+ | </ | ||
We also represent the separation vector using a cylindrical unit vector, too: | We also represent the separation vector using a cylindrical unit vector, too: | ||
$$\vec{r} = \vec{r}_{\text{obs}} - \vec{r}_{\text{source}} = 0 - R\hat{s} = -R\hat{s}$$ | $$\vec{r} = \vec{r}_{\text{obs}} - \vec{r}_{\text{source}} = 0 - R\hat{s} = -R\hat{s}$$ | ||
- | Now, we combined | + | Now, we combine |
$$\text{d}\vec{l} \times \vec{r} = (R\text{d}\phi\hat{\phi}) \times (-R\hat{s}) = R^2 \text{d}\phi (-\hat{\phi} \times \hat{s}) = R^2 \text{d}\phi \hat{z}$$ | $$\text{d}\vec{l} \times \vec{r} = (R\text{d}\phi\hat{\phi}) \times (-R\hat{s}) = R^2 \text{d}\phi (-\hat{\phi} \times \hat{s}) = R^2 \text{d}\phi \hat{z}$$ | ||
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. | 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, | + | In order to set up the integration, |
\begin{align*} | \begin{align*} | ||
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&= \frac{\mu_0 I}{2R} \hat{z} | &= \frac{\mu_0 I}{2R} \hat{z} | ||
\end{align*} | \end{align*} | ||
+ | <WRAP TIP> | ||
+ | ===Assumption=== | ||
+ | Assuming the current is steady allows us to take the $I$ out of the above integral, if this were not the case we would have to find out how current varies in the direction of $\phi$ and leave it in the integral. | ||
+ | </ | ||
We show the visual result below. | We show the visual result below. | ||
- | {{ 184_notes: | + | {{ 184_notes: |