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184_notes:examples:week10_current_ring [2017/11/03 16:44] – [Challenge: Magnetic Field from a Ring of Current] tallpaul | 184_notes:examples:week10_current_ring [2021/07/07 17:41] – schram45 | ||
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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: | ||
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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: |