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184_notes:examples:week10_current_ring [2021/07/07 17:33] – schram45 | 184_notes:examples:week10_current_ring [2021/07/07 17:42] – 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: 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. | * 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: |
* 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. | * 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. | ||
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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. | ||
$$\text{d}\vec{l} = R\text{d}\phi\hat{\phi}$$ | $$\text{d}\vec{l} = R\text{d}\phi\hat{\phi}$$ | ||
+ | |||
+ | <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: |