184_notes:examples:week10_current_ring

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184_notes:examples:week10_current_ring [2021/07/07 17:35] – schram45184_notes:examples:week10_current_ring [2021/07/07 17:52] (current) – [Solution] schram45
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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.  (Note: We could also express $\hat{\phi}$ in Cartesian coordinates as $\hat{\phi} = -\sin(\phi) \hat{x} + \cos(\phi) \hat{y}$. This approach would yield the same solution, but we'll stick with cylindrical coordinates for this solution.) The length of our $\text{d}\vec{l}$ is $R\text{d}\phi$ which comes from the [[https://en.wikipedia.org/wiki/Arc_length|arc length]] formula. We can therefore write 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.  (Note: We could also express $\hat{\phi}$ in Cartesian coordinates as $\hat{\phi} = -\sin(\phi) \hat{x} + \cos(\phi) \hat{y}$. This approach would yield the same solution, but we'll stick with cylindrical coordinates for this solution.) The length of our $\text{d}\vec{l}$ is $R\text{d}\phi$ which comes from the [[https://en.wikipedia.org/wiki/Arc_length|arc length]] formula. We can therefore write
 $$\text{d}\vec{l} = R\text{d}\phi\hat{\phi}$$ $$\text{d}\vec{l} = R\text{d}\phi\hat{\phi}$$
 +
 +<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.
 +</WRAP>
  
 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.
 +</WRAP>
  
 We show the visual result below. We show the visual result below.
  
 {{ 184_notes:10_ring_with_b.png?300 |Ring of Current, With Magnetic Field at Center}} {{ 184_notes:10_ring_with_b.png?300 |Ring of Current, With Magnetic Field at Center}}
 +
 +Since the observation point is in the same plane as the ring and at the center, applying RHR to a few different points on the ring will always yield a result in the +z direction at our observation point. This aligns with our answer, which is great. We also would expect the B-field to decrease as the ring got bigger (R increases), as each chunk of current carrying wire would be further away, and increase as it got smaller, as each chunk of current carrying wire would be closer to our observation point. Similarly, we would expect the B-field to increase as the current increases as you would have charges moving faster producing a stronger B-field. These are all things reflected by our equation, so we should be happy with our solution.
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  • Last modified: 2021/07/07 17:35
  • by schram45