184_notes:examples:week10_current_ring

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
Next revision
Previous revision
184_notes:examples:week10_current_ring [2017/10/31 22:56] tallpaul184_notes:examples:week10_current_ring [2021/07/07 17:52] (current) – [Solution] schram45
Line 1: Line 1:
-=====Magnetic Field from a Ring of Current=====+=====Challenge Example: Magnetic Field from a Ring of Current=====
 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?
  
Line 10: Line 10:
  
 ===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===
Line 26: Line 26:
 {{ 184_notes:10_ring_dl.png?400 |Ring of Current, Broken Down}} {{ 184_notes:10_ring_dl.png?400 |Ring of Current, Broken Down}}
  
-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://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:
 $$\vec{r} =  \vec{r}_{\text{obs}} - \vec{r}_{\text{source}} = 0 - R\hat{s} = -R\hat{s}$$ $$\vec{r} =  \vec{r}_{\text{obs}} - \vec{r}_{\text{source}} = 0 - R\hat{s} = -R\hat{s}$$
  
-Now, we combined the two vectors in their cross product:+Now, we combine the two vectors in their cross product:
 $$\text{d}\vec{l} \times \vec{r} = (R\text{d}\phi\hat{\phi}) \times (-R\hat{s}) = R^2 \text{d}\phi (-\hat{\phi} \times \hat{s}) = R^2 \text{d}\phi \hat{z}$$ $$\text{d}\vec{l} \times \vec{r} = (R\text{d}\phi\hat{\phi}) \times (-R\hat{s}) = R^2 \text{d}\phi (-\hat{\phi} \times \hat{s}) = R^2 \text{d}\phi \hat{z}$$
  
 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, we also need the magnitude of the separation vector, which conveniently is just $R$ -- you should be able to convince yourself of this. Lastly, we need the bounds on our integration. We will be integrating over $\text{d}\phi$. We want to account for every piece of the ring, so it will sufficient to set our integration from $\phi=0$ to $\phi=2\pi$.+In order to set up the integration, we also need the magnitude of the separation vector, which conveniently is just $R$. This should make some intuitive sense because everywhere on the ring would be the same distance away from the center. Lastly, we need the bounds on our integration. We will be integrating over $\text{d}\phi$. We want to account for every piece of the ring, so it will sufficient to set our integration from $\phi=0$ to $\phi=2\pi$.
  
 \begin{align*} \begin{align*}
Line 45: Line 50:
         &= \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?400 |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.
  • 184_notes/examples/week10_current_ring.1509490587.txt.gz
  • Last modified: 2017/10/31 22:56
  • by tallpaul