Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision Next revisionBoth sides next revision | ||
184_notes:examples:week12_changing_shape [2017/11/10 02:44] – tallpaul | 184_notes:examples:week12_changing_shape [2017/11/12 17:55] – dmcpadden | ||
---|---|---|---|
Line 15: | Line 15: | ||
===Representations=== | ===Representations=== | ||
* We represent magnetic flux through an area as | * We represent magnetic flux through an area as | ||
- | $$\Phi_B = \int \vec{B} \bullet | + | $$\Phi_B = \int \vec{B} \bullet d\vec{A}$$ |
* We represent the steps with the following visual: | * We represent the steps with the following visual: | ||
- | {{yeas}} | + | {{ 184_notes: |
====Solution==== | ====Solution==== | ||
- | Since the magnetic field has a uniform direction, and the area of the loop is flat (meaning $\text{d}\vec{A}$ does not change direction | + | Since the magnetic field has a uniform direction, and the area of the loop is flat (meaning $d\vec{A}$ does not change direction |
- | $$\vec{B} \bullet | + | $$\int \vec{B} \bullet d\vec{A} = \int BdA\cos\theta$$ |
+ | |||
+ | Since $B$ and $\theta$ do not change for different little pieces ($dA$) of the area, we can pull them outside the integral: | ||
+ | |||
+ | $$\int BdA\cos\theta =B\cos\theta \int dA = BA\cos\theta$$ | ||
- | Since $B$ and $\theta$ do not change for different little pieces ($\text{d}A$) | + | It will be easier to concern ourselves with this value, rather than try to describe the integral calculation each time. At the beginning |
- | $$\int B\text{d}A\cos\theta =B\cos\theta \int \text{d}A = BA\cos\theta$$ | + | **Step 1:** As soon as we begin to stretch out our circle, we can imagine that its area begins to decrease, much like when you pinch a straw. We don't change its orientation with respect to the magnetic field, but since its area decreases, we expect that the flux through the loop will also decrease. |
- | Area for a square is just $A = L^2$, and $\theta$ is different for each loop: | + | **Step |
- | \[ | + | **Step 3:** As we rotate the stretched loop again, we are rotating it in such a way that the area vector also rotates. In fact, the area vector becomes less and less aligned with the magnetic field, which indicates that $\cos \theta$ will be decreasing during this motion. This causes us to expect that the magnetic flux through the loop will decrease during this rotation. Alternatively, |
- | \Phi_B = \begin{cases} | + | |
- | | + | |
- | BL^2\cos 90^\text{o} = 0 & \text{Loop 2} \\ | + | |
- | | + | |
- | | + | |
- | \] | + | |
- | Notice that we could' | + | If the loop were to continue rotating in the last step, eventually |