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184_notes:examples:week12_changing_shape [2017/11/10 02:44] – tallpaul | 184_notes:examples:week12_changing_shape [2018/04/11 19:39] – [Solution] pwirving | ||
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===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: | ||
+ | |||
+ | $$\int BdA\cos\theta =B\cos\theta \int dA = BA\cos\theta$$ | ||
+ | |||
+ | It will be easier to concern ourselves with this value, rather than try to describe the integral calculation each time. At the beginning of the motion, the loop is just a circle. Its area vector and the magnetic field are aligned (parallel), so it has some nonzero magnetic flux. | ||
+ | |||
+ | **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. | ||
- | Since $B$ and $\theta$ | + | **Step 2:** As we rotate the stretched loop, notice that the area vector |
- | $$\int B\text{d}A\cos\theta | + | **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 $\cos \theta$ |
- | Area for a square is just $A = L^2$, and $\theta$ | + | If the loop were to continue rotating in the last step, eventually we would have zero magnetic flux, and as it rotates back around the other way, we could imagine that the flux would then be defined as " |
- | \[ | ||
- | \Phi_B = \begin{cases} | ||
- | | ||
- | | ||
- | | ||
- | | ||
- | \] | ||
- | Notice that we could' |