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184_notes:examples:week12_changing_shape [2017/11/10 02:29] – created tallpaul | 184_notes:examples:week12_changing_shape [2017/11/10 03:42] – [Flux Through a Changing, Rotating Shape] tallpaul | ||
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- | ===== Flux through | + | ===== Flux Through |
- | Suppose you have a magnetic field $\vec{B} = 0.6 \text{ mT } \hat{x}$. Three identical square loops with side lengths | + | Suppose you have a magnetic field directed in the $-\hat{z}$-direction, |
- | + | ||
- | {{ 184_notes: | + | |
===Facts=== | ===Facts=== | ||
- | | + | * The magnetic field is directed into the page. |
- | | + | * The steps for changing and rotating the loop are outlined in the problem statement. |
- | * The length of a square' | + | |
===Lacking=== | ===Lacking=== | ||
- | * The magnetic flux through each loop. | + | * A description of the magnetic flux. |
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * The loops have flat faces. | + | * The magnetic field is constant in time, and the same everywhere. |
- | * The magnetic field does not change with time, and is uniform | + | * The steps for changing |
===Representations=== | ===Representations=== | ||
* We represent magnetic flux through an area as | * We represent magnetic flux through an area as | ||
$$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A}$$ | $$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A}$$ | ||
- | * We represent the situation | + | * We represent the steps with the following visual: |
+ | {{ 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 $\text{d}\vec{A}$ does not change direction |
- | $$\vec{B} \bullet \text{d}\vec{A} = B\text{d}A\cos\theta$$ | + | $$\int \vec{B} \bullet \text{d}\vec{A} = \int B\text{d}A\cos\theta$$ |
Since $B$ and $\theta$ do not change for different little pieces ($\text{d}A$) of the area, we can pull them outside the integral: | Since $B$ and $\theta$ do not change for different little pieces ($\text{d}A$) of the area, we can pull them outside the integral: | ||
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$$\int B\text{d}A\cos\theta =B\cos\theta \int \text{d}A = BA\cos\theta$$ | $$\int B\text{d}A\cos\theta =B\cos\theta \int \text{d}A = BA\cos\theta$$ | ||
- | Area for a square | + | 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. | ||
+ | |||
+ | **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 |