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184_notes:examples:week12_flux_examples [2017/11/08 14:38] – [Solution] tallpaul | 184_notes:examples:week12_flux_examples [2018/08/09 18:08] (current) – curdemma | ||
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===== Review of Flux through a Loop ===== | ===== Review of Flux through a Loop ===== | ||
Suppose you have a magnetic field $\vec{B} = 0.6 \text{ mT } \hat{x}$. Three identical square loops with side lengths $L = 0.5 \text{ m}$ are situated as shown below. The perspective shows a side view of the square loops, so they appear very thin even though they are squares when viewed face on. | Suppose you have a magnetic field $\vec{B} = 0.6 \text{ mT } \hat{x}$. Three identical square loops with side lengths $L = 0.5 \text{ m}$ are situated as shown below. The perspective shows a side view of the square loops, so they appear very thin even though they are squares when viewed face on. | ||
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===Facts=== | ===Facts=== | ||
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* 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 with the given representation in the example statement above. | + | * We represent the situation with the given representation in the example statement above. Below, we also show a side and front view of the first loop for clarity. |
+ | [{{ 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 either), then we can simplify the dot product: | 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 either), then we can simplify the dot product: | ||
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\Phi_B = \begin{cases} | \Phi_B = \begin{cases} | ||
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\] | \] | ||
- | It remains to find the direction of the force, | + | Notice that we could' |
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- | This means that the net force on the loop is $0$, the loop's center of mass won't move! However, the opposing forces on opposite sides will cause the loop to spin -- there is a torque! The calculation | + | |
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- | The calculation is here: | + | |
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- | $$\vec{\tau} = \vec{r} \times \vec{F} = \vec{r}_\text{left} \times \vec{F}_\text{left} + \vec{r}_\text{right} \times \vec{F}_\text{right} = \left(-\frac{L}{2} \hat{x}\right) \times \left(IBL \hat{z}\right) + \left(\frac{L}{2} \hat{x}\right) \times \left(-IBL \hat{z}\right) = IBL^2 \hat{y}$$ | + |