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184_notes:examples:week14_b_field_capacitor [2017/11/26 23:41] – [Solution] tallpaul | 184_notes:examples:week14_b_field_capacitor [2021/07/22 13:49] – schram45 | ||
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===== Magnetic Field from a Charging Capacitor ===== | ===== Magnetic Field from a Charging Capacitor ===== | ||
- | Suppose you have a parallel plate capacitor that is charging with a current $I=3 \text{ A}$. The plates are circular, with radius $R=10 \text{ m}$ and a distance $d=1 \text{ cm}$. What is the magnetic field in the plane parallel to but in between the plates? | + | Suppose you have a parallel plate capacitor that is charging with a current $I=3 \text{ A}$. The plates are circular, with radius $R=10 \text{ m}$ and a distance $d=1 \text{ cm}$ apart. What is the magnetic field in the plane parallel to but in between the plates? |
- | {{ 184_notes: | + | [{{ 184_notes: |
===Facts=== | ===Facts=== | ||
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===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
* We are only concerned about a snapshot in time, so the current is $I$, even though this may change at a later time as the capacitor charges. | * We are only concerned about a snapshot in time, so the current is $I$, even though this may change at a later time as the capacitor charges. | ||
- | * The electric field between the plates is the same as the electric field between infinite plates. | + | * The electric field between the plates is the same as the electric field between infinite plates |
- | * The electric field outside the plates is zero. | + | * The electric field outside the plates is zero: This also ties back to having two big plates separated by a small distance. |
===Representations=== | ===Representations=== | ||
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* We represent the situation with the following visual: | * We represent the situation with the following visual: | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
====Solution==== | ====Solution==== | ||
- | We wish to find the magnetic field in the plane we've shown in the representations. | + | We wish to find the magnetic field in the plane we've shown in the representations. |
- | {{ 184_notes: | + | Due to the circular symmetry of the problem, we choose a circular loop in which to situate our integral $\int \vec{B}\bullet\text{d}\vec{l}$. We also choose for the loop to be the perimeter of a flat surface, so that the entire thing lies in the plane of interest, and there is no enclosed current (so $I_{enc} = 0$ - there is only the changing electric field). We show the drawn loop below, split into two cases on the radius of the loop. |
- | Below, we also draw the direction of the magnetic field along the loops. We know the magnetic field is directed along our circular loop -- if it pointed in or out a little bit, we may be able to conceive of the closed surface with magnetic flux through it, which would imply the existence of a magnetic monopole. This cannot be the case! We also know that the field is directed counterclockwise, | + | [{{ 184_notes: |
- | {{ 184_notes: | + | Below, we also draw the direction of the magnetic field along the loops. We know the magnetic field is directed along our circular loop (since the changing electric flux creates a curly magnetic field) -- if it pointed in or out a little bit, we may be able to conceive of the closed surface with magnetic flux through it, which would imply the existence of a magnetic monopole. This cannot be the case! We also know that the field is directed counterclockwise, |
+ | |||
+ | [{{ 184_notes: | ||
We are pretty well set up to simplify our calculation of the integral in the representations, | We are pretty well set up to simplify our calculation of the integral in the representations, | ||
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Next, we need to find the changing electric flux in our loop. Since our loop was described with a flat surface, and the electric field is directed parallel to the area-vector of the loop, we can write electric flux as $\Phi_E = \vec{E} \bullet \vec{A} = EA$. This formula will need to be split up for parts of the surface inside the plates versus outside, since the electric field is different. | Next, we need to find the changing electric flux in our loop. Since our loop was described with a flat surface, and the electric field is directed parallel to the area-vector of the loop, we can write electric flux as $\Phi_E = \vec{E} \bullet \vec{A} = EA$. This formula will need to be split up for parts of the surface inside the plates versus outside, since the electric field is different. | ||
- | $$\Phi_\text{E, | + | $$\Phi_\text{E, |
- | $$\Phi_\text{E, | + | $$\Phi_\text{E, |
Now, if we wish the find the change in flux, we will take a time derivative. Notice that all the terms in the flux expressions above are constant, except for $Q$, which is changing with time as dictated by $I$. | Now, if we wish the find the change in flux, we will take a time derivative. Notice that all the terms in the flux expressions above are constant, except for $Q$, which is changing with time as dictated by $I$. | ||
- | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}}{\epsilon_0\pi r^2} = \frac{I}{\epsilon_0\pi r^2} \text{, inside, } r<R$$ | + | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}r^2}{\epsilon_0 |
- | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}}{\epsilon_0\pi R^2} = \frac{I}{\epsilon_0\pi R^2} \text{, outside, } r>R$$ | + | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}}{\epsilon_0} = \frac{I}{\epsilon_0} \text{, outside, } r>R$$ |
+ | |||
+ | We can now connect the pieces together (remember, $I_{enc}=0$, | ||
+ | |||
+ | $$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 \frac{Ir^2}{R^2} \text{, inside, } r<R$$ | ||
+ | $$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 I \text{, outside, } r>R$$ | ||
+ | |||
+ | We are ready to write out the magnetic field. | ||
- | When the rings begin to move towards one another, you can imagine that the second ring experiences an increase in magnetic flux, since the magnetic field is stronger closer to the ring with current. When we say the flux " | + | \[ |
+ | B(r) = \begin{cases} | ||
+ | \frac{\mu_0 I r}{2\pi R^2} &&& | ||
+ | \frac{\mu_0 I}{2\pi r} &&& | ||
+ | | ||
+ | \] | ||
- | Since the flux is increasing, we expect | + | Notice, the distance between |
- | {{ 184_notes:12_coils_induced_current.png?400 |Induced Current}} | + | [{{ 184_notes:14_capacitor_b_field_graph.png?400 |B-Field Strength, Graphed}}] |
- | In fact, this induced current and resulting magnetic field from the second ring will actually cause a changing flux through the first ring that actually increases its currently momentarily. | + | We have enough information |
+ | $$B_{\text{max}} = \frac{\mu_0 I}{2\pi R} = \frac{4\pi \cdot 10^{-7} \text{Tm/A} \cdot 3\text{ A}}{2\pi \cdot 10 \text{ m}} = 60 \text{ nT}$$ |