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184_notes:examples:week14_b_field_capacitor [2017/11/27 00:52] – [Solution] tallpaul | 184_notes:examples:week14_b_field_capacitor [2021/07/22 13:51] (current) – 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. Making this assumption allows us to simplify down our equations when calculating the flux through our surface. |
===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 $I$. | ||
- | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}}{\epsilon_0} = \frac{I}{\epsilon_0} \text{, inside, } r<R$$ | + | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}r^2}{\epsilon_0 |
- | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}r^2}{\epsilon_0R^2} = \frac{Ir^2}{\epsilon_0R^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$$ |
We can now connect the pieces together (remember, $I_{enc}=0$, | We can now connect the pieces together (remember, $I_{enc}=0$, | ||
- | $$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 I \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 \frac{Ir^2}{R^2} |
- | $$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0\frac{Ir^2}{R^2} | + | $$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. | We are ready to write out the magnetic field. | ||
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\[ | \[ | ||
B(r) = \begin{cases} | B(r) = \begin{cases} | ||
- | \frac{\mu_0 I}{2\pi | + | \frac{\mu_0 I r}{2\pi |
- | \frac{\mu_0 I r}{2\pi | + | \frac{\mu_0 I}{2\pi |
| | ||
\] | \] | ||
+ | |||
+ | Notice, the distance between the plates has no effect on the magnetic field calculation. Also, the amount of the charge on the plates at a given time does not matter -- we only care about how fast the charge is changing (the current!). Also, it is interesting that outside the plates, the magnetic field is the same as it would be for a long wire. This would be just as if the capacitor were not there, and the wire were connected. Below, we show a graph of the magnetic field strength as a function of the distance from the center of the capacitor. | ||
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+ | [{{ 184_notes: | ||
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+ | We have enough information to find the maximum B-field, which is at the edge of the plates: | ||
+ | $$B_{\text{max}} = \frac{\mu_0 I}{2\pi R} = \frac{4\pi \cdot 10^{-7} \text{Tm/A} \cdot 3\text{ A}}{2\pi \cdot 10 \text{ m}} = 60 \text{ nT}$$ |