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184_notes:examples:week14_step_down_transformer [2017/11/29 16:28] – tallpaul | 184_notes:examples:week14_step_down_transformer [2021/07/22 13:56] (current) – [Solution] schram45 | ||
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===== Designing a Step-down Transformer ===== | ===== Designing a Step-down Transformer ===== | ||
Recall the [[184_notes: | Recall the [[184_notes: | ||
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* The high-voltage line carries $240 \text{ kV}$. | * The high-voltage line carries $240 \text{ kV}$. | ||
* We want the low-voltage line to carry $120 \text{ V}$. | * We want the low-voltage line to carry $120 \text{ V}$. | ||
- | * We know the step-up transformer was created by wrapping two solenoids around an iron torus, and connected the high-voltage line to one solenoid, and the low-voltage line to the other. The low-voltage solenoid had fewer turns. A more detailed representation is shown below. | + | * We know the step-up transformer was created by wrapping two solenoids around an iron ring, and connected the high-voltage line to one solenoid, and the low-voltage line to the other. The low-voltage solenoid had fewer turns. A more detailed representation is shown below. |
===Lacking=== | ===Lacking=== | ||
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===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * We have access to the same materials as we did for the step-up transformer. | + | * We have access to the same materials as we did for the step-up transformer: This allows us to use some of the same relationships from the step-up transformer solution. |
* The step-down transformer we are building will have a similar design to the step-up transformer. | * The step-down transformer we are building will have a similar design to the step-up transformer. | ||
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* We represent the step-up transformer, | * We represent the step-up transformer, | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
====Solution==== | ====Solution==== | ||
- | It makes sense intuitively to just flip the step-up transformer for our design. If the step-up transformer brings voltage down, we could just reverse it to bring the voltage up. Our design is shown below, where the number of turns in the primary solenoid is much greater than the number of turns in the secondary solenoid -- it's just the flipped step-up transformer, | + | It makes sense intuitively to just flip the step-up transformer for our design. If the step-up transformer brings voltage down, we could just reverse it to bring the voltage up. Our design is shown below, where **now the number of turns in the primary solenoid is much greater than the number of turns in the secondary solenoid** -- it's just the flipped step-up transformer, |
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- | {{ 184_notes: | + | |
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- | We wish to find the magnetic field in the plane we've shown in the representations. We know from the notes that a changing electric field should create a curly magnetic field. Since the capacitor plates are charging, the electric field between the two plates will be increasing and thus create a curly magnetic field. We will think about two cases: one that looks at the magnetic field inside the capacitor and one that looks at the magnetic field outside the capacitor. | + | |
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- | 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. | + | |
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- | {{ 184_notes: | + | |
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- | 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, | + | |
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- | {{ 184_notes: | + | |
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- | We are pretty well set up to simplify our calculation of the integral in the representations, | + | |
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- | $$\int \vec{B} \bullet \text{d}\vec{l} = \int B(r) \text{d}l = B(r) \int \text{d}l = 2\pi r B(r)$$ | + | |
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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. | + | |
- | $$\Phi_\text{E, in} = EA = \frac{Q/ | + | [{{ 184_notes: |
- | $$\Phi_\text{E, | + | |
- | Now, if we wish the find the change | + | As with the step-up transformer, the two solenoids are wrapped around |
- | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}r^2}{\epsilon_0 R^2} = \frac{Ir^2}{\epsilon_0 R^2} \text{, inside, } r<R$$ | + | [{{ 184_notes: |
- | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}}{\epsilon_0} = \frac{I}{\epsilon_0} \text{, outside, } r>R$$ | + | |
- | We can now connect | + | The iron is able to align its atoms with the magnetic field much faster than the current alternates between directions, which is why we draw the magnetic field the same everywhere. The iron also greatly amplifies the magnetic field that the primary solenoid would produce in air, so even though |
- | $$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 \frac{Ir^2}{R^2} \text{, inside, } r<R$$ | + | Since the magnetic field in the iron ring is changing with the alternating current, |
- | $$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 I \text{, outside, | + | $$-V_{S}=\frac{d\Phi_{B_{S}}}{dt}$$ |
- | We are ready to write out the magnetic field. | + | The rest of our calculation follows just as it would for the [[184_notes: |
+ | $$\Phi_{B_{S}}=\int \vec{B}_S \bullet \vec{dA}_S$$ | ||
+ | where the area here would be the cross-sectional area of the iron ring (since this is where the magnetic field is). The direction of the magnetic field would always be perpendicular to the area of the cylinder - so the flux would simplfy to: | ||
+ | $$\Phi_{B_{S}}=B_S A_S$$ | ||
+ | But this is the magnetic flux through one loop in the secondary solenoid - if we want the total flux through all the loops then we have to multiply by the number of loops: | ||
+ | $$\Phi_{B_{S}}=B_S A_S N_S$$ | ||
+ | We can then plug this into the voltage equation we wrote above: | ||
+ | $$-V_{S}=\frac{d}{dt}\biggl(B_S A_S N_S\biggr)$$ | ||
- | \[ | + | Since $N_S$ and $A_S$ are constant with respect to time (not adding/ |
- | B(r) = \begin{cases} | + | $$-V_{S}=N_S A_S \frac{d}{dt}\biggl(B_S\biggr)$$ |
- | | + | //__If we assume that the iron ring has a constant cross-sectional area__//, then $A_P=A_S$ and we already said that $B_P=B_S$. This means we can rewrite the flux through the secondary solenoid in terms of the magnetic field and area of the primary solenoid. |
- | \frac{\mu_0 I}{2\pi r} &&& | + | $$-V_{S}=N_S A_P \frac{d}{dt}\biggl(B_P\biggr)$$ |
- | \end{cases} | + | |
- | \] | + | |
- | Notice, the distance between the plates has no effect on the magnetic field calculation. Also, the amount of the charge on the plates | + | If we look at the $A_P\frac{d}{dt}\biggl(B_P\biggr)$ term, this looks very much like the changing |
+ | $$\frac{d\Phi_{B_{P}}}{dt}=\frac{d}{dt}\biggl(B_P A_P\biggr)$$ | ||
+ | If we use Faraday' | ||
+ | $$-V_P=N_P \frac{d}{dt}\biggl(B_P A_P\biggr)$$ | ||
+ | $$\frac{d}{dt}\biggl(B_P A_P\biggr)=\frac{-V_P}{N_P}$$ | ||
- | {{ 184_notes:14_capacitor_b_field_graph.png? | + | Finally, We can plug this result into the $V_S$ equation above to get the potential in the secondary solenoid as it relates to the potential in the primary solenoid: |
+ | $$-V_S = N_S \frac{-V_P}{N_P}$$ | ||
+ | $$V_S=V_P \frac{N_S}{N_P}$$ | ||
- | We have enough information | + | Remember, we need to step down from the $240 \text{ |
- | $$B_{\text{max}} = \frac{\mu_0 I}{2\pi R} = \frac{4\pi \cdot 10^{-7} \text{Tm/A} \cdot 3\text{ A}}{2\pi \cdot 10 \text{ m}} = 60 \text{ nT}$$ | + |