Differences

This shows you the differences between two versions of the page.

--- 184_notes:examples:week14_step_down_transformer [2017/11/28 21:53] – tallpaul
+++ 184_notes:examples:week14_step_down_transformer [2021/07/22 13:56] (current) – [Solution] schram45
@@ Line 1: / Line 1: @@
+[[184_notes:ac|Return to Changing Flux from an Alternating Current notes]]
 ===== Designing a Step-down Transformer =====
-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?
+Recall the [[184_notes:ac#Voltage_Transformer|discussion on voltage transformers]]. We designed a step-up transformer in the notes, which is used to convert small voltages from a generator into high voltages, which get carried long distances to residential areas. High-voltage power lines are dangerous, though, because the potential difference between the power lines and the ground is so enormous. Before the lines enter a residential area, they are sent to a transformer, from which low-voltage lines carry power to the residential area. How can you design a step-down transformer to convert a line that has a $240 \text{ kV}$ potential difference to ground, into a power line that has a $120 \text{ V}$ potential difference to ground?
-{{ 184_notes:14_capacitor_picture.png?400 |Charging Capacitors}}
 ===Facts===
-  * The capacitor is a parallel plate capacitor with circular plates.
+  * The high-voltage line carries $240 \text{ kV}$.
-  * $R=10 \text{ m}$
+  * We want the low-voltage line to carry $120 \text{ V}$.
-  * $d=1 \text{ cm}$
+  * 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.
-  * The capacitor is charging with a current  $I=3 \text{ A}$ .
 ===Lacking===
-  * A description of the magnetic field.
+  * We need a design for the step-down transformer.
 ===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 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 electric field between the plates is the same as the electric field between infinite plates (we'll ignore the electric field at the edges of the capacitor).
+  * The step-down transformer we are building will have a similar design to the step-up transformer.
-  * The electric field outside the plates is zero.
 ===Representations===
-  * We represent the electric field in a parallel plate capacitor as $$\vec{E} = \frac{Q/A}{\epsilon_0} \hat{x}$$ where $Q$ is the charge on a plate, $A$ is the area of the plate, and $\hat{x}$ is directed from one plate to the other.
+  * We represent magnetic flux as $$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A}$$
-  * We can represent the magnetic field from a changing electric field as
+  * We can represent induced voltage as $$V_{ind} = -\frac{\text{d}\Phi}{\text{d}t}$$
-$$\int \vec{B}\bullet \text{d}\vec{l} = \mu_0 I_{enc} + \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t}$$
+  * We represent the step-up transformer, as described in the Facts, with the following visual:
-  * We represent the situation with the following visual:
-{{ 184_notes:14_capacitor_side_view.png?300 |Plane in which we wish to find B-field}}
+[{{ 184_notes:week14_7.png?600 |Step-up Transformer}}]
 ====Solution====
-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.
+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, as described.
-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.
-{{ 184_notes:14_capacitor_loops.png?600 |Circular Loops}}
-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, due to the increasing electric field into the page. (This comes from an extension of Lenz's Law, but will not needed for this course).
-{{ 184_notes:14_capacitor_b_field_loops.png?600 |Circular Loops, with B-field shown}}
-We are pretty well set up to simplify our calculation of the integral in the representations, since the B-field is parallel to the loop's perimeter. Below, we show the integral calculation, where the magnetic field at a radius  $r$  is displayed as  $B(r)$ .
- $\int \vec{B} \bullet \text{d}\vec{l} = \int B(r) \text{d}l = B(r) \int \text{d}l = 2\pi r B(r)$
-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.
-$$\Phi_\text{E, in} = EA = \frac{Q/A_{\text{plate}}}{\epsilon_0}A_{\text{loop}} = \frac{Q}{\epsilon_0\pi R^2}\pi r^2 = \frac{Qr^2}{\epsilon_0 R^2}$$
+[{{ 184_notes:14_step_down_design.png?600 |Step-down Transformer}}]
- $\Phi_\text{E, out} = EA = E_\text{in}A_\text{in} + E_\text{out}A_\text{out} = \frac{Q/A_{\text{plate}}}{\epsilon_0}A_{\text{plate}} + 0 = \frac{Q}{\epsilon_0\pi R^2}\pi R^2 = \frac{Q}{\epsilon_0}$
-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  $I$ .
+As with the step-up transformer, the two solenoids are wrapped around the same iron ring. When current exists in the primary solenoid, it creates a magnetic field inside the solenoid, aligned along the iron ring. The magnetic field from the solenoid will cause all of the atoms within the iron to align with the magnetic field from the primary solenoid. Because iron atoms are very responsive to magnetic fields, even the atoms that are outside the primary solenoid will align with this magnetic field (largely because they are feeling the effects of their neighboring iron atoms). Because the magnetic field in the primary solenoid is oscillating (due to the alternating current), this means that the magnetic field in all of the iron ring is also constantly changing. For snapshot in time, the field may look like this:
-$$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}r^2}{\epsilon_0 R^2} = \frac{Ir^2}{\epsilon_0 R^2} \text{, inside, } r<R$$
+[{{ 184_notes:14_step_down_b_field.png?600 |Step-down Transformer with B-field}}]
- $\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}}{\epsilon_0} = \frac{I}{\epsilon_0} \text{, outside, } r>R$
-We can now connect the pieces together (remember, $I_{enc}=0$, so we omit it below). We can write:
+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 $B_P$ contains magnetic field contributions from the primary solenoid //and// from the iron, the contribution from the iron is far greater. For this reason, we approximate the magnetic field as the same at all locations in the iron. By this approximation,  $B_P = B_S$ .
-$$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 \frac{Ir^2}{R^2} \text{, inside, } r<R$$
+Since the magnetic field in the iron ring is changing with the alternating current, (in the primary solenoid), this will induce a voltage ($V_S$) in the secondary solenoid. We can use Faraday's Law to write this as:
-$$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 I \text{, outside, } r>R$$
+$$-V_{S}=\frac{d\Phi_{B_{S}}}{dt}$$
-We are ready to write out the magnetic field.
+The rest of our calculation follows just as it would for the [[184_notes:ac#Voltage_Transformer|analysis of the step-up transformer]]. We can rewrite the flux  $\Phi_{B2}$  using the flux definition:
+ $\Phi_{B_{S}}=\int \vec{B}_S \bullet \vec{dA}_S$
+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/taking away loops or increasing/decreasing the area), these terms can come out of the derivative:
-B(r) = \begin{cases}
+$$-V_{S}=N_S A_S \frac{d}{dt}\biggl(B_S\biggr)$$
-          \frac{\mu_0 I r}{2\pi R^2} &&& r<R \\
+//__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} &&& r>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 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.
+If we look at the $A_P\frac{d}{dt}\biggl(B_P\biggr)$ term, this looks very much like the changing magnetic flux (or induced voltage) through a single loop of the //primary// solenoid:
+ $\frac{d\Phi_{B_{P}}}{dt}=\frac{d}{dt}\biggl(B_P A_P\biggr)$
+If we use Faraday's Law, we can rewrite the flux through a single loop of the primary solenoid in terms of the number of coils and the induced voltage in the whole solenoid:
+ $-V_P=N_P \frac{d}{dt}\biggl(B_P A_P\biggr)$
+ $\frac{d}{dt}\biggl(B_P A_P\biggr)=\frac{-V_P}{N_P}$
-{{ 184_notes:14_capacitor_b_field_graph.png?400 |B-Field Strength, Graphed}}
+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 to find the maximum B-field, which is at the edge of the plates:
+Remember, we need to step down from the $240 \text{ kV} $power line to a$ 120 \text{ V} $line. This is a factor of 2000. One way to achieve this would be to set$ N_P = 20000 $, and$ N_S = 10$. Due to the huge step down, it may be even easier to design a series of step-down transformers, so that we don't have to have such a large number of turns for the secondary solenoid. Maybe apply a factor of $N_S/N_P=1/40 $for one transformer, and then$ N_S/N_P=1/50$ for a second transformer. You should be able to convince yourself that this would be physically equivalent to just one step-down transformer with $N_S/N_P=1/2000$.
-$$B_{\text{max}} = \frac{\mu_0 I}{2\pi R} = \frac{4\pi \cdot 10^{-7} \text{Tm/A} \cdot 3\text{ A}}{2\pi \cdot 10 \text{ m}} = 60 \text{ nT}$$