184_notes:examples:week14_step_down_transformer

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Recall the 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?

Facts

  • The high-voltage line carries $240 \text{ kV}$.
  • 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.

Lacking

  • We need a design for the step-down transformer.

Approximations & Assumptions

  • We have access to the same materials as we did for the step-up transformer.
  • The step-down transformer we are building will have a similar design to the step-up transformer.

Representations

  • We represent magnetic flux as $$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A}$$
  • We can represent induced voltage as $$V_{ind} = -\frac{\text{d}\Phi}{\text{d}t}$$
  • We represent the step-up transformer, as described in the Facts, with the following visual:

Step-up Transformer

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.

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.

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).

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.

$$\Phi_\text{E, in} = EA = \frac{Q/A_{\text{plate}}}{\epsilon_0}A_{\text{loop}} = \frac{Q}{\epsilon_0\pi R^2}\pi r^2 = \frac{Qr^2}{\epsilon_0 R^2}$$ $$\Phi_\text{E, out} = EA = E_\text{in}A_\text{in} + E_\text{out}A_\text{out} = \frac{Q/A_{\text{plate}}}{\epsilon_0}A_{\text{plate}} + 0 = \frac{Q}{\epsilon_0\pi R^2}\pi R^2 = \frac{Q}{\epsilon_0}$$

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 $I$.

$$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}r^2}{\epsilon_0 R^2} = \frac{Ir^2}{\epsilon_0 R^2} \text{, inside, } r<R$$ $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}}{\epsilon_0} = \frac{I}{\epsilon_0} \text{, outside, } r>R$$

We can now connect the pieces together (remember, $I_{enc}=0$, so we omit it below). We can write:

$$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 \frac{Ir^2}{R^2} \text{, inside, } r<R$$ $$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 I \text{, outside, } r>R$$

We are ready to write out the magnetic field.

\[ B(r) = \begin{cases} \frac{\mu_0 I r}{2\pi R^2} &&& r<R \\ \frac{\mu_0 I}{2\pi r} &&& r>R \end{cases} \]

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.

B-Field Strength, Graphed

We have enough information to find the maximum B-field, which is at the edge of the plates: $$B_{\text{max}} = \frac{\mu_0 I}{2\pi R} = \frac{4\pi \cdot 10^{-7} \text{Tm/A} \cdot 3\text{ A}}{2\pi \cdot 10 \text{ m}} = 60 \text{ nT}$$

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  • Last modified: 2017/11/29 16:20
  • by tallpaul