184_projects:power_lines_sol

powerlines.jpeg

Boar tigers are everywhere. Last night, they came through Lakeview and tore down all the power lines, including the transformer for the incoming transmission line from the power plant on the edge of town. All the wires are chewed up and torn, and there are deep claw marks on the utility poles, which luckily are still standing. The residents of Lakeview are pretty spooked. Lakeview needs its power back as soon as possible.

There is an incoming transmission line on the edge of town from which you need to set up the power lines that will run through all of Lakeview. The manager of the power plant, Dr. Erma Cürd, has supplied you with some specifications: Each power line will be erected to connect the transformer from the incoming transmission line to the homes and businesses in Lakeview. Each line is $5 \text{ km}$ long and is made of a metal alloy with $0.008 \text{ $\Omega$/m}$ of resistance.

The most important decision in this reconstruction process is to determine which transformer to install at the incoming transmission line. A given transformer will create a specified voltage drop from the transmission to the residential area. However, there are some risks associated with your choice. One risk is that the electric field along the line will heat up the wire and cause it to melt, which will happen when the electric field reaches $3 \text{ kV/m}$. Another risk is that the line may create a magnetic field on the ground that is dangerous for people walking around and may interfere with portable electronics. The safety limit for the magnetic field is $10 \text{ mT}$.

You have three options for your choice of transformer. The voltage drop on the line from the functioning transmission line to the residential area can be $1 \text{ MV}$, $10 \text{ MV}$, or $100 \text{ MV}$. Evaluate each decision and produce a recommendation based on the safety concerns and the power that the transformer will produce for Lakeview.

Learning Goals

  • Use Ampere's Law to calculate the magnetic field outside of a current-carrying wire.
  • Explain why you pick your Amperian loop and how it helps you simplify your calculations.
  • Explain the general steps that you take when using Ampere's Law.
  • Explain what would change about your solution if the wire were coaxial (this part is extra).

Learning Issues

  • The main goals of this problem are accomplished by crunching numbers and calculating values. There isn't a lot of conceptual discussion surrounding the way in which the students are solving the problem. Be aware of how much focus is being devoted to the numbers and make sure to touch on the more conceptual points.
  • This is the first time the students are doing a lot of work with Ampere's Law. They should really focus on making sure that they understand the general steps to working with Ampere's Law and the assumptions that they need to be viable in order to even consider this method.
  • The co-axial cable portion of this problem is a little bit tricky. The common solution that students come to are two wires side-by-side or stacked on top of one another. This fulfills the $I_{enc}$ part of Ampere's Law, but the symmetry is broken on the $\oint \vec{B} \cdot \vec{dl}$ side of the equation. The only way that this works is if the currents are nested, one wire inside of another.
  • The equations for the co-axial cable part of the problem can get dicey. Make sure the students are drawing out good representations of the problems and clearly defining all of their variables. If they make it this far, they may need some additional guidance and help keeping the equations straight.

Timing Issues

This problem has the potential to go very quickly if students understand Ampere's law and can do the math well enough. Just keep an eye on this make sure to emphasize the concepts and talk through the general steps for using Ampere's Law with your students.

Students might go a few different ways with this. You should guide them to make a couple assumptions that will simplify the problem and actually allow them to perform a calculation with Ampere's Law:

Assumptions

  • Question: What need to be made in order to simplify the problem?
  • Answer: Here are some possible assumptions and approximations:
    • The wire is straight.
    • The wire is the same distance from the ground at all points.

A drawing of the situation may look something like the following:

Power Line Representation

If students are able to make the assumptions listed above, they should be pretty well set to apply Ampere's Law to find the magnetic field from the wire near the ground. These are the steps:

  1. Figure out the general shape of the magnetic field.
  2. Choose an Amperian loop that a) follows the magnetic field (to simplify the dot product) and that b) has a constant magnetic field along the length of the loop (to pull the B out of the integral)
  3. As a check, it may help from students to draw both the magnetic field vectors and the dl-vectors at a few points along their loop.
  4. Find the current enclosed by the loop. This can be found with Ohm's Law for this particular project.
  5. Solve for the magnitude of the magnetic field and double check the direction using the right hand rule.

An Amperian loop for the wire may look like this:

Ampere's Law is defined as: $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$$

For the loop drawn above, we are happy to see that the magnetic field is parallel to the dl-vectors for all points along the loop. So our dot-product simplifies to a product of magnitudes:

$$\oint B\text{d}l = \mu_0 I_{enc}$$

Another thing is that the magnetic field's magnitude is the same for all points along the loop. This is because the points along the loop are all the same distance from the wire. If students move the loop a little off of the center of the wire, this won't be true anymore! Since $B$ is a constant for our particular loop, we can pull it out of the integral:

$$B\oint \text{d}l = \mu_0 I_{enc}$$

And, now a simple integral of $\text{d}l$ is the length of the loop:

$$B\cdot 2\pi r = \mu_0 I_{enc}$$

The variable $r$ is the radius of the Amperian loop. This will change as the size of the loop changes. Which means the magnetic field is different if we have a different distance away from the current-carrying wire. In our case, we are interested in what the magnetic field at the ground is, so we will use the height, $h$, of the powerline as the $r$ of our Amperian loop. This is perfectly okay because we can draw our amperian loop as big or small as we want. If students solve for $B$ at this point, they will find a magnitude:

$$B = \frac{\mu_0 I_{enc}}{2\pi h}$$

The last bit is to notice that $I_{enc} = I$, since the Amperian loop encloses the whole current. So we get:

$$B = \frac{\mu_0 I}{2\pi h}$$

The electric field magnitude along the line is $$E = \frac{\Delta V}{L},$$ where $\Delta V$ is the voltage drop and $L$ is 5 km.

We can find $I$ with Ohm's Law, $$I=\frac{\Delta V}{R}$$. $R$ comes from the resistance per meter multiplied by the 5 km wire-length (total resistance should be $40\text{ }\Omega$. When we put it all together, using $h = 15 \text{ m}$, we get: \[ E = \begin{cases} 0.2 \text{ kV/m} & \Delta V=1 \text{ MV} \\ 2 \text{ kV/m} & \Delta V=10 \text{ MV} \\ 20 \text{ kV/m} & \Delta V=100 \text{ MV} \\ \end{cases} \] and \[ B = \begin{cases} 0.33 \text{ mT} & \Delta V=1 \text{ MV} \\ 3.33 \text{ mT} & \Delta V=10 \text{ MV} \\ 33.3 \text{ mT} & \Delta V=100 \text{ MV} \\ \end{cases} \]

From the problem statement, $10 \text{ mT}$ is the level of dangerous magnetic field, and $3 \text{ kV/m}$ is the level of dangerous electric field.

According to these results, the largest voltage drop option produces both a dangerous magnetic field and a dangerous electric field. We should choose the second option as this will put more power on the lines. Students don't need to compute the power, but maybe they want to: $$P=I\Delta V=\frac{\Delta V^2}{R}=25 \text{ GW}$$

Tutor Questions
  • Question: What does the electric field look like in your model?
  • Expected Answer: Exists along (inside) the wire in the direction of the wire
  • Question: What do the magnetic field vectors look like outside the wire? How do you know they point in that direction?
  • Expected Answer: They form a circle around the wire. Right Hand Rule!
  • Question: How does your model differ from the real world?
  • Expected Answer: Power line not infinitely long, probably zig-zags, sags, resistance of wire may change with weather, boar tigers do not actually exist.

Part 2: Coaxial Cable

There are two extension questions that you can ask that will get students to think about a more complicated situation:

  • Question: How can you redesign the power line so that there is no magnetic field outside the wire, but the wire still carries current?
  • Expected Answer:We would need to enclose an equal amount of current going in the opposite direction
  • Question: Start by finding, what is the magnetic field inside the wire?
  • The solution to this question depends on how the students choose to model the coaxial cable. A pretty general solution is shown below.

Coaxial Power Line

Model the coaxial cable as in infinitely long wire, with an inner wire having radius $a$, and an outer wire occupying the space at radius $b<r<c$. A current $I$ runs oppositely in the outer wire as in the inner wire. We create a circular Amperian loop, shown below in the cross-section.

Amperian Loop Shown in Green

Ampere's Law gives us: $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$$

Students should be able to give an argument about circular symmetry of the magnetic field for how to simplify the integral as they did in the first part of this problem. The integral should simplify to the magnitude of the magnitude of the magnetic field times the circumference of the Amperian loop: $$B\cdot 2 \pi r = \mu_0 I_{enc}$$

When the magnetic field is isolated, we get: $$B(r) = \frac{\mu_0}{2 \pi r} I_{enc}$$

To find $I_{enc}$, students should be able to use simple geometric arguments, which requires that they assume that the current density is uniform in each part of the coaxial cable.

$r < a$: For the inside of the first wire, we can use Ampere's Law using the same assumptions as Part 1.

$$\oint B\text{d}l = \mu_0 I_{enc}$$

$$B\oint \text{d}l = \mu_0 I_{enc}$$

$$B 2 \pi r = \mu_0 I_{enc}$$

The amount of current we enclose now is the fraction of the total cross-sectional area of the inner cable times the current in the inside cable, or $I_{enc} = I \frac{\pi r^2}{\pi a^2}$. Plugging this in we get

$$B 2 * \pi r = \mu_0 I \frac{\pi r^2}{\pi a^2}$$ $$B = \mu_0 I\frac{r}{2 \pi a^2}$$

$a < r < b$: This solution is the exact same as the solution from Part 1 as we enclose all of the current on the inside wire.

$$B = \frac{\mu_0 I}{2 \pi r }$$

$b < r < c$: This case is a little more complicated because we enclose all of the inside wire's current but we also enclose a fraction of the outside wire's current, which goes in the opposite direction and reduces our overall enclosed current. WE can write the amount of current we enclose as $I_enc = I_{r <= a} + I_{b < r <c} = I - I \frac{\pi r^2 - \pi b^2}{\pi c^2 - \pi b^2}$. Notice in the second term, there is a difference in areas in the numerator to figure out what fraction of the second wire's area we enclose with an Amperian Loop of radius $r$, which is going to be the area enclosed in the loop, $\pi r^2$, minus the area of the co-axial cable which has a radius $r < b$, $\pi b^2$. This represents the amount of area that our loop encloses just in the exterior wire. The denominator is the area of the outer wire, which is the area of the circle with radius $c$ minus the area of a circle with radius $b$, $\pi c^2 - \pi b^2$. Plugging this all in and simplifying, we get

$$\oint B\text{d}l = \mu_0 I_{enc}$$

$$B\oint \text{d}l = \mu_0 I_{enc}$$

$$B 2 \pi r = \mu_0 I_{enc}$$

$$B 2 \pi r = \mu_0 I - I (\frac{\pi r^2 - \pi b^2}{\pi c^2 - \pi b^2})$$

$$B = \frac{\mu_0 I}{2 \pi r} \frac{c^2 - r^2}{c^2 - b^2}$$

$r > c$: outside of the co-axial cable, drawing an Amperian loop encloses an equal amount of the current in both wires, which cancel out. Since $I_{enc} = 0$, $\oint \vec{B} \cdot \vec{dl} = 0$, so there is no magnetic field.

$$B = 0$$

We could make this argument from enclosed currents alone, which are summarized below.

\[ I_{\text{enc}} = \begin{cases} \frac{r^2}{a^2} I & r<a \\ I & a<r<b \\ \frac{c^2-r^2}{c^2-b^2}I & b<r<c \\ 0 & c<r \\ \end{cases} \]

Tutor Questions
  • Question: What do the magnetic field vectors look like inside the wire? How do you know they point in that direction?
  • Expected Answer: They form a circle around the wire. Right Hand Rule!
  • Question: Make a graph of the B-field vs r. Why does it look the way it does?
  • Expected Answer: B-field builds linearly as we encompass more current with our Amperian Loop, then falls inversely linear when we are between the wires. Then we encompass less and less current as we expand our loop from b to c so the B-field decreases both linearly and inversely linearly. The B-field is 0 outside the wire. See below.

B-field vs r

  • Question: Explain how you were able to simplify your integral for $\vec{B}\bullet \text{d}\vec{l}$.
  • Expected Answer: In Amperian Loop, $B$ is constant, and always parallel to loop if loop is a circle. So we can pull $B$ out of the integral and then evaluate $\int \text{d}l$ as the length of the loop.
  • Question: How do the steps for Ampere's Law compare to the steps for Gauss's Law? What is similar/different?
  • Expected Answer: Require similar kinds of symmetry and steps are very similar - 1) Imagine surface/loop, 2) pick a loop that satisfies the symmetry (field parallel to dl/dA and field is constant), 3) find the Qenc or Ienc, and 4) find the E/B field. Different because looking at different fields, Q vs I, and closed surface vs closed loop.
  • Question: Any lingering questions?
  • 184_projects/power_lines_sol.txt
  • Last modified: 2021/08/19 19:14
  • by dmcpadden