184_projects:s20_project_11_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?

As the last of the new power lines were assembled the town can finally try to enact its escape plan. The townsfolk with the support of the S.P.A.R.T.A.N task force have been secretly constructing a subway in order to evacuate the town from the omnipresence of the storm cloud.

Now that the power is back on, the people of Lakeview are itching to use this subway system to escape. Before the town opens up access to the subway system, Lakeview's general manager of transportation, Dr. Hawk Natkins, has asked you to go underground and make sure everything seems safe.

You and your team comply and descend into the subway system. You notice a bright blue glow coming from deep in one of the subway tunnels… Yup, it's definitely a pack of hungry EM-boar tigers, ready to munch on a subway car or something. You immediately radio up to Dr. Natkins to see what you should do. He explains that from their previous experiences with EM-boar tigers they become frightened by a magnetic field of 5mT. He also tells you the wiring embedded in the walls of the subway tunnel are arranged just like a big solenoid.

The tunnel is 1 km long, and Dr. Natkins estimates there are about 8,000 coils of wire from one end of the tunnel to the other. Based on from what you can see right now, the tunnel looks to be about 10 meters wide. Dr. Natkins says that if you can tell him a value for the current needed in the solenoid's wire, he will relay the value to Dr. Cürd over at the power plant, and she will be able to set the current to the desired value to produce a magnetic field. Your job is to figure out what value of current you'd like to relay to Dr. Natkins, and why this will produce a big enough magnetic field to scare off the EM-boar tigers.

Uh-Oh

Learning Goals

  • Use Ampere's Law to calculate the magnetic field inside a solenoid.
  • 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.
  • Practice using a differently shaped Amperian loop, or figuring out the enclosed current for more than one wire.

Learning Issues

  • The goal of this problem is to intuitively build up the magnetic field for a solenoid using Ampere's Law. However, once students know they are trying to setup the magnetic field for a solenoid, they will probably just Google it and write down the answer. Push your students to not only have and use the equation, but to show detailed and documented steps of the derivation to get said equation.
  • The intuition for getting the magnetic field for a solenoid can be a little tricky. It is important to this problem because in using Ampere's Law, we have to have some intuition for what the magnetic field “looks like” first. See steps in solution for how to build this up possibly.

Timing Issues

  • If the students don't know they're supposed to be modeling a solenoid by about 30 minutes in, step in and try and talk with them about why they should be considering a solenoid and why that matches this problem.
  • There are a couple of extension problems for the students to work through. This one is pretty short, so likely that they will get to some of those.

We are going to be using Ampere's Law to find the magnetic field inside the giant subway solenoid.

Step 0: Figure out what the shape of the magnetic field should look like for the solenoid. If we assume that the solenoid is very long compared to its diameter and the solenoid's coils are evenly spaced apart, then the following picture is reasonable for the magnetic field.

The magnetic field inside of the solenoid is constant and the magnetic field outside of the solenoid is negligible and approximately zero. How can we build up this intuition? Let's start by considering the magnetic field for a single current carrying loop.

If we split the loop vertically, that is take a slice right down the middle, we can look at what the magnetic field for just that loop.

For a single loop, the magnetic field points straight in the middle of the loop, buy then starts to curve around the wires on the edges. What would the magnetic field look like for two loops evenly spaced apart?

The field is strongest in the region between the two loops where it points straight through. Because of superposition, the magnetic fields in this region add together and become larger. On the outside of the loops near the edges, superposition also dictates that the magnetic field has to get smaller. This is because contributions from one loop start cancelling out with the other loop, reducing the overall magnetic field. This can be determined using the RHR. If we look at the magnetic field from the left loop at the location in the upper right hand corner, the RHR tells us that the field should point diagonally up and to the left. The loop on the right at the same location has a magnetic field which points down and to the left. The resultant of these two vectors is one that points mostly down and to the left (because it is closer to the loop on the right and thus that contribution to the magnetic field is greater) but with a smaller magnitude.

We continue building up this model with increasingly more loops and the trend continues.

The magnetic field in the inside of the loops (the region “in the middle”) is strong and remaining relatively constant while the magnetic field near the outside edges is getting smaller and smaller.

We will assume then for large enough number of loops that the solenoid has a constant magnetic field in the inside which points along the direction that we stack more loops.

Step 1: We want to pick an Amperian loop with sides that are either parallel or perpendicular to the magnetic field so that we can simplify $\int \vec{B} \cdot \vec{dl}$ in Ampere's law. To do this, we will choose a square oriented as such.

Step 2: We need to start by finding $\int \vec{B} \cdot \vec{dl}$. For sides 1 and 3, the magnetic field is perpendicular to $\vec{dl}$, so $\vec{B} \cdot \vec{dl} = 0$. On side 2, there is no magnetic field because we assumed an infinitely long solenoid, so $\vec{B} \cdot \vec{dl} = 0$. Thus, we only have to worry about side 4.

On side 4, $\vec{B}$ and $\vec{dl}$ are parallel, so $\vec{B} \cdot \vec{dl} = B dl$. Ampere's law can be rewritten as $\int B dl = B \int dl = Bl$ where $l$ is the length of our Amperian loop (the length of the rectangle inside of the solenoid). This side of the equation is proportional to the amount of current enclosed, or $Bl = \mu_0 I_{enc}$.

Step 3: now that we have an expression for the magnetic field (almost), we need to find an expression for the amount of current enclosed. This is a little tricky because if we make the Amperian loop longer, we enclose more current because our rectangle encompasses more current carrying loops. Since we assumed the coils are evenly spaced out, we can set up a proportion for the number of loops per unit length of our Amperian loop: $$ \frac{N_{tot}}{L_{tot}} = \frac{N_{enc}}{l}$$ Here $N_{tot}$ is the total number of coils in the solenoid, $L$ is the length of our solenoid, and $N_{enc}$ is the number of coils that we enclose in our Amperian loop of length $l$. Therefore, the number of coils we enclose is given as $$N_{enc} = \frac{N}{L} l$$

The amount of current we enclose in our Amperian loop is the current in any one given loop times the number of loops enclosed or $I_{enc} = I N_{enc} = I \frac{N}{L} l$.

Step 4: Now we just have to put this all together. Subbing our expression for the amount of enclosed current back in, we get $$Bl = \mu_0 I \frac{N}{L} l$$ $$B = \mu_0 I \frac{N}{L}$$

Since the Earth's magnetic field is about 50 $\mu$T and that seems to be fine, we probably want something on order 1000 times as strong to affect the boar tigers. So, we solve for the current need to supply 50 mT given our current solenoid setup.

$$\frac{BL}{\mu_0 N} = I$$

$$\frac{5 \cdot 10^{-3} 1000}{4 \pi \cdot 10^{-7} 8000} = I = 497.4 A$$

This is a massive amount of current needed to power the solenoid, but not totally unreasonable. Have students do some Googling to see what other things draw that kind of current if they are unconvinced.

Discussion Prompts:

  • Question: Does the magnetic field depend on the size of your Amperian loop that you chose?
  • Answer: No, the $l$ canceled out in the derivation. This is because we assumed that the solenoid was basically infinitely long, so distance away from it shouldn't matter. This is also good because we imagined the loop in the first place, so hopefully that choice doesn't change the outcome of the problem.
  • Question: What direction does the magnetic field point? Can you tell the direction using Ampere's Law?
  • Answer: We can't tell from Ampere's Law. We lose that information when we take the dot product because then we are only concerned with magnitudes. The direction of the magnetic field can be determined using the RHR and depends on how the students defined the current. It should point along the solenoid.
  • Question: How did you choose your Amperian loop? Would you want to use a circle in this case as we have previously?
  • Answer: We pick our Amperian loop so that $\vec{B}$ is either parallel or perpendicular to $d\vec{l}$ and constant along our Amperian loop. We wouldn't want to use a circle in this case because the direction of $\vec{B}$ changes relative to $\vec{dl}$ inside of the solenoid. A circle doesn't fulfill our symmetry requirements.

Evaluation Questions

  • Question: How big is a “big” magnetic field?
  • Answer: Earth's magnetic field is about 50 $\mu$T, $mT$ is the magnitude of a fridge magnet, 1-5 $T$ is an MRI makes. The biggest magnets in the world are about 45 T. 100 T is totally unreasonable. Probably want something between 1 mT and 1 T for it to be “safe.”
  • Question: What assumptions did you have to make in this problem and why?
  • Answer: We assumed that the solenoid was very long compared to the radius of the coils, making it practically infinitely long. The coils are evenly spaced, which ensures that the magnetic field is constant in magnitude and direction inside the solenoid. The magnetic field outside of the solenoid is zero. This means that the magnetic field we found is only good for finding magnetic fields at the center of a long solenoid (not good at the ends of the solenoid or for shorter length solenoids).

Extension Questions

  • Question: If Dr. Curd calls you back and says the power plant can only supply 300 A, what would you change to make the same magnetic field?
  • Answer: We have to change $N$ or $L$. We could either make the solenoid shorter or add more turns per unit length. We need to make the density of coils larger.
  • Question: If the boar tigers have an excess charge of 50 $C$ and are running down the tunnel toward you at a speed of 13.6 m/s, what would you expect to happen when Dr. Curd increases the current to your desired level? Boar tigers have a mass of 130 kg.
  • Answer: If the boar tiger is running straight at you, then they are running parallel to the magnetic field, so $\vec{F} = q \vec{v} \times \vec{B} = 0$. If, however, they run perpendicular to the magnetic field, they will experience a force given by $\vec{F} = q \vec{v} \times \vec{B}$. Moving perpendicular to the field would throw the boar tiger into the wall of the tunnel.
  • 184_projects/s20_project_11_sol.txt
  • Last modified: 2020/03/25 15:08
  • by curdemma