184_projects:s21_project_11_sol

Case A

1) Induced current direction:

  • B-field is into the page
  • Pick the area vector to be between the rails and into the page (note:we pick this)
  • $Φ_i$ is into the page and small
  • $Φ_f$ is into the page and larger $Φ_f>Φ_i$
  • As the bar slides the area increases and changes the magnetic flux through the open surface
  • The change in magnetic flux is into the page
  • The induced current generates a magnetic field that will be in the direction that opposes the change in magnetic flux, therefore, the opposite direction of the change in flux is out of the page.

2) Calculate $V_ind$:

$$V_{ind}=-\frac{d \Phi_B}{dt}$$ $$=-\frac{d}{dt}(\oint \vec{B}\cdot d\vec{A})$$ $\vec{B}$ is parallel to $d\vec{A}$ so can dot simplify the dot product: $$=-\frac{d}{dt}(\oint B dA)$$ $\vec{B}$ is constant with area: $$=-\frac{d}{dt}( B)(\oint dA)=-\frac{d}{dt}(BA)$$ $\vec{B}$ is constant with time: $$=- B\frac{d}{dt}(A)$$ Because open surface is a rectangle $A=lw$, where w is constant and l is changing, at a constant rate: $$V_{ind}=- B\frac{d}{dt}(wl)=- Bw\frac{dl}{dt}$$ $$l=l_0+vt$$ Assuming velocity is constant: $$\frac{dl}{dt}=v$$ $$V_{ind}=-Bwv$$ $$|V_{ind}|=(1.2)(.25)(.5)=0.15V$$

3) Flux versus time graph

FIXME (made with matlab is there a better software to use?)

4) Induced voltage vs time graph ($V_{ind}$=(-)slope of $Φ$)

5) Extension:What direction is the force on the rail?

It is in the opposite direction as the velocity. So the magnetic force on the bar with the induced current would actually work to slow the bar down. If this negative sign were not present, the induced current would flow clockwise instead of counter-clockwise, thus making the magnetic force on the bar point in the same direction as the velocity. This would act to speed up the bar and give us a way to get more energy than we started with, which would violate the conservation of energy.

Case B

1) Induced current direction:

  • B-field is out of the page
  • Pick the area vector to be between the rails and out of the page (we pick this)
  • $Φ_i$ is out of the page and big
  • $Φ_f$ is into the page and smaller $Φ_f<Φ_i$
  • As the bar slides the area decreases and changes the magnetic flux through the open surface
  • The change in magnetic flux is into the page
  • The induced current generates a magnetic field that will be in the direction that opposes the change in magnetic flux, therefore, the opposite direction of the change in flux is out of the page.

2) Calculate $V_ind$: $$V_{ind}=-\frac{d \Phi_B}{dt}$$ $$=-\frac{d}{dt}(\oint \vec{B}\cdot d\vec{A})$$ $\vec{B}$ is parallel to $d\vec{A}$ so can dot simplify the dot product: $$=-\frac{d}{dt}(\oint B dA)$$ $\vec{B}$ is constant with area: $$=-\frac{d}{dt}( B)(\oint dA)=-\frac{d}{dt}(BA)$$ $\vec{B}$ is constant with time: $$=- B\frac{d}{dt}(A)$$ Because open surface is a rectangle $A=lw$, where w is constant and l is changing: $$V_{ind}=- B\frac{d}{dt}(wl)=- Bw\frac{dl}{dt}$$ Assume l is changing at a constant rate, where $l_0$≡initial length and I $vt$≡the amount the length is decreasing $$l=l_0-vt$$ Assuming velocity is constant: $$\frac{dl}{dt}=-v$$ $$V_{ind}=-Bw(-v)=Bwv$$ $$V_{ind}=(1.2)(.5)(.4)=0.24V$$

3) Flux versus time graph (flux is decreasing)

4) Induced voltage vs time graph ($V_{ind}$=(-)slope of $Φ$)

5) Extension: What direction is the force on the rail?

It is in the opposite direction as the velocity (the force is to the right).

6) Extension: What if the moving rail kept moving beyond the rail that is stationary?

If the moving rail continued past the stationary rail the area to be between the rails would transition from increasing to decreasing, therefore the direction of the change in magnetic flux would be out of the page. Induced current would also change direction (clockwise). (In orange on graphs)

Case C

1) Induced current direction:

  • Assume current direction from power supply (see diagram above).
  • The current from the power supply is increasing in the same direction overtime, therefore, the magnetic field produced by that current is increasing.
  • $Φ_f<Φ_i$
  • The change in magnetic flux is in the direction of the increasing magnetic field
  • The induced current generates a magnetic field that will be in the direction that opposes the change in magnetic flux, therefore, the induced current is in the opposite direction (in purple above).

2) Calculate $V_ind$: $$V_{ind}=-\frac{d \Phi_B}{dt}$$ $$=-\frac{d}{dt}(\oint \vec{B}\cdot d\vec{A})$$ $\vec{B}$ is parallel to $d\vec{A}$ so can simplify the dot product: $$=-\frac{d}{dt}(\oint B dA)$$ Assume $\vec{B}$ is constant through area because area of the induced coil is smaller than area of coil connected the to power supply: $$V_ind=-\frac{d}{dt}( B)(\oint dA)=-\frac{d}{dt}(BA_{sm})$$ IMPORTANT: $A_{sm}$ is the area of the small coil in which magnetic flux changes, B is the magnetic field from the large coil.

Magnetic field equation for an observation point at the center of the circular coil, out of the plane of the ring: $$B=\frac{\mu_0 I}{2} \frac{R^2}{(z^2+R^2)^{3/2}}$$ $$V_{ind}=-\frac{d}{dt}(BA_{sm})=-\frac{d}{dt}\frac{\mu_0 I}{2} \frac{R^2}{(z^2+R^2)^{3/2}}A_{sm}$$ Everything is constant except I: $$\frac{dI}{dt}=\frac{d}{dt}(0.1+0.03t)=.03$$ $$|V_{ind}|=\frac{\mu_0}{2} \frac{R^2}{(z^2+R^2)^{3/2}}πr^2\frac{dI}{dt}$$ $$|V_{ind}|=\frac{\mu_0}{2} \frac{R^2}{(z^2+R^2)^{3/2}}πr^2(.03)$$ $$|V_{ind}|=1.01*10^{-11}V$$

3) Flux versus time graph (flux is decreasing)

4) Induced voltage vs time graph ($V_{ind}$=(-)slope of $Φ$)

5) What would be is the force on the coil?

The force on the coil would be radially inward at all points on the small coil (where induced current is flowing).

Case D

1) Induced current direction:

  • Assume current direction from power supply (see diagram above).
  • The current from the power supply is decreases and eventually changes direction overtime, therefore, the magnetic field produced by that current decreases in magnitude until the supplied current flips directions.FIXME (ask for phrasing advice)
  • $Φ_f<Φ_i$
  • The change in magnetic flux is in the opposite direction of the area vector.
  • The induced current generates a magnetic field that will be in the direction that opposes the change in magnetic flux, therefore, the induced current is in the opposite direction (in purple above).

2) Calculate $V_ind$: $$V_{ind}=-\frac{d \Phi_B}{dt}$$ $$=-\frac{d}{dt}(\oint \vec{B}\cdot d\vec{A})$$ $\vec{B}$ is parallel to $d\vec{A}$ so can simplify the dot product: $$=-\frac{d}{dt}(\oint B dA)$$ Assume $\vec{B}$ is constant through area because area of the induced coil is smaller than area of coil connected the to power supply: $$V_ind=-\frac{d}{dt}( B)(\oint dA)=-\frac{d}{dt}(BA_{sm})$$ IMPORTANT: $A_{sm}$ is the area of the small coil in which magnetic flux changes, B is the magnetic field from the large coil.

Magnetic field equation for an observation point at the center of the circular coil, out of the plane of the ring: $$B=\frac{\mu_0 I}{2} \frac{R^2}{(z^2+R^2)^{3/2}}$$ $$V_{ind}=-\frac{d}{dt}(BA_{sm})=-\frac{d}{dt}\frac{\mu_0 I}{2} \frac{R^2}{(z^2+R^2)^{3/2}}A_{sm}$$ Everything is constant except I: $$\frac{dI}{dt}=\frac{d}{dt}(5-0.1t)=-0.1$$ $$V_{ind}=-\frac{\mu_0}{2} \frac{R^2}{(z^2+R^2)^{3/2}}πr^2\frac{dI}{dt}$$ $$V_{ind}=-\frac{\mu_0}{2} \frac{R^2}{(z^2+R^2)^{3/2}}πr^2(-0.1)$$ $$V_{ind}= 3.37*10^{-11}V$$

3) Flux versus time graph (flux is decreasing)

4) Induced voltage vs time graph ($V_{ind}$=(-)slope of $Φ$)

5) What would be is the force on the coil?

The force on the coil would be radially outward at all points on the small coil (where induced current is flowing).

Discussion Prompts:

  • Question: How did you simplify the dot product in your flux calculation?
  • Answer: We assumed the area vector was parallel to the magnetic field at all points on the open surface.
  • Question: What is the area you care about? What is the B-field?
  • Answer: In cases A and B we care about the external B-field and the area surrounded by the rails. In cases C and D we care about the B-field produced from the current in the circuit connected to the power source. The areas we care about is the area of the small loop where current is induced.
  • Question: For your problem, what is the thing that is changing with time?
  • Answer: Case A: area, Case B: area, Case C: current, Case D: current

Alternative Energy Source

The people of Lakeview are starting to get used to living life in the magnetic field, now that it seems to be doing a pretty good job keeping EM-boar tigers from entering the town and causing mayhem. Local alternative energy scientist Alysson Watersa has even installed a source of clean energy that is made possible by the magnetic field. She has constructed a water mill with her bare hands and situated it into the side of the cliff, conveniently below a waterfall so that the motion of the water turns the mill at a frequency of $15 \text{ rpm}$. Dr. Watersa also fixed a board along the axle of the mill, and she has laid a loop of 12-Gauge copper wire (with a radius of $1 \text{ m}$) on the board. One end of the loop becomes a wire that comes out the side of the board and is grounded into the cliff. The other end of the loop becomes a wire that runs out the other side of the mill, all the way back to Lakeview. As the mill turns, so does the loop, causing a current to exist in the wire, which can be used as a small convenience for the good people of Lakeview.

Unfortunately, local superstitious person Manny Callabero is causing a ruckus about the new energy source. He claims that it's all hocus pocus, and Dr. Watersa is just using the water mill as a ploy to hide her stash of jellybeans in the face of the cliff in order to avoid sharing. He won't listen to Dr. Watersa for even a second, so it's up to you to explain to Manny how the water mill works and why it produces the current that it does. It may be helpful for Manny to see a diagram of the mill and a graph of the current produced. Be as detailed as possible, as Manny can be very stubborn about changing his mind. Dr. Watersa's blueprints, shown below, may be helpful.

Diagram

Learning Goals

  • Calculate the induced current from a changing magnetic flux
  • Use the right hand rule to determine the direction of the induced current
  • Explain why induced current is different from the current in a circuit (in other words, what does it mean for a current to be “induced”?)
  • Explain why there is a negative sign in Faraday's Law

Learning Issues

  • Picturing this in 3D is tough - draw a loop on a piece of paper and then use your clipboard to try and play out the scenario with your students.
  • When talking about the magnetic flux in this problem, it helps to get students to work from the formal version of the flux to simplified version of the flux, which they may find useful $$\Phi_{B} = \int \vec{B} \cdot d\vec{A} = \int |\vec{B}| |\vec{dA}| \cos \theta = B \cos \theta \int dA = BA \cos \theta$$. You should talk through this progression and have your students work through this so they understand how the flux varies with the angle (something that is important in this problem!!). Then, ask your students what is it about the flux that is changing because we care about the change in magnetic flux when talking about induced currents/voltages. The magnetic field and area of the loop are both unchanging because of the problem setup, so it must be $\theta$.
  • Writing out $\theta$ in terms of time is also going to be tricky for students because it pulls on some Physics I concepts. The equation is $\theta = \omega t$ where $\omega$ is the angular velocity. In this problem, we know that the board spins at 15 revolutions per minute. Using units, we can deduce that 1 revolution is $2 \pi$ radians in 60 seconds. So, $\omega = \frac{15 2 \pi}{60} = \frac{\pi}{2}$ radians/second. If we multiply by time, we get radians back, which are the units of $\theta$, so our equation checks out.
  • Remember how to calculate resistance from materials properties: $R = \frac{\rho L}{A}$ where $L$ is the length of wire and $A$ is the cross sectional area of the wire.
  • When talking about the direction of the induced current, be careful with language about vectors/scalars. Flux is a scalar quantity, so saying the flux “points up” does not make sense. However, flux can be positive and negative, like energy, depending on if $\vec{B}$ and $d\vec{A}$ point in the same direction or opposite directions. Here's an example of how we can describe what is going on:

Now, from Faraday's Law, we know that the induced current opposes the change in flux, so the induced current should create a positive flux. From our picture of the initial situation, we defined a positive flux as the magnetic field pointing down through the loop. We need to find the direction of the current that would give a magnetic field pointing downward. From the right-hand rule, this would be clockwise around the loop (looking top down). The induced current in the loop is clockwise in this example. Be conscientious of the word choice you are using when describing this type of stuff because it can get confusing very quickly. Also, make sure you are defining clockwise and counter-clockwise consistently. As the loop rotates, the same definition for CW and CCW needs to be maintained.

Alternate Solutions

Quick math overview: We know $\Phi_B = B A \cos \theta$. Using $\theta = \omega t$, we can write the flux as $\Phi_B = B A \cos(\omega t)$. The change in flux is the negative induced current, so if we assume that the flux is constantly changing (the waterfall is always going), then $\frac{d \Phi_B}{dt} = - I_{ind} R$, which, if we take a time derivative of the flux equation we just derived, we get $\frac{d}{dt} (BA\cos(\omega t) = - I_{ind} R$. Solving for the induced current gives us $\frac{ BA \omega \sin(\omega t)}{R} = I_{ind}$.

This problem is pretty tough, so we basically give students their representation. It shouldn't take them very long to figure out what the motion of the water mill looks like.

The students may need to look up or change the units on a few things:

  • Resistivity of copper is $\rho = 1.72 \cdot 10^{-8} \Omega \text{m}$.
  • Radius of 12-Gauge wire is $1 \text{ mm}$, they need to use this to find area.
  • Alternatively, students may look up that the resistance per length of 12-Gauge copper wire is $5.2 \cdot 10^{-3} \Omega\text{/m}$.
  • The frequency of $15 \text{ rpm}$ rotation is $\omega = \pi/2 \text{ s}^{-1}$.
  • From the previous projects, the magnetic field in the Lakeview area is $B = 5 \text{ mT}$, which is directed into the ground.

Students wish to find current, and it may take them a while to figure out how to do this. If they are stuck, it may be worthwhile directing them to the notes on curly electric field or magnetic flux. The current is induced in the loop because a changing magnetic flux produces a curly electric field. An electric field along the loop creates a current. Eventually, they should be able to realize that Faraday's Law is needed, especially the form that includes induced current:

$$I_{ind}R = -\frac{\text{d}\Phi_B}{\text{d}t}$$

At this point, students should know that they need to find flux. We use the following definition, since $\vec{B}$ is constant, and area is flat:

$$\Phi_B = \left| \vec{B} \right| \left| \vec{A}_{\text{loop}} \right| \cos \theta = BA_{\text{loop}}\cos\theta$$

We can visualize theta with the image below. If the board is rotating at a frequency $\omega$, then we could rewrite $\theta = \omega t$ (assuming no phase shift). Students may not realize this at first. Since they must eventually take the time derivative of magnetic flux, they will have to represent something in terms of time. For this problem, that thing is $\theta$.

Theta

So we can write: \begin{align*} \Phi_B &= BA_{\text{loop}}\cos(\omega t) \\ -\frac{\text{d}\Phi_B}{\text{d}t} &= BA_{\text{loop}}\omega\sin(\omega t) \end{align*}

In order to find current, we also need resistance. Students will probability look up resistivity and radius of 12-Gauge copper wire, and use that information to find resistance:

$$R = \frac{\rho L}{A_{\text{cross section}}} = \frac{1.72 \cdot 10^{-8} \Omega \text{m } \cdot 2\pi \cdot 1 \text{ m}}{\pi (1 \cdot 10^{-3} \text{ m})^2} = 0.034 \Omega$$

At this point, students should be able to express the induced current:

$$I_\text{ind} = \frac{1}{R}\left(-\frac{\text{d}\Phi_B}{\text{d}t}\right) = \frac{BA_{\text{loop}}\omega\sin(\omega t)}{R}$$

This is alternating current! We get $I_\text{max} = 0.726 \text{ A}$, and a period of $4$ seconds. A graph is shown below.

Current vs. Time

The math is not hard once the students get to this point, but there are a lot of conceptual struggles that they are going to have to work through. Be prepared to have a lot of conversations about direction of the induced current based on changing magnetic flux as outlined in the Learning Issues.

Discussion Prompts

  • Question: What direction does the induced current flow? For the first half? For the second half? How do you know?
  • Expected Answer: Have students walk you through the right hand rule - see example in learning issues. Make sure they understand why the current changes direction - that the flux decreases as the board spins then increases so current changes directions.
  • Question: Why do we need a negative sign in Faraday's Law? (Also called Lenz's Law.)
  • Expected Answer: There needs to be a negative sign in Faraday's Law for energy conservation to still hold. If the negative sign were not there, there would be situations where you get more energy than you started with (aka you can get infinite energy!)
  • Question: What creates this current? What is pushing the charges around the loop?
  • Expected Answer: The electric field produced by the changing magnetic flux creates this current. (It is NOT due to surface charges.) This electric field would point in the same direction as the induced current (curly around the loop and alternating directions with time).
  • Question: What does it mean for the current to be induced? How is it different from the current in a circuit?
  • Expect Answer: When we say induced current we mean the current that is produced by a curly electric field (which comes from a changing magnetic flux). This is different from the current in a circuit because it has a different source. Both currents are produced from electric fields; however in a circuit, the electric field is created by surface charges whereas the induced current comes from the curly electric field.

Evaluation Questions

  • Question: Is the current produced enough to power the town? What could you do to increase the current?
  • Expected Answer: No the current is not very large. We could make B bigger, make A bigger, make it rotate faster, or use more loops.
  • Question: How did you simplify this problem? What assumptions did you make in this problem?
  • Expected Answer: B-field is constant, board rotates uniformly, wire is ohmic,

Extension Questions

  • Question: This is very similar to how real electrical generators work. As an engineering problem, how do you think they prevent the wires from getting twisted up?
  • Expected Answer: Either the ends of the loop are cut and left as conductive brushes which then are free to spin on a conductive plate, or you can rotate the magnetic instead of the loop.
  • Do you have any lingering questions? Anything that doesn't make sense?
  • Expected Answer: TBD
  • 184_projects/s21_project_11_sol.txt
  • Last modified: 2021/04/09 20:30
  • by dmcpadden