WEBVTT physics/ap-physics-c-electricity-magnetism/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I'm Dan Fullerton and in this lesson we are going to talk about Faraday’s law and Lenz’s law, some of my favorite topics in E and M.
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Is it really starts to tie the entire course together.
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Our objectives include recognizing situations in which change in flux through a loop
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will cause an induced EMF or current in the loop.
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Calculating the magnitude and direction of the induced EMF in current loop of wire or a conducting bar
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where the magnitude of a related quantity such as magnetic field or area of the loop
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as specified is a nonlinear function of time, the change in magnetic fields or areas.
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Analyzing the forces that act on induced currents to determine the chemical consequences of those forces.
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Let us talk about what Faraday’s law tells us.
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Faraday’s law talks about the induced EMF, the induced potential due to a change in magnetic field.
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It says it is equal in magnitude, the rate of change of the magnetic flux through a surface bounded by a circuit.
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Moving charges create magnetic fields, change in magnetic fields move charges by creating potential.
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The direction of the induced current is given by Lenz’s law.
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You usually use Faraday’s law and Lenz’s law together.
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Here is Faraday’s law, the induced EMF typically written as script D
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is the opposite of the time rate of change of the magnetic flux through that surface or the magnetic flux.
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If you expanded that out, it is the opposite of the derivative of the integral / the open surface of B ⋅ DA.
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The EMF is the integral/ the closed loop of E ⋅ DL.
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That is your EMF, that is your potential around a closed loop.
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If you do not have any change in magnetic flux that actually becomes your Kirchhoff's voltage law.
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Lenz’s law is this negative sign right there.
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That is where the Lenz’s law piece comes in.
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We have our surface, we have magnetic field through it and the area is changing or the magnetic field strength is changing,
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it leads to an induced potential difference in the circuit.
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The direction of that, whether it is going clockwise or counterclockwise, that piece comes from Lenz’s law.
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Let us go to Lenz’s law here.
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The current induced by a change in magnetic flux creates a magnetic field opposing the change in flux.
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Or the way I like to think about it is, if something is changing Lenz’s law, the induced current
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or the induce potential difference wants to counteract that.
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As an example, if we had this magnet being moved down into this loop of wire so the magnetic flux into this loop of wire is increasing.
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The induced potential, the induced current wants to oppose that.
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If it is increasing down into that loop, it wants to create a magnetic field that opposes that up out of the loop.
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And as I do that with my right hand rule, if the magnetic field strength is coming out of the loop, is what it is trying to oppose,
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wrap the fingers of your right hand around that and you will see that you will start to get the current in this direction,
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that counterclockwise direction.
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You can also look at it from the perspective, let us draw another one over here.
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We have got a north, I will put a south right beside it.
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I will draw our loop of wire.
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If this time, we have our magnetic flux coming through there but now we are moving our magnet away
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from that loop of wire, Lenz’s law says that it wants it to stay the same.
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It does not want to reduce.
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It is going to create a current in the wire that is going to want to maintain that.
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It is going to increase the magnetic field strength down through the wire.
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By the right hand rule, that means our current must be flowing that way if we are moving our magnet up out of that loop.
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That is how Lenz’s law work.
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It is just a matter of practice.
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Here we go, if we have a sample like this.
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If the magnetic field is increasing, we are going to induce a clockwise current.
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Why? As we take our thumb and point it out of the page, if it is increasing we want to oppose that.
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To oppose that, we go the opposite direction, wrap the fingers of your right hand and you get clockwise.
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If that magnetic field strength there is decreasing, Lenz’s law you want to work to maintain where it does not want to decrease.
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That means you need to add magnetic field coming out of that loop so point your finger coming out of that loop.
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Wrap the fingers of your right hand around that and you are going to get the counterclockwise current.
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More examples for Lenz’s law and how you apply it.
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What we have really done now is, we have put together Maxwell’s equations.
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We started with Gauss’s law, the integral / the close surface of E ⋅ DA is the enclosed charge ÷ ε₀.
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This is always true, it is a law.
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It is useful however in situations of planar, spherical, or cylindrical symmetry.
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Gauss’s law for magnetism, the integral / close surface of B ⋅ DA is 0, which you can also read as there are no magnetic monopoles.
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We went to ampere’s law next, the integral / the closed loop of B ⋅ DL = μ₀ I.
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That is a way to find the magnetic field strength when you have situations of symmetry that is a lot simpler than the Biot-Savart.
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And faraday's law now, the integral/ the closed loop of E ⋅ DL, my voltage is a round the close path is - D / DT,
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integral over the open surface of B ⋅ DA.
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Or - the time rate of change of the magnetic flux due to surface.
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By the way, note the * here for ampere’s law.
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There is a little bit more to ampere’s law than just this that we are going to get to at the very end of the course.
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We are going to come back to that and add to it just a touch but you have got the basics for ampere’s law for now.
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Let us go to some examples, a magnetic field of strength B of T given by 3T² -2T + 1
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is directed out of the plane of the circular loop of wire as shown.
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A lamp somewhere is part of that loop.
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Find the generated electro motor force as a function of time.
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We are going to go right to faraday’s law here.
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The induce EMF is the opposite of the time rate of change of the magnetic flux which is going to be - D / DT integral / the open surface of B ⋅ DA.
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Which you will notice that as we do our integral, the area is not changing.
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The area is constant.
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Integral of B ⋅ DA with their variable of integration DA is just going to be BA.
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This becomes - D/ DT × BA, which implies then that B is a function of time but the area is not.
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Our induced EMF, we can pull our A out of the derivative, is - A × the derivative of B with respect to T,
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which is - A × the derivative of our function B, which is 3T² -2T + 1.
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Which implies then that our electromotor force is going to be, our area is - π R² and the derivative of this is going to be 6T-2.
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There is our function for induced EMF.
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Alright now that we have that, let us see if we can find the current through the 100 ohm lamp as a function of time.
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We know the EMF, current is potential ÷ resistance.
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Ohms law which is - π R² × 6T-2, we just found that in the previous page, ÷ 100 ohms, that easy.
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How about the direction of our current flow at time T = 5s?
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Let us take a look, the derivative of B with respect to T at time T is 6T -2.
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At 5s, that is going to be T = 5s DB DT = 6 × 5s -2 is going to be 28.
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We have got a positive rate of change of magnetic flux.
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As we do that, the positive we are getting bigger this way so it wants to oppose that
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down into this plane of the page, it is going to be clockwise to Lenz’s law.
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Let us do a second example here.
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We have got a circuit in which a current carrying rod on rails is moved to the left with some constant velocity V.
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If the circuit is perpendicular to a constant magnetic field shown here in blue, determine the induced EMF in the circuit.
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We will use Faraday’s law again, E = – D φ BDT which is - D/ DT.
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The integral / the open surface of B ⋅ DA which is going to be, B can come out and DA, A is going to be changing but only one piece is.
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That is going to be - DL and that is coming out we have got this changing X, that will be DX DT.
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DX DT is the definition of V.
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Since V = DX DT, we can write this then as E = -B LV, which may be
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a formula you have memorized if you have a previous algebra based physics course.
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Best way to get better at these I think is to do a lot of practice.
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What I'm going to do is a bunch of old AP problems from old AP exams, the free response problems and we will see how they go.
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Let us start with a 2012 E and M exam free response question number 3.
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You can download that right up here, take a look at it.
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Print it out if you want, give it a shot, and then we will come back and do it together.
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Just like in our example 2, we have one of these situations
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of a cross bar that can be moving as you have different current running through the system.
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Taking a look at part A, it asks us to determine the magnitude of the magnetic flux through the loop when the crossbar is at the position shown.
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That is pretty straightforward, it is not moving.
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Our total magnetic flux is just the integral / the open surface of B ⋅ DA, which is going to be, we have got B0.
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The integral / the open surface of DA, if everything is being nice and constant is just going to be our length L × our width or height A to 0.
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Let us take a look at part B, now the ×bar is released from rest and it slides with negligible friction down without losing electrical contact.
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Indicate the direction of the current and the ×bar as it falls.
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We have got a ×bar and as it falls, the flux through the loop is decreasing.
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Therefore, the induced current wants an inward magnetic flux to oppose that change.
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Because of that, that requires a clockwise current due to Lenz’s Law and the right hand rule.
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If this is the top of that loop at clockwise current is going to have current going to the right.
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To the right would be the direction of current due to Lenz’s law and the right hand rules.
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And I would explain that in words, it will probably be a great idea do not just leave the drawing there when it asks you to justify your answer.
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C, find the magnitude of the current and the crossbar as it falls to the function of speed.
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Current is potential difference ÷ resistance.
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It is going to be – D φ B/ DT ÷ our resistance is going to be - D/ DT of, we have B₀ LH all ÷ R.
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Pulling out our constants, we can pull out - B₀ L/ R.
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We have got DH DT, but DH DT is just going to be our velocity.
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Since velocity = DH DT, we can keep going here and write this as current = - B₀ L × our velocity ÷ R.
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Since it asks for the magnitude, we will state that the magnitude of I = B₀ LV/ R.
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We will plug in through this one, let us give ourselves some more room on another page.
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As we go to part D, derive a differential equation to determine the speed of the crossbar as a function of time.
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The way I would start here is with a free body diagram.
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We have the magnetic force up and we have gravity down.
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Let us call down our positive Y direction as we set our axis.
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As we look at this from the Lenz’s Newton's second law, net force in the Y direction is going to be MG –
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the magnetic force FB and all of that must be equal to MA.
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Find the magnetic force though, recall our magnetic force that is going to be the integral of I × DL × B.
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Or in this case, IL B0.
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We can write this then as MG – IL B0 = MA.
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Which implies then, we know that I = B₀ LV/ R, we did that previously.
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We also know that acceleration is just the time rate of change of the velocity, the derivative of velocity.
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We can write this as, let us see, we have got MG – I, substituting that in there
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we are going to get a B₀² L² V/ R and that must be equal to M DV DT.
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There we have our differential equation.
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We have velocity and the derivative of velocity in the same equation.
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If you wanted to, you could go clean this up a bit more, putting it to standard form.
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It would be fine if you just left it that way, I will do the one last step, in case you want to see it.
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I might write it as DV DT + B₀² L²/ M RV - G = 0.
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As you go through some manipulations but you have already done that differential equation once you get to that step.
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Looking at part E, determine the terminal speed of the crossbar.
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The trick here is recognizing what it said its terminal speed, its terminal velocity, the magnetic force and the gravitational force must balance.
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There is no longer any acceleration so the magnetic force must equal the gravitational force MG.
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Therefore, we said the magnetic force was IL B₀ must equal MG.
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We figured out that I was B₀ LV/ R.
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Substituting that in the left hand side, B₀² L² V/ R = MG.
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Or solving for V which is the terminal velocity, that is just going to be MG R/ B₀² L².
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Finally part F, if the resistance of the crossbar is increased, does the terminal speed increase? Decrease? Or remain the same?
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As we look right here, if resistance increases none of these other factors change along with it.
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It is pretty easy to see that the terminal velocity is going to increase.
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It asked us to give a physical justification for answer in terms of the forces on the ×bar.
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To do the physical justification, I would mention something like the terminal velocity increases
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since currents going to decrease as R increases.
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Since I = B/ R, current decreases R increases and with decreased current, that magnetic force decreases
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so the bar has to fall faster to generate enough current to create a magnetic force that is strong enough to balance the force of gravity.
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Something like that or another physical justification that explains why that is going to increase.
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And that covers the 2012 question.
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Let us take a look at 2010 E and M question number 3 in the free response section.
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Link to download it, print it out, give it a try.
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Here this looks like some of the sample problems we have been doing.
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At least parts of it.
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I got a long straight wire which carries a current I to the right and it varies as a function of time.
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We also have a rectangular loop that has a light bulb with resistance R in there.
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Indicate the direction of the current in the loop for part A.
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As I look at that, the magnetic field through the loop is coming out of the page and it is decreasing.
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You can see by the function I = I₀ – KT.
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The induced current must oppose that by creating flux out of the page or out of the screen.
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By Lenz’s law which by the right hand rule is going to be counterclockwise current.
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Counterclockwise and then explain your answer in terms of Lenz’s law.
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Do not just write Lenz’s law, make sure you talk about the flux that are changing.
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What that is going to do to the current, all of that.
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Part B, indicate whether the light bulb gets brighter, dimmer, or stays the same brightness over the time period of interest.
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As we look at that one, I varies with T.
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The change in flux, the derivative of that is going to be constant.
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As long as that is constant, the induced current is going to be constant.
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And brightness depends on power which is I² R.
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If I is constant and R is constant then the brightness must be constant.
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That must be constant or I suppose you should write remains the same.
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And of course, give a detailed explanation to justify your answer.
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Taking a look at part C, determine the magnetic field at T equal 0 due to current in the long wire, the distance R from the wire.
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B = μ₀ I/ 2 π R.
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Since we are at time T equal 0 that means current = I0.
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Plugging T in for our current equation, therefore B just = I₀/ 2 π R.
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Let us go on to D and I'm going to go to the next page to make sure we have got plenty of room here.
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As we look at D, derive an expression for the magnetic flux through the loop as a function of time.
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The magnetic flux φ B is going to be the integral / the open surface of B ⋅ DA should be the integral from R = D to R = D + A of μ₀ I / 2 π R × B DR.
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As we cut that up to the little horizontal slices and add them all up to get our total flux.
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We can pull out our constants from the integral, μ₀ IB/ 2 π.
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Integral from R = D to D + A of DR/ R, which is going to be the natural log.
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Μ₀ I B/ 2 π log of D + A / D, which implies then that the total flux φ B is going to be, we got μ₀,
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we have got B / 2 π log of D + A/ D.
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We also have our current I, but I is a function of time so let us write that as I₀ – KT.
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We are off to part E, derive an expression for the power dissipated by the light bulb.
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Let us see, the resistance is not going to change, that is constant.
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We are going to need to know either current or voltage, we can use power = V² / R, we will use the induced EMF there.
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First thing to know that induced EMF that is just going to be the opposite of the time rate of change of the flux with respect to time.
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Faraday’s law is going to be - the derivative with respect to T of μ₀ B/ 2 π log of D + A / D × I₀ – KT.
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And that looks like quite a bit but realize the only part of this is a function of T is that - KT over here.
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What we are going to end up with is our EMF going to be K μ B/ 2 π × the log of D + A/ D.
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To the find our power, all we have to do is put all that together to be EMF²/ R or we have μ₀ KB / 2 π log of D + A / D,
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that whole piece squared and then ÷ the resistance R.
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Let us move on and take a look at another problem.
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Let us go to the 2009 exam E and M free response number 3.
00:26:11.600 --> 00:26:15.200
You have not noticed there a lot of these, that means on the E and M test,
00:26:15.200 --> 00:26:21.600
chances are mighty high you are going to see a Faraday’s law, Lenz’s law, free response question.
00:26:21.600 --> 00:26:27.500
Taking a look at the 2009 exam free response number 3.
00:26:27.500 --> 00:26:33.100
We have this square loop of wire with a magnetic field going through it and a couple light bulbs on either side.
00:26:33.100 --> 00:26:42.600
It looks like the magnetic field is a function of time, derive an expression for the magnitude of the EMF generated in the loop.
00:26:42.600 --> 00:26:46.200
Right away makes me think Faraday’s law.
00:26:46.200 --> 00:26:56.600
A, our induced EMF is - time rate of change, our flux derivative of our magnetic flux.
00:26:56.600 --> 00:27:05.800
Which is - D/ DT of, our flux is just going to be, area that is not changing with respective to time, × B.
00:27:05.800 --> 00:27:17.800
That will be - A DB DT, which would be- A × the derivative with respect to time of our function
00:27:17.800 --> 00:27:26.200
for our magnetic field strength AT + B is just going to be - A × a.
00:27:26.200 --> 00:27:30.700
Our area there is a square, it is just going to be L².
00:27:30.700 --> 00:27:44.700
Since A = L², we can say that the magnitude of E is just going to be a L².
00:27:44.700 --> 00:27:48.100
Since we are after magnitude, we do not need worry about our negative sign there.
00:27:48.100 --> 00:27:55.100
E = a L².
00:27:55.100 --> 00:27:58.400
Let us take a look here at B.
00:27:58.400 --> 00:28:02.400
Determine an expression for the current through bulb 2.
00:28:02.400 --> 00:28:09.400
The current through bulb 2 is just going to be the potential ÷ the resistance, which is going to be,
00:28:09.400 --> 00:28:12.500
we got a potential drop those are identical.
00:28:12.500 --> 00:28:20.400
It is going to be half the total potential a L² / 2 × our resistance R0.
00:28:20.400 --> 00:28:22.900
I think that is it.
00:28:22.900 --> 00:28:27.700
Indicate on the diagram the direction of the current through bulb 2.
00:28:27.700 --> 00:28:34.700
As we look at that through bulb 2, it looks like we have an increasing magnetic field in the page.
00:28:34.700 --> 00:28:43.900
As we have that increasing, we want to oppose that, that means you are going to have by Lenz’s law at counterclockwise current.
00:28:43.900 --> 00:28:51.500
Through bulb 2, it is going to be heading up, counterclockwise current through our loops.
00:28:51.500 --> 00:29:02.900
Taking a look at part C then, find an expression for the power dissipated in bulb 1.
00:29:02.900 --> 00:29:09.400
The power is going to be current × voltage.
00:29:09.400 --> 00:29:25.000
If we find the total, it is going to be, we got a L² / 2 R0 is the current and our potential is going to be,
00:29:25.000 --> 00:29:37.800
our total potential is a L² so the potential drop a× our single bulb is going to B ⋅ ÷ 2, since they are identical.
00:29:37.800 --> 00:29:56.900
I come up with, we have a² down to the 4/ 4 R0.
00:29:56.900 --> 00:29:59.700
Now we are going to switch things up a bit.
00:29:59.700 --> 00:30:06.900
It looks like for part D, we have added another identical bulb 3 in parallel with bulb 2,
00:30:06.900 --> 00:30:10.000
completely outside the magnetic field as shown.
00:30:10.000 --> 00:30:14.200
How does the brightness of one compared to what was in the previous circuit?
00:30:14.200 --> 00:30:24.000
Let us see, if we have 2 and 3 in parallel, the equivalent of 2 and 3 over there is going to have less resistance than it had previously.
00:30:24.000 --> 00:30:30.400
Therefore, the total current in the circuit is going to go up so the current through I1 must go up.
00:30:30.400 --> 00:30:32.600
We got the same potential different.
00:30:32.600 --> 00:30:36.300
It got the current going up, if I1 is going up.
00:30:36.300 --> 00:30:43.300
Therefore, we should say that that should get brighter.
00:30:43.300 --> 00:30:50.000
EMF is going to be the same but 2 and 3 are in parallel so as current through 1 goes up, the whole thing gets brighter.
00:30:50.000 --> 00:30:53.900
Brightness is related to power.
00:30:53.900 --> 00:31:04.600
We have got a part E, E says now, we are going to remove that bald 3 but add a wire down the middle.
00:31:04.600 --> 00:31:08.800
How does the brightness of one compared to what was in the first circuit?
00:31:08.800 --> 00:31:14.100
If you put that wire right in the middle, you have that change in flux,
00:31:14.100 --> 00:31:18.300
what you are going to do is each loop is going to create a counterclockwise current.
00:31:18.300 --> 00:31:24.200
In the middle, there going to each to go in opposite directions, completely cancel each other out since it is symmetric circuit and
00:31:24.200 --> 00:31:27.800
the whole thing it acts as if you do not have that wire there in the middle at all.
00:31:27.800 --> 00:31:34.500
It is the exact same circuit function as you had initially so that is going to be the same.
00:31:34.500 --> 00:31:38.200
Again, make sure you justify your answer, explain how you came to that reasoning.
00:31:38.200 --> 00:31:42.400
There are couple different path ways you can take to get there.
00:31:42.400 --> 00:31:46.900
There is the 2009 E and M number 3.
00:31:46.900 --> 00:31:53.600
For the next one, we are going to go to one of the more challenging questions I have seen on an AP exam.
00:31:53.600 --> 00:32:00.800
This is going to the 2008 exam free response 3 which is really mostly a Biot-Savart question but
00:32:00.800 --> 00:32:04.200
it does have some Faraday’s law here at the end.
00:32:04.200 --> 00:32:07.000
This one is a bit involved.
00:32:07.000 --> 00:32:15.600
We have got a circular loop of wire in figure 1 that has a radius R and carries some current I and point P is R/ 2 above the center.
00:32:15.600 --> 00:32:24.300
First off, find the direction of the magnetic field at point P due to the current in the loop.
00:32:24.300 --> 00:32:29.100
By the right hand rule, that should be heading upwards as you go around the circle.
00:32:29.100 --> 00:32:34.700
A1 should be up.
00:32:34.700 --> 00:32:39.700
2 however, calculate the magnitude of the magnetic field B1 at point P.
00:32:39.700 --> 00:32:42.800
This is going to take a little bit of work, I think.
00:32:42.800 --> 00:32:48.400
As we look at 2, let us make a diagram of what we have over here.
00:32:48.400 --> 00:33:00.100
We got our loop of wire, we have got a point P up here, and we have done Biot-Savart problems that are similar.
00:33:00.100 --> 00:33:07.700
We have got our radius R, we got our current flowing through our loop I.
00:33:07.700 --> 00:33:15.300
This distance here is R/ 2, let us pick some point over on the side.
00:33:15.300 --> 00:33:26.200
We will call that our DL and we will draw our line for the distance from P to DL.
00:33:26.200 --> 00:33:43.000
That is our R vector, where it looks like R is going to be √ of R² + R/ 2², which when we simplify that,
00:33:43.000 --> 00:33:50.500
that is going to come out to be the magnitude of that is R/ 2 √ φ.
00:33:50.500 --> 00:33:56.700
There is our R, we will define this angle as φ.
00:33:56.700 --> 00:33:59.200
I think we have got a pretty good setup.
00:33:59.200 --> 00:34:08.200
As we come here from here to there, we are going to have our DB in that direction.
00:34:08.200 --> 00:34:13.300
It is pretty easy to see we do are not going to have to worry about the vertical component of this.
00:34:13.300 --> 00:34:20.600
Let me just continue up this line a little bit because if that is φ, it is usually helpful to know that that angle there is φ as well.
00:34:20.600 --> 00:34:25.100
Let us see where we can get with what we have so far.
00:34:25.100 --> 00:34:46.300
Biot-Savart, DB =, we have got μ₀ I/ 4 π × DL × R vector ÷ R³.
00:34:46.300 --> 00:34:54.500
We can state that DB = μ₀ I/ 4 π R³.
00:34:54.500 --> 00:35:05.100
DL × R that is going to be the DL R sin θ.
00:35:05.100 --> 00:35:25.500
Θ is 90° between those so that is going to be 1, which implies that DB is going to be equal to μ₀ I/ 4 π R² DL.
00:35:25.500 --> 00:35:30.600
We are only worried about the vertical component, everything else is going to cancel out here.
00:35:30.600 --> 00:35:37.300
Let us see if we can integrate to find our magnetic field strength.
00:35:37.300 --> 00:35:43.200
The vertical component here is going to be the integral of DB vertical which will be
00:35:43.200 --> 00:35:59.700
the integral of DB cos φ which is equal to, as we look at that cos φ is going to be SOHCAHTOA.
00:35:59.700 --> 00:36:08.100
Adjacent / hypotenuse that is going to be R / r.
00:36:08.100 --> 00:36:28.500
This is going to be the integral of R / r DB, which implies then that vertical component is going to be the integral of R / r.
00:36:28.500 --> 00:36:41.400
We already found out the DB is μ₀ I/ 4 π R² DL.
00:36:41.400 --> 00:36:52.300
It is just going to be, if we pull out all our constants, you have μ₀ IR/ 4 π r².
00:36:52.300 --> 00:36:57.100
I realize that we could take r³ since we know what r is in terms of R and simplify it.
00:36:57.100 --> 00:36:59.600
We just leave it like this for now.
00:36:59.600 --> 00:37:19.500
× the integral around our entire wire of DL is going to be μ₀ IR/ 4 π r³ × 2 π R,
00:37:19.500 --> 00:37:33.000
which implies the vertical component of our magnetic field is going to be μ₀ I R² / our π
00:37:33.000 --> 00:37:37.500
are going to cancel out, we are going to get 2 R³.
00:37:37.500 --> 00:37:48.100
We will substitute in, it is going to be since we know that r is R / 2 × √ φ.
00:37:48.100 --> 00:38:02.100
I’m going to have for our vertical component, it is going to be equal to μ₀ IR² / 2 ×, let us see.
00:38:02.100 --> 00:38:09.900
We are going to end up with R/ 2 3 is going to be R³/ 8.
00:38:09.900 --> 00:38:19.700
The √ of φ³ is going to be φ √ φ, so that is going to be μ₀ I.
00:38:19.700 --> 00:38:35.900
Our R² can cancel R² to that, over 1/4 × φ, φ/ 4 of √ φ × R.
00:38:35.900 --> 00:38:58.900
If I bring that 4 to the top, we are going to have a vertical component to our magnetic field of 4 μ₀ I/ φ √ φ R.
00:38:58.900 --> 00:39:06.100
One of the tougher pieces of a problem I have seen on AP exam.
00:39:06.100 --> 00:39:13.700
Let us go to our next page to take a look at part B.
00:39:13.700 --> 00:39:20.700
For part B, it says determine the magnitude of the net magnetic field B at that point P.
00:39:20.700 --> 00:39:24.100
We have now that the second ring here.
00:39:24.100 --> 00:39:32.700
B at point P is just going to be twice what we had before, because the magnetic field due to both of those is going to end up giving you,
00:39:32.700 --> 00:39:37.500
you are going to add together because they are in the same direction.
00:39:37.500 --> 00:39:48.100
That is 2, what we had for B vertical is just going to be 8 μ₀ I/ φ √ φ R.
00:39:48.100 --> 00:39:54.600
That is what we are going to call B net.
00:39:54.600 --> 00:39:59.300
We have got a square loop of wire between those of length S on each side.
00:39:59.300 --> 00:40:06.300
We are asked to determine the magnetic flux due to the square loop in terms of B₀ and S.
00:40:06.300 --> 00:40:20.000
For part C, magnetic flux is the integral of B ⋅ DA is just going to be, in this case just BA.
00:40:20.000 --> 00:40:30.700
The area is not changing so that is 8 μ₀ I/ φ √ φ R S².
00:40:30.700 --> 00:40:34.000
It wants it in terms of B net, that is just B net.
00:40:34.000 --> 00:40:49.400
Simply, that is just B net S².
00:40:49.400 --> 00:40:56.100
Part B, a square loop is now rotated about an axis in its plane at angular speed ω.
00:40:56.100 --> 00:41:04.500
In terms B net S and ω, find the induced EMF as a function of time and assume it is horizontal T = 0.
00:41:04.500 --> 00:41:10.400
Induced EMF, we will use Faraday’s law – D φ BDT.
00:41:10.400 --> 00:41:22.200
It will be - the derivative with respect to T of what we had for our flux B net S².
00:41:22.200 --> 00:41:24.200
But it is rotating now at time T = 0, it is horizontal.
00:41:24.200 --> 00:41:38.700
We can use a cos factor to model that of cos ω T.
00:41:38.700 --> 00:41:47.100
As we take the derivative of this, we can pull the B and the S is out, those are constants.
00:41:47.100 --> 00:41:58.000
That is going to be – B net S² × the derivative with respect to T of cos ω T,
00:41:58.000 --> 00:42:04.400
which implies then that EMF is going to be, derivative of cos is opposite of sign.
00:42:04.400 --> 00:42:06.900
Those negatives will cancel out.
00:42:06.900 --> 00:42:26.800
We will have ω × our net magnetic field S² sin ω T.
00:42:26.800 --> 00:42:33.200
It is the end of that question.
00:42:33.200 --> 00:42:37.200
Let us go to the 2007 exam now free response number 3.
00:42:37.200 --> 00:42:41.700
We have seen something similar to this by now as well.
00:42:41.700 --> 00:42:50.300
We have got another rod on nails with a chrome wire but they bent into a 3 sided rectangle.
00:42:50.300 --> 00:42:54.800
We have got the moving rod, that is moving to the right.
00:42:54.800 --> 00:42:59.800
All in the uniform magnetic field in the plane of the screen.
00:42:59.800 --> 00:43:06.000
As we do all this, indicate the direction of the current induced in the circuit as it moves to the right.
00:43:06.000 --> 00:43:12.200
As it move to the right, you are increasing the flux into this page or screen so the induced magnetic field has to oppose this out of the screen.
00:43:12.200 --> 00:43:22.200
The induced current by the right hand rule and Lenz’s law must be counterclockwise.
00:43:22.200 --> 00:43:25.900
Counterclockwise and justify your answer.
00:43:25.900 --> 00:43:28.400
I will leave you guys to explain that one in your own papers.
00:43:28.400 --> 00:43:32.300
Probably it is good to practice that, do not just leave it as counterclockwise.
00:43:32.300 --> 00:43:38.200
Do not just say Lenz’s law or right hand rule, walk through the explanation.
00:43:38.200 --> 00:43:47.100
Part B, derive an expression for the magnitude of the induced current as a function of time T.
00:43:47.100 --> 00:43:49.300
We have already solved this problem before.
00:43:49.300 --> 00:43:51.600
You might have seen in previous physics courses.
00:43:51.600 --> 00:44:00.300
The induced EMF is just BLV and we know that the resistance, we can figure out with a little bit of geometry.
00:44:00.300 --> 00:44:07.300
We have got the resistance prior to length λ × the total length.
00:44:07.300 --> 00:44:15.100
We are going to have L + the length of the top and bottom pieces which is going to be 2 VT.
00:44:15.100 --> 00:44:20.300
If we want the current by Ohm’s law, it is V/ R.
00:44:20.300 --> 00:44:34.900
That is going to be BLV / λ × L + 2 BT.
00:44:34.900 --> 00:44:38.200
Λ × O + 2 BT should work.
00:44:38.200 --> 00:44:47.500
Taking a look at parts C now, derive an expression for the magnitude of the magnetic force on the rod as a function of time.
00:44:47.500 --> 00:45:04.000
The force on that rod is just going to be ILB and we just found I, that is going to be BLV/ λ × L + 2 VT.
00:45:04.000 --> 00:45:30.800
And we have still got our LB which is going to be B² L² V all ÷ λ × (L + 2 VT).
00:45:30.800 --> 00:45:37.800
On to part D, we are going to graph so let us give ourselves some more room for part D.
00:45:37.800 --> 00:45:41.000
On the axis below ,sketch a graph of the external force as a function of time that
00:45:41.000 --> 00:45:48.300
must be applied to the rod to keep it moving at constant speed.
00:45:48.300 --> 00:46:03.400
Graph a force vs. Time, label this is our external force, here is our time.
00:46:03.400 --> 00:46:09.800
The trick here is, in order to move at constant speed, we know that the external force must match,
00:46:09.800 --> 00:46:14.600
must absolutely oppose the magnetic force so that they are balanced, equal.
00:46:14.600 --> 00:46:19.100
When we are at equilibrium, we are moving at a constant speed, no acceleration.
00:46:19.100 --> 00:46:37.700
We know what time 0, that force was to B² LV/ λ because FB at T equal 0 is B² L² V/ λ L.
00:46:37.700 --> 00:46:47.200
Our L cancel out, B² V/ λ and over time, that is going to steadily decrease.
00:46:47.200 --> 00:46:55.300
It is going to look something like that, I would expect.
00:46:55.300 --> 00:47:01.200
Finally E, what do we have for part E?
00:47:01.200 --> 00:47:04.400
The force pulling the rod is now removed, my goodness.
00:47:04.400 --> 00:47:11.500
Indicate whether the speed of the rod increases, decreases, or remains the same, and explain why.
00:47:11.500 --> 00:47:21.000
If there is no external force to keep it moving at the constant speed, the magnetic force is opposing the motion.
00:47:21.000 --> 00:47:35.300
If it is opposing motion, by Newton’s second law which says that F = MA, the rod must be accelerating
00:47:35.300 --> 00:47:51.000
but it is accelerating in the direction opposite its velocity, it is slowing down.
00:47:51.000 --> 00:48:03.300
Therefore, our answer must be decreases.
00:48:03.300 --> 00:48:12.800
Let us take a look now, go to the 2006 exam free response number 3.
00:48:12.800 --> 00:48:14.500
Here is a kind of a unique one.
00:48:14.500 --> 00:48:17.800
We now have our circuit that is hanging from the spring.
00:48:17.800 --> 00:48:24.800
I wonder how they add up some of these but we will deal with it anyhow.
00:48:24.800 --> 00:48:32.400
Part A, on a diagram of the loop, indicate the direction of the magnetic forces of any that act on each side of the loop.
00:48:32.400 --> 00:48:38.500
Let us draw the loop in there, we have got our source of potential difference.
00:48:38.500 --> 00:48:45.000
We got our switch and looks like it is closed.
00:48:45.000 --> 00:48:48.100
There is our loop up.
00:48:48.100 --> 00:48:54.500
As we look at the different portions of the loop, its top part is not in the magnetic field so we do not have to worry about that.
00:48:54.500 --> 00:48:57.600
Our current is going to be flowing this way.
00:48:57.600 --> 00:49:04.000
Over here on the left hand side, by the right hand rule current flowing toward the top of the screen,
00:49:04.000 --> 00:49:09.900
positive charges point the fingers of your right hand for the top of the screen and then into the screen the direction of the magnetic field.
00:49:09.900 --> 00:49:15.500
We should see a force to the left.
00:49:15.500 --> 00:49:20.300
Similarly on the right side, we are going to see a force to the right.
00:49:20.300 --> 00:49:26.200
On the bottom, we are going to see a force down.
00:49:26.200 --> 00:49:32.500
Left, right, and down, that should cover part A.
00:49:32.500 --> 00:49:38.100
For part B, the switch is open and the loop eventually comes to rest at new equilibrium position
00:49:38.100 --> 00:49:40.300
at distance X from its former position.
00:49:40.300 --> 00:49:44.800
Find an expression for the magnitude of the uniform magnetic field.
00:49:44.800 --> 00:49:47.900
Let us draw a free body diagram there.
00:49:47.900 --> 00:49:55.700
We have got the force of the spring KX by Hooke’s law up and we have the magnetic force down.
00:49:55.700 --> 00:50:00.800
If it is coming to rest at new equilibrium position, these two have to be equal.
00:50:00.800 --> 00:50:06.600
The spring force must equal the magnetic force in magnitude.
00:50:06.600 --> 00:50:16.100
Therefore, KX the force of the spring must equal the force on our loop of wire in a magnetic field which by now we know as ILB.
00:50:16.100 --> 00:50:22.600
Therefore, B is going to be equal to KX / IL.
00:50:22.600 --> 00:50:27.900
The one trick here, our length is W in this problem.
00:50:27.900 --> 00:50:38.300
I would write this as KX / IW to be consistent with the values and variables that they have given us.
00:50:38.300 --> 00:50:47.500
Part B, moving on to part C of the question.
00:50:47.500 --> 00:50:53.600
The spring in the loop place with a loop and the same dimensions of resistance but no battery and no switch.
00:50:53.600 --> 00:50:58.400
The new loop is pulled upwards at constant speed V0.
00:50:58.400 --> 00:51:03.400
On the diagram, indicate the direction of the induced current loop as it moves upward.
00:51:03.400 --> 00:51:07.400
Lenz’s Law problem.
00:51:07.400 --> 00:51:20.200
We have got our loop right here, we have magnetic field into the plane of the screen.
00:51:20.200 --> 00:51:26.700
And we are pulling that thing upward for some unknown reason with velocity V0.
00:51:26.700 --> 00:51:34.800
As we are pulling that up, the flux into the plane of the screen, the magnetic flux is decreasing.
00:51:34.800 --> 00:51:39.000
It is going to want to oppose that by Lenz’s law by increasing that flux.
00:51:39.000 --> 00:51:41.100
Point the thumb of your right hand into the plane of the screen.
00:51:41.100 --> 00:51:52.300
The direction you want increase that by Lenz’s law and you get a clockwise induced current.
00:51:52.300 --> 00:51:58.100
There is C1, let us take a look at C2.
00:51:58.100 --> 00:52:00.900
Derive an expression for the magnitude of this current.
00:52:00.900 --> 00:52:05.100
I could not just leave it without telling you the direction.
00:52:05.100 --> 00:52:13.800
EMF is - D φ BDT, we will start there.
00:52:13.800 --> 00:52:24.700
That is going to be - D / DT and our flux is the integral of the B ⋅ DA thought that open surface.
00:52:24.700 --> 00:52:30.300
That is going to be -, we can pull B out, that is constant for this problem.
00:52:30.300 --> 00:52:34.800
The derivative with respect to T of the integral of DA.
00:52:34.800 --> 00:52:44.000
That is going to be – B, the integral of DA is just going to be A, that is not changing with the variable as you integrate / variable DA.
00:52:44.000 --> 00:52:58.000
That is going to be - B DA DT, which will be - B × the derivative with respect to time of our area which is W our width.
00:52:58.000 --> 00:53:03.900
Our Y coordinate, Y and W you can come out of there, that is not a constant that is not changing.
00:53:03.900 --> 00:53:19.900
- W DY DT, but DY DT is just our V0 so then we can say that E, our induced EMF is – BW V0.
00:53:19.900 --> 00:53:26.100
If we want the current, ohms law I = E / R.
00:53:26.100 --> 00:53:30.500
I suppose we are after the magnitude of the current so let us put it that way.
00:53:30.500 --> 00:53:52.300
It is going to be the magnitude of - BW V₀/ R or just BW V₀ / R.
00:53:52.300 --> 00:54:02.100
Moving on to D, find the power dissipated in the loop as it pulls its constant speed out of the field.
00:54:02.100 --> 00:54:12.700
Power is current × voltage, we just found our current is BW V₀ ÷ R.
00:54:12.700 --> 00:54:21.100
We found our potential magnitude of that as BW V₀.
00:54:21.100 --> 00:54:32.200
When I put all these together, I come up with B² W² V₀² / R.
00:54:32.200 --> 00:54:36.400
That one was not so bad.
00:54:36.400 --> 00:54:45.300
We are back to part E, suppose the magnitude of the magnetic field is increased.
00:54:45.300 --> 00:54:52.000
Does the external force required to pull the loop at speed V₀ increase, decrease, or remain the same?
00:54:52.000 --> 00:55:04.900
If you have a stronger magnetic field, you are going to get a stronger magnetic force which is equal to ILB.
00:55:04.900 --> 00:55:16.400
The larger induced current which was BW V₀ / R, as B goes up I induced goes up.
00:55:16.400 --> 00:55:22.800
Both of those are going to lead to a stronger force applied.
00:55:22.800 --> 00:55:31.200
All lead to stronger F external.
00:55:31.200 --> 00:55:41.000
We can say that that is going to increase.
00:55:41.000 --> 00:55:43.500
That is the end of that question.
00:55:43.500 --> 00:55:48.900
Let us do one last practice problem here.
00:55:48.900 --> 00:55:55.000
We will go to the 2013 exam free response number 3.
00:55:55.000 --> 00:55:58.400
We have got a loop of wire in a magnetic field.
00:55:58.400 --> 00:56:06.200
It gives us the area, it gives us a nice little plot of the magnetic field strength as a function of time.
00:56:06.200 --> 00:56:08.300
There is something funky about that graph too.
00:56:08.300 --> 00:56:11.600
If you look at the X axis, the time in seconds.
00:56:11.600 --> 00:56:20.000
We have even intervals of 4, 8, 12, 16 but at the last interval looks like it should be 4s, but it is only 2 to get the 18
00:56:20.000 --> 00:56:24.300
so that is not drawn to scalar, somebody made a mistake when they drew up the question.
00:56:24.300 --> 00:56:29.100
Do not think that is going to affect us though but we will keep an eye for that.
00:56:29.100 --> 00:56:39.500
Alright part A1, derive an expression for the magnitude of the induced EMF in the loop as a function of time from 0 to 8s.
00:56:39.500 --> 00:56:45.100
In that curvy part of the graph where we are given that function for magnetic field strength.
00:56:45.100 --> 00:56:59.600
We will go to Faraday’s law E = – D φ B DT, which is - the derivative with respect to time of AB which,
00:56:59.600 --> 00:57:12.700
since A is a constant, will be - A DB DT which is - A × the derivative with respect to time of our function
00:57:12.700 --> 00:57:23.400
for B given to us on that graph is 1.8 E⁰.05 T.
00:57:23.400 --> 00:57:34.500
This implies then that E = - A and the derivative of that is going to be, the derivative of E ⁺U is E ⁺UDU.
00:57:34.500 --> 00:57:45.500
We will have our 1.8 E⁰.05 T × -0.05.
00:57:45.500 --> 00:57:50.900
Putting all that together that is going to be our negative area 0.25 m².
00:57:50.900 --> 00:58:16.300
We got our 1.8 E⁰.05 T, we have got our -0.05, which implies then that our total induce EMF is going to be 0.0225 E⁰.05 T.
00:58:16.300 --> 00:58:22.400
There is our induced EMF as a function of time from 0 to 8s.
00:58:22.400 --> 00:58:32.500
I guess that is A2, calculate the magnitude of the induced current at 4s.
00:58:32.500 --> 00:58:48.400
Current is E/ R V/ R by Ohm’s law, is just going to be 0.0225 E⁰.05 T/ 12.
00:58:48.400 --> 00:59:09.400
Since, we said T is = 4, I plugged into my calculator and I come up with a value of about 0.00154 amps or 1.54 milliamps.
00:59:09.400 --> 00:59:13.500
Let us take a look now at part B here.
00:59:13.500 --> 00:59:14.400
Give ourselves some room.
00:59:14.400 --> 00:59:17.400
It looks like it is a graph in question.
00:59:17.400 --> 00:59:34.500
We will draw our graph here first.
00:59:34.500 --> 00:59:42.300
We have current in amps vs. Time in seconds.
00:59:42.300 --> 00:59:47.800
We got 4, 8, 12, 16, and then they have another evenly spaced.
00:59:47.800 --> 00:59:50.900
I’m going to draw mine right and say that is 18.
00:59:50.900 --> 00:59:57.000
And 4, 8, 12, 16, 18s for our graph.
00:59:57.000 --> 01:00:05.400
Sketch a graph of the induced current of loop as a function of time assuming a counterclockwise current is positive.
01:00:05.400 --> 01:00:10.500
This is counterclockwise up here, this is clockwise.
01:00:10.500 --> 01:00:15.200
If I were to draw that graph, I'm going to start with the easy parts first.
01:00:15.200 --> 01:00:23.400
Recognizing that where we do not have any change in magnetic flux it is going to be 0 so from 8 to 12 s, it is got to be 0.
01:00:23.400 --> 01:00:26.400
We do not have any DB DT is 0.
01:00:26.400 --> 01:00:31.500
And beyond 16s, 16 to 18 is 0.
01:00:31.500 --> 01:00:37.900
Now, let us also look in the region from 0 to 8s.
01:00:37.900 --> 01:00:46.300
It is going to be a clockwise current Lenz’s law and it is going to go something like that.
01:00:46.300 --> 01:00:59.500
And from 12 to 16s, we have got a strong negative slope but it is constant so we should have a strong negative or clockwise current right there.
01:00:59.500 --> 01:01:03.200
That should lineup with 12 and 16s.
01:01:03.200 --> 01:01:14.700
I would do that for our graph, part 2 B2 says for the time interval from 12 to 16s, justify the direction of the current that you have indicated.
01:01:14.700 --> 01:01:18.600
By Lenz’s law, the magnetic field is in the page and it is decreasing.
01:01:18.600 --> 01:01:27.800
The induced current creates a field into the page that is opposing that change which is clockwise by the right hand rule.
01:01:27.800 --> 01:01:32.300
There is part 2, write that out somehow in your own words, of course.
01:01:32.300 --> 01:01:38.400
For part C, let us take a look at C here.
01:01:38.400 --> 01:01:43.800
Calculate the total energy dissipated in the loop during the first 8s.
01:01:43.800 --> 01:01:57.500
Energy is going to be the integral of power with respect to time so that is going to be the integral of E²/ R DT
01:01:57.500 --> 01:02:14.800
to be the integral from T = 0 to 8s of 0.0225 E⁰.05 T all of that squared ÷ our 12 ohms DT.
01:02:14.800 --> 01:02:31.700
Pulling out our constants, that will be 0.0225²/ 12 integral from 0 to 8s of E⁰.1 T DT.
01:02:31.700 --> 01:02:44.300
This implies then that our energy is going to be, all of that comes out to about 4.22 × 10⁻⁵.
01:02:44.300 --> 01:02:51.200
We get to have that E ⁺U DU, we need to have a DU there so it is going to be -0.1.
01:02:51.200 --> 01:03:07.100
We got to pullout a 1/ -0.1 infinity integral sign, integral from 0 to 8 of E⁰.1 T × -0.1 DT.
01:03:07.100 --> 01:03:12.500
We can fit the form E ⁺U DU integral of E ⁺U DU is just E U.
01:03:12.500 --> 01:03:22.500
That is going to be negative 4.22 × 10⁻⁴, as we multiply those together,
01:03:22.500 --> 01:03:41.500
× E⁰.1 T which we are going to evaluate from 0 to 8s, which implies then that E = -4.22 × 10⁻⁴.
01:03:41.500 --> 01:03:54.400
E⁰.8 - E⁰, E⁰ is 1 so this is just going to be -4.22 × 10⁻⁴.
01:03:54.400 --> 01:04:15.400
E⁰.8 is 0.449 – E⁰ is 01, put that all into my calculator and I come up with about 2.32 × 10⁻⁴ J.
01:04:15.400 --> 01:04:24.600
I think that finishes out our look at Faraday’s law and Lenz’s law with a bunch of sample problems.
01:04:24.600 --> 01:04:29.600
Keep practicing these, you will see a lot of the same patterns and forms come up again and again and again.
01:04:29.600 --> 01:04:32.200
Thank you so much for watching www.educator.com.
01:04:32.200 --> 01:04:33.000
Make it a great day everyone.