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 final lesson we are going to continue our work on the 1998 practice exam,
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this time focusing on the free response questions.
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Let us dive right in and make sure you have got the test printed out.
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I highly recommend that you take a shot at these questions before coming back to the video.
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Starting with number 1, let us look at part A.
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We have that sphere B1 and a point/ A and that B1 sphere is held in place.
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We area asked to find the charge on sphere B1.
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As I look at this, the first thing I think I'm going to do is draw a free body diagram for B1 to see all the forces that are on it.
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Let us take a look here.
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If there is our sphere, we will put our Y axis and X axis.
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And labeling their forces, we have an electrical force repulsion to the right.
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We have a force of tension from our string, we will call that T.
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We have the weight of the sphere MG.
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Using Newton’s second law, we can write that the net force in the Y direction is going to be T sin 70°.
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If 70° here, if that is 20 to match what they have on our diagram would give us 4 θ, - MG = 0,
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which implies then that T must equal MG/ sin 70° which is our mass 0.025 × the acceleration
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due to gravity on the surface of the Earth 9.8 m / s² / the sin of 70° or about 0.26 N.
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We try the same thing in the X direction.
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The net force in the X direction, we have the electrical force to the right - T cos 70, the X component of that tension.
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Those all have to be = to 0 because it is in equilibrium, it is not accelerating.
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Which implies that the electrical force = T cos 70° is going to be, we just found T was 0.26 N cos 70° or 0.089 N.
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If we know the electrical force, we should be able to back out that charge using in Coulomb’s law.
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Our electrical force is K QA QB / R² which implies that the charge on sphere B is just going to be our electrical force × R² ÷ K QA.
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Our electrical force we just found was 0.089 N.
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The distance between the centers of our charged particles is 1.5 m, K 9 × 10⁹ N meters²/ C² which is the same as 1/ 4 π ε₀.
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The charge on A is 120 µc.
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Put that all into my calculator and I find the value for QB of 1.86 × 10⁻⁷ C.
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Part A, check.
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Part B asks, suppose this b1 is replaced by a second suspended sphere B2 that has the same mass but this one is conducting.
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If we establish equilibrium again, what happens to that equilibrium angle?
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That angle has going to have to be less, that angle θ is going to be less than 20°
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because the charges are going to move in the conductor leaving more positive charge on the far side of your B2 sphere
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and less on the near side near A.
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Effectively, that distance between charges increases so the electrical force between them is going to decrease by Coulomb’s law.
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Electrical force decreases, it is going to come down a little bit.
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I would say that that is going to be less than 20°.
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Here for part C, the sphere B2 is now replaced by a very long horizontal none conducting tube.
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The tube is hollow with thin walls, gives us the radius and the uniform positive charge per unit length λ.
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Use Gauss’s law to show that the electric field is given by that expression.
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I’m really good at Gauss’s law, so we will try that.
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Here is our tube and the Gaussian surface I'm going to choose is a cylinder around that.
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Choose a Gaussian surface over here at some distance R from its center.
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The radius of the little one inside this R, the radius of our Gaussian surface is r.
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We can write Gauss’s law, integral / the closed surface of E ⋅ DA is equal to the total enclosed charge ÷ ε 0.
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The left hand side electric field should be constant because we chose this with symmetry, that in mind.
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And the area of our close surface is going to be 2 π R × its length L and
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that is going to be equal to the enclosed charge which is the linear charge density × our length ÷ ε₀.
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Therefore, our electric field is going to be λ L / 2 π ε₀ RL.
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Our L's make a ratio of 1 so that is just going to the λ / 2 π ε₀ R.
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And if we start putting our values in here, λ is 0.1 × 10⁻⁶ C / m, 2 π ε₀ is 8.54 × 10 ⁻12 C²/ N-m².
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And we also have our R down here.
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This implies then that the electric field is going to be, I come up with about 1797/ R N-m/ C,
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which is approximately 1800/ assuming R is in meters, N/ C with R in meters.
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We prove that 1800/ R N/ C.
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Very good, moving on to part B.
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A small sphere A with charged 120 µc is now brought in the vicinity of the tube and held at distance of 1.5 m.
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Find the force the tube exerts on the sphere.
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That should be pretty easy now that we have the electric field strength.
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The force is just charge × electric field which is going to be our 120 × 10⁻⁶ C × our 1800 N/ C / R 1.5 m or 0.144 N.
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And part E, let us take a look at E here.
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Calculate the work done against the electrostatic repulsion to move sphere A toward
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the tube from a distance R = 1.5 m to a distance R= 0.3 m from the tube.
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That should be a pretty straightforward calculation of work.
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Work = the integral of F ⋅ DR, which is going to be the opposite of the integral as we go from R = 1.5 m to 0.3 m
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throughout the work done against the electrostatic repulsion of, we have our 120 × 10⁻⁶ our charge × our 1800 N/ C ÷ R
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to give us our force QE DR, which implies then that the work is going to be equal to -0.216, when I pull the constants out.
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Integral from R = 1.5 to 0.3 of DR/ R.
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Integral of DR/ R is nat log of R.
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W = -0.216 log of R evaluated from 1.5 to 0.3, which is -0.216 log of 0.3 - log of 1.5,
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which is when I plugged into my calculator about 0.348 J.
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That finishes up free response problem number 1.
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Let us move on to number 2, a circuit problem with the capacitor and inductor.
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As we look here at number 2, we are going to start off with the circuit, the switches initially open,
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the capacitor is uncharged, and there is a voltmeter but they are not showing the measure of potential difference across R1.
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On the diagram above, draw the voltmeter with a proper connections for measuring the potential difference.
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That was nice and simple, what you will only do is you take your drawing that you have there, there is your 20V.
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We have R1 just put your voltmeter in parallel V with R1.
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Taking a look at part B, at time T = 0, the switch is moved to position A so we now have the capacitor in the circuit.
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Find the voltmeter reading for the time right after you do.
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The trick here is the moment you close that, that capacitor initially is going to act like a wire.
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Effectively, you have a circuit that looks like this.
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There is R1, there is R2.
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R1 is 10 ohms, R 2 is 20 ohms, and if we want to know the voltmeter reading,
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we could do this as a voltage divider and probably not too tough to say that it is going to drop 1/3 of the voltage across,
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1/3 of the resistance to 10 ohms which will be 1/3 of 20 or 6.67V.
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But more formally, we can do this by saying the total current in the circuit is E/ R which is going to be 20 V/ your total resistance 30 ohms or 0.667 amps.
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The voltage drop across R1 is just going to be current × resistor 1 which is 0.667 amps × our 10 ohms which is about 6.67 V.
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Moving on to part C, after a long time the measurement of potential difference across R1 is again taken,
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determine for this later time the voltmeter reading.
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After a long time, the capacitor is going to act like an open.
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That is going to break your circuit, you are not going to have any current flowing through R1.
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Therefore, there is no voltage drop across R1 so the voltmeter reading is going to be equal to 0.
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And C2, the charge on the capacitor.
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The entire voltage will be across the capacitor at that point.
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If C = Q/ V, that implies that Q = CV, which is going to be your capacitance 15 µf or 15 × 10⁻⁶ F ×
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your potential difference across your capacitor 20 V or 3 × 10⁻⁴ C.
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There is part C, let us take a look at D.
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At a still later time t = T, the switch is moved to position B.
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Determine the voltmeter reading right after that happens.
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Now we have got a circuit that looks kind of like this, for 2D we got out 20V.
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At that next position for our switch, there is R1.
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Here is our inductor and we have R2 in the mix, that is a 2 H conductor.
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Initially, the inductor poses current flowing and acts like an open so potential across R1 is going to be 0.
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Let us take a look at 2 E, a long time after t = T, the current in R1 reaches some constant final value I final.
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Determine what that is going to be.
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After a long time, that inductor just acts like a wire so I final is going to be equal to the potential ÷ the resistance in our circuit.
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Which is again, 20 V/ 30 ohms or 0.667 amps.
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Part 2, determine the energy stored in the inductor.
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The energy in the inductor is given by ½ LI² and it is going to be ½ × our inductance 2 H × the square of our current 0.667 amps² or about 0.445 J.
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We have got a part F, write but do not solve the differential equation for the current in R1 as a function of time.
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We can use Faraday’s law to do this, the integral/ the closed loop of E ⋅ DL = - the time rate of change of the magnetic flux.
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As we go around here, remember no electric field in there so we have got - E + I R1.
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We have got the voltage drop across I R2 and all of that has to equal their change in flux - L DI DT.
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There is no electric field here so no contribution to the left hand side from your inductor.
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We got ID IDT so I suppose we have already written that differential equation.
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But just to clean it up a touch, let us call that E – I R1 + R2 - L DI DT = 0.
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There is the differential equation and we do not have to solve it this time.
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That is a straightforward number 2 and I think we are going to make up for here in number 3.
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This is one of the trickier free response problems, more involved free response problems I have seen on AP exam.
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Here we have a conducting bar of mass M that is placed on some conducting rails
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at distance L apart that is at an angle with respect to the horizontal.
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We have got a magnetic field coming out of that ramp.
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We are asked to determine the current on the circuit when the bar has reached the constant final speed, so 3A.
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First thing I'm going to do is I'm going to make a free body diagram for that bar and I'm going to tilt my axis
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so it is a little easier for me to see what is going.
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On our bar we are going to have a magnetic force up the ramp and we are going to have gravitational force pulling it down.
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We know that that force is going to be up the ramp, we can use the right hand rule Q × V × B
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to find the force on the charges in the bar creating the current.
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And then the charges in the bar move to find the force on the bar which is going to be up the incline.
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As I do this, as I take a look here, determine the current in the circuit when the bar is reached a constant final speed.
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Let us see, net force in the X direction, Newton’s second law point our free body diagram
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is going to be the X component of its weight MG sin θ - FB the magnetic force, all has to equal MA.
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Since V is constant, A = 0.
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Therefore, we can say that the magnetic force has to equal MG sin θ, its constant final speed.
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The magnetic force, my current flowing through the wire is going to be I LB.
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Therefore, we can state that MG sin θ must equal I LB or the current must be equal to MG sin θ/ LB.
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For part B, we are asked to determine the constant final speed of the bar.
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For B, if current is potential / resistance which is – D φ B DT/ R.
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Let us see here, we could do that and all of that must equal MG sin θ/ LB.
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It would be helpful to know what this D φ B term is.
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Φ B is going to be B × A or in this case B LX are geometry.
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D φ B with respect to time is just going to be BL and X is the only thing that changes with time.
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That is DX DT which is going to B BL × its velocity.
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I can put that in up there for that.
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This implies then that, we have B LV/ R = MG sin θ/ LB.
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And solving just for B 4V, V would then equal, we will have MG R sin θ/ L² B².
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Plugging away through this, part C.
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Part C says, determine the rate at which energy is being dissipated when its reach its constant final speed.
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Varying at which energy is dissipated.
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A lots of fancy way of saying find the power.
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We know I, we know R, so power = I² R is just going to be this² multiplied by R.
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We will have M² G² R sin² θ / L² B².
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Not too bad on C.
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D, express the speed of the bar as a function of time T from the time it is released at T equal 0.
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This one looks a bit more involved.
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Let us see, we know MG sin θ - the magnetic force = MA Newton’s second law.
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We said the magnetic force was I LB, we also know that A is the time rate of change of velocity DV DT.
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MG sin θ – I LB must equal M DV DT.
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But we also know that I = B LV / R, therefore, MG sin θ -, substituting in here for FB, B LV/ R.
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We have still got our LB again, LV LB = M DV DT.
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Let us move that M by all sides by M.
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I get on the left hand side G sin θ - L² B² V/ MR = DV DT.
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Which implies then, it is time to do our separation of variables.
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This might get a little bit ugly, DV/ L² B² V/ MR - G sin θ = – DT.
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We are going to try and integrate this.
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We are going to integrate, the composite.
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The integral from V equal 0 to some final value of V of, we have got DV / L² B² V / MR -G sin θ.
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If I want that = the integral from t equal 0 to T of DT with a negative sign, I want this to fit the formula DU/ U.
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I have got DV, I got V here.
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I have got all this other stuff there as well.
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I’m going to have to have that at the top to make that fit the form so that is L² B² / MR.
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If I multiplying the left by L² B²/ MR, I have to multiply the right by L² B²/ MR × negative integral from 0 to T of DT.
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That fits the form of DU/ U.
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This implies then that this should be a nat log of our U which is down here,
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L² B² V/ MR -G sin θ all evaluated from V = 0 to V, must equal.
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This thankfully is little bit easier to integrate, - L² D² T/ MR.
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We can expand this in the left as we substitute in our 0 and V.
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The left hand side will be, we get L² B² V/ MR - G sin θ - the log of - G sin θ = - L² B² T/ MR.
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We can compress this right hand side.
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The difference of logs is the log of the quotient.
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That is going to imply that the log of, we have L² B² V/ M R - G sin θ/ -G sin θ = - L² B² T/ MR.
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Where do we go from here?
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Let us see, it looks like we are going to have to do some simplifications.
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I think we can get rid of that G sin θ or at least incorporate it a little bit better.
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Which implies then that the log of - L² B² V/ MR G sin θ + 1 = - L² B² T/ MR, raising both sides to E
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to get rid of that natural log, which implies in the left hand side becomes - L² B² V/ M R G sin θ +1
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must equal E ^- L² B² T/ MR, which implies then, moving on to the next page.
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L² B² V/ MR G sin θ = 1 - E ^- L² B² T/ MR.
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Since we are solving for V, we can do one last step.
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Get VL by itself and say that V =, we will have MR G sin θ/ L² B² × (1 - E ^- L² B² T/ MR).
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Let us get that a nice big 3D box because good heaven knows we certainly earned that on that problem.
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A lot involved on part D there.
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Quite a tricky bit of work and a lot of math.
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Let us finish off with part E, so 3 even.
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Suppose the experiment is performed again, this time with the second identical resistor connecting the rails at the bottom.
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Will this affect the final speed and how? Justify your answer.
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If you have 2 resistors in parallel, your total equivalent resistance is going to go down.
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We just said that the final velocity is MG R sin θ/ B² L².
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If R decreases, our final velocity must go down.
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I would say that the final speed increases because R is decreasing there.
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Hopefully that gets you a great start to AP Physics C Electricity and Magnetism.
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Thank you so much for joining us at www.educator.com.
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Make it a great day everybody.