1. Physics 4 – April 11, 2017
Do Now – A loop of conducting wire is placed in the horizontal plane within a decreasing magnetic field that is directed upward. Determine the direction of the induced current in the wire as viewed from above.
Objective:
Review
Assignment: Study for test
Agenda:
Omelet review
Some thermodynamics B.2

2. Capacitors
A capacitor can be formed by any two conductors separated by a distance.
Capacitors are used to store electrical energy.
In practice, capacitors are two identical rectangular parallel plates with an area, A, separated by a distance d with a vacuum between the plates.
If another substance is placed between the plates it is called a dielectric material.
Capacitors are described by their capacitance: C
Unit: Farad, F 1 F = 1 C/V
𝐶= 𝑄 𝑉
𝑄=𝐶𝑉

3. Factors affecting capacitance
The capacitance of a capacitor can be calculated with 𝑪=𝝐 𝑨 𝒅
Where C is the capacitance,  is the permittivity of a vacuum = 8.85 x F/m, A is the area of the plates and d is the separation of the plates.
A higher C can be obtained with larger plate areas, smaller separation of plates, and the introduction of a dielectric material.  for all substances in greater than  for a vacuum.
Recall something else we know about parallel plates: V = Ed with a uniform electric field present between the two plates when a potential V is applied.

4. Sample problem
Consider a parallel plate capacitor with the area for each plate of m2 separated 3.00 mm by a vacuum. Calculate the charge on one plate of the capacitor when it is connected to a 24 V potential.

5. Capacitors in Parallel
Just like resistors in parallel, the potential for each element is the same.
The charge on each capacitor in parallel adds.
QT = Q1+Q2= C1V +C2V = (C1 + C2) V = CT V
For capacitors in parallel, CT = C1 + C2
Notice this is backward from the way resistors are added.
You can solve circuits with capacitors in them using a Q=CxV table just like we used V=IxR tables.

6. Capacitors in Series
For capacitors in series charge for each element is the same.
The potential on each capacitor in series adds.
VT =V1 + V2 = Q/C1+Q/C2 = (1/C1 + 1/C2) V = Q/CT
For capacitors in series, 1/CT = 1/C1 + 1/C2
This is also backward from the way resistors are added.
We won’t look at very complicated circuits with capacitors. Just these two basic kinds.

7. Energy stored in Capacitors
The purpose of a capacitor is to store electrical energy. So how much energy does it store??
For a graph of V vs Q for a capacitor is a straight line.
The slope of the line is 1/C.
Recall W = qV
The area under the curve is the energy stored in the capacitor. E = ½ QV Q = CV
𝐸= 1 2 C 𝑉 2

8. Charging Capacitors
Consider a circuit that connects a capacitor to an electromotive force. Ex: when the switch is placed to A in this circuit.
Charges will build up on the capacitor until the potential across the capacitor is equal to the potential of the battery.
Current starts out at V/r, but decreases to zero.

9. Discharging capacitors
The switch is not flipped to B and the capacitor discharges through the resistor R.
The current, I, potential, V and charge, Q on the capacitor all decrease exponentially at the same rate.

10. Time constant, 
The time constant, , is a time that reflects when the conditions have dropped to 37% of their initial values.
 = RC and has the units of seconds. (Verify this unit)
The time constant is related to the half-life of the decay, the time needed for half to decay
After three , the capacitor is mostly discharged.
τ= 𝑡 ln 2

11. Exponential decay
Current, potential and charge all decay exponentially in the same way.
Each of these equations allow the quantity to be calculated at any given time after the capacitor begins to discharge.
NOTE: they apply ONLY to discharge, not charging.
Initial charge = CV for the charged capacitor.
Initial current = V/R or Q/ (Show how equal)
Initial potential = potential for the charged capacitor.
Recall when solving these equations that ex and ln x are inverse operations.

12. Sample problem
A 45 μF capacitor is charged using a 12 V battery. The battery is disconnected and the capacitor is connected to a 32k resistor through which it discharges. Determine the current, potential and charge for the capacitor after 1.00 second.

13. Magnetism Another fundamental property of matter.
Demonstrated by electrons and protons.
Has a polarity – a north and south pole, a direction
Exists as a field – a directional vector field
Sources include
Particles
Bar magnets
Current carrying wires and solenoids (coiled cylinder of wire)

14. Magnetic fields around sources

15. Direction of magnetic field
Fixed: From south to north
Current in wire: Right hand grip rule – Thumb points in direction of current, fingers curl around wire in direction of circular magnetic field
Solenoid: Right hand grip rule – Fingers curl in direction of wire current around the solenoid, thumb points in direction of the magnetic field
Can be determined in lab using a simple compass.

16. Magnetic Force
Magnetic fields produce a force on moving charges.
F = qvBsin Where q is the charge, v is the speed of the charge, B is the magnetic flux density, and  is the angle between the direction of the charge and the direction of the magnetic field.
Magnetic flux density – measures the strength of a magnetic field.
Symbol: B
Unit: Tesla, T A 1 T magnetic flux density produces a 1 N force on a 1 C charge moving at 1 m/s at right angles to the field.

17. Direction of the magnetic force
The Right Hand Rule: 3 equivalent versions:
Flat hand
Thumb =charge velocity, Fingers = mag. field, Palm = force
Perpendicular 3 fingers
Thumb = charge velocity, First = mag. Field, Middle = force
Finger curl motion
Fingers point = charge velocity, Palm/Curl fingers = mag.field, Thumb = force

18. Magnetic force on a current wire
F = BILsin Where B is the magnetic flux density, I is the current, L is the length of wire in the field, and  is the angle between the current and the magnetic field.
Direction of force given by the same right hand rule as for moving charges, a current is a moving charge.
Force directions are defined for a positive moving charge or current
Recall that electrons move in the opposite direction to a current and will experience the magnetic force in the opposite direction.

19. Force between parallel wires
Two wires can combine their magnetic fields in regular vector field addition, just like we saw with electric fields.
Two wires can be parallel if the currents flow in the same direction or anti-parallel if the currents flow in opposite directions.
Parallel currents create an attractive force
Antiparallel create a repulsive force

20. Definition of the Ampere
Remember we use the ampere as the fundamental SI base units for all things electrical.
Why? Because it can be standardized through the magnetic force between current carrying wires:
If the force on a 1 m length of two wires that are 1 m apart and carrying equal currents is 2 x 10-7 N, then the current in each wire is defined to be 1 Ampere.

21. Motional Electromotive Force (Emf)
We’ve seen that a current carrying wire creates a magnetic field.
The opposite cause and effect can also occur. A magnetic field can cause charges in a wire to move.
How? By moving the wire through the magnetic field.
Right hand rule shows that the magnetic field pushes protons up and electrons down.
The result is a separation of charge, which is a potential difference V… an induced EMF

22. Magnitude of induced EMF 
 = BvL where B is the magnetic field (T), v is the speed of the wire moving through the field, and L is the length of the wire located within the field.
Called a motional emf because it requires motion to create the emf.
If there are multiple loops of wire, you can multiply this emf by the number of loops, N, to find the overall emf of the solenoid.
 = BvLN Both of these forms are in the data packet

23. Power of Induced emf
Connecting this emf to a simple circuit, you can determine the current through a resistor, R. IR = BvL then solve for I = BvL/R
The now current carrying wire generates a force F = BIL = B(BvL/R)L = B2vL2 /R
Recall Power = Fv = W/t
Power of the force generated by the induced emf: P = (B2vL2 /R)v = B2v2L2 /R
Remembering  = BvL , P= 2/R This is equivalent to P = V2/R we saw before in Ch 5.2.

24. Magnetic Flux, 
Motion is required for an induced emf. We can move the magnetic field through a stationary solenoid as shown.
Equivalent to moving a wire through a magnetic field. Only relative motion is required.
Observation of this system show larger currents with:
Greater relative speed
Stronger magnetic field
Higher number of turns
Greater area of loop
Max value when magnet and turns are perpendicular

25. Magnetic Flux, 
The amount of magnetism flowing normally (perpendicular) over an area is called the magnetic flux, .
The magnitude of magnetic flux is  = BAcos where  is the angle relative to the normal to the area.
Unit = weber Wb, 1 Wb = 1 T m2
Useful analogies: Hair within a rubberband or noodles in a loop => magnetic field lines within a given area. Flux is the density of hair, noodles or field lines.
When the magnetic field passes through many loops of a solenoid, it is call a magnetic flux linkage and  = NBAcos

26. Faraday’s Law
The induced emf is equal to the rate of change of the magnetic flux linkage:  = N /t.
 = BAcos and  is the angle to the normal to the area
Note: A changing flux creates an induced emf, not necessarily a current. For a current it has to be placed in a circuit with a resistor.
Ex: The magnetic field through a single loop of area 0.20 m2 is changing at a rate of 4.0 T/s. What is the induced emf?

27. Lenz’s Law
The induced emf will be in such a direction as to oppose the change in the magnetic flux that created the current.
Consider an increase as directed in the direction of the magnetic field. Consider a decrease as directed opposite the magnetic field.
Pretend the loop is a solenoid with one turn and use the Right Hand Grip rule to determine the direction of a magnetic field for the loop that will oppose the “change direction”
Consider a loop in a field directed into board a) increasing, b) decreasing
Consider a loop in a field directed out of board a) increasing b) decreasing

28. Generators
Nearly all of the power stations you read about in Ch 8 were creating a rotational energy for a generator. Here we see how that rotation is turned into electrical energy.
When the loop rotations in the constant magnetic field, the magnetic flux varies because the angle of the area exposes to the magnetic field varies.
The changes in the magnetic flux vary according to the cosine of the angle between the normal to the area and the direction of the magnetic field.

29. Alternating Potential
From Faraday’s law we know that a changing magnetic flux will create an induced electromotive force (emf or voltage, V)
Given that the flux can be modeled with cosine: 𝚽=𝑵𝑩𝑨𝒄𝒐𝒔(𝝎𝒕) where  is the angular speed of rotation of the loop,
Period will be T = 2/, Frequency = f and =2fT = t

30. Alternating Potential
Because potential is the negative derivative of magnetic flux, and 𝚽=𝑵𝑩𝑨𝒄𝒐𝒔(𝝎𝒕)
V=𝝎𝑵𝑩𝑨𝒔𝒊𝒏(𝝎𝒕)
If you haven’t had calc, just accept it. If you have, apply the chain rule to derive.
Notice that when the flux is at a maximum the potential is zero and vice versa.

31. Alternating Current (AC)
When this potential is used in a circuit, the resulting current alternates between forward and backward.
Electrons no longer drift as they do in DC. Instead they wiggle back and forth.
Notice that the Current is exactly in phase with the potential.
𝑰= 𝑰 𝟎 𝒔𝒊𝒏(𝝎𝒕) where 𝑰 𝟎 = 𝑽 𝟎 𝑹

32. RMS Average Voltage and Current
Because these two quantities vary evenly between positive and negative values, one cannot do a simple average. You always get 0.
The solution is what we saw with statistical mechanics: Use a root mean square average.
First square all quantities, then find the average, then square root the answer.
End result is that 𝑰 𝒓𝒎𝒔 = 𝑰 𝒐 𝟐 and 𝑽 𝒓𝒎𝒔 = 𝑽 𝒐 𝟐

33. AC Power
Recall that P = IV. So Pave = 𝑰 𝒐 𝟐 x 𝑽 𝒐 𝟐 = P0/2 or Pave =IrmsVrms = Po/2
Using P = I2/R:
𝑷= 𝝎𝑵𝑩𝑨 𝟐 𝑹 𝒔𝒊𝒏 𝟐 𝝎𝒕
Notice power is always a positive value.

34. Transformers
The purpose of a transformer is to change the voltage for AC electrical lines.
Transformers will not work on DC current because they require a variable magnetic flux.
A transformer has two coils a primary coil which represents the input circuit and a secondary coil which represent the output circuit.
If the voltage of the secondary circuit is higher, it is a step up transformer.
If the voltage of the secondary circuit is lower, it is a step down transformer.
The ratio of the secondary and primary voltages depend only on the ratio of the number of turns for the primary and secondary coil. 𝑽 𝑷 𝑽 𝑺 = 𝑵 𝑷 𝑵 𝑺 = 𝑰 𝑺 𝑰 𝑷

35. Types of Processes
A thermodynamic process is one that takes a system from one state to another.
There are several types of processes dependent on the conditions for the change. (We’ve already seen most with the simple gas laws.)
Isothermal process – Temperature is constant e.g. Boyle’s law
Isobaric process – Pressure is constant e.g. Charles’ law
Isovolumetric process – Volume is constant e.g. Gay-Lussac’s Law
Adiabatic process – No heat transfer

36. PV Diagrams
Pressure vs Volume graphs are the most common way to describe the state of an ideal gas and the processes it undergoes.
Isobaric and isovolumetric processes are horizontal and vertical lines respectively.
Isothermal processes are described by a strict inverse relationship.
Recall Boyles law PV = constant when T is constant.
The closer a line is to the origin, the lower the temperature.

37. Adiabatic Processes
In an adiabatic process, Q = 0. From this, the relationship between P and V can be determined.
The resulting equation means
Its graph on a PV diagram will be similar to an isothermal line, but steeper.
Problems similar to Bolye’s law problems can be solved using this equation if an adiabatic process is described.
Ex: An ideal gas initially at 105 kPa expands adiabatically from 3.5 x 10-4 m3 to 8.9 x 10-4 m3. What is the new pressure?

38. Adiabatic Processes
Notice that for an adiabatic process, P, V and T change.
The initial and final states of an adiabatic process are located on different isothermal lines.
Using the ideal gas law to find the VT relationship, solve PV = nRT for P.
Substituting this into the adiabatic equation and collecting all constant terms. (n, R and constant)
If a gas expands adiabatically, the temperature drops.
If a gas compresses adiabatically, the temperature rises.

39. PV Work Derived Part 1 – Piston h signs
Work is defined as a force applied through a distance.
Recall that work done on a system is positive and work done by a system is negative.
Therefore, the work done by an external force for a compression will be positive and the work done by an external force for an expansion will be negative.
Consider a piston on a canister holding a gas. The piston creates an eternal force. Pick a height = 0 point as the initial piston location.
For an expansion, the h of the gas sample is positive (h – 0)
For a compression, the h of the gas sample is negative (-h – 0)
h=h
h=0
h=-h

40. PV Work Derived – Part 2 Work external
Workext for expansion is negative Workext for compression is positive
Expansion, h positive Compression, h negative
Recall that Work requires the displacement to be in the same direction as the force.
Here, our external force is 180 from the direction of h.
Cos 180 = – 1 . Therefore Wext = – F h
Wext, exp = – F h would be negative, Wext,comp = – F h would be positive.
Results as expected above. Verify this.

41. PV Work Derived – Part 3 Gas Pressure
To transform our Wext expression into PV work, we first recognize that the force exerted by the gas pressure is an equal and opposite reaction force to the external applied force.
W done by the gas is the opposite of the Wext
To convert force into P and h into V, multiply by A/A as a form of 1.
Sort the factors and replace by the definitions of P and V.
Ta Da. That’s why the work done by a gas is called PV work.
W for an expansion is positive and W for a compression is negative

42. PV work on a PV diagram
P is the vertical axis and V is the horizontal, so PV corresponds to the area under the curve for a process.
If the process is not isobaric, the PV work is still the area under the curve.
Sometimes this is easy to calculate, but for isothermals and adiabatic processes, it requires calculus.
Area sweeps to the right are positive work (V is positive)
Area sweeps to the left are negative work (V is negative)

43. Work done for a PV diagram cycle
Many processes that are shown on a PV diagram are cycles that return the state of the system to the initial conditions.
The energy change for a cycle is zero because energy is a state function.
Work is not a state function.
Work is the area within the loop created by the processes that make up the cycle.
CW loop is pos. W. CCW loop is neg. W.

44. First Law of Thermodynamics
When heat is added to a gas, that heat may increase the temperature of the sample of gas or it may cause the gas to expand doing some PV work, or some combination of the two options.
Q = U + W
This is the First Law of Thermodynamics.
Or: The change in internal energy of a closed system is equal to the amount of heat added to the system less the work done by the system.
U = Q – W
Notice in this form, a state function is dependent on two non-state functions.
Also called the Law of Conservation of Energy.

45. Sign conventions for First Law
W > 0 for expansion
W< 0 for compression W = PV
Q > 0 for heat added to system (endothermic)
Q < 0 for heat removed (exothermic) U = Q – W
U > 0 if temperature increases.
U < 0 if temperature decreases. U = 3/2 nRT

46. First Law Problems Remember to think about the sign conventions.
Ex: 5000 J of heat are added to two moles of an ideal monatomic gas, initially at a temperature of 500 K, while the gas performs 7500 J of work. a) What is the change in internal energy of the gas? b)What is the final temperature of the gas?

47. First Law Consequences of Processes
Isothermal process T = 0 So U=0 0 = Q – W and Q = W
Isobaric process W = PV nothing special
Isovolumetric process W = 0 U = Q
Adiabatic process Q = U = – W

48. Second Law of Thermodynamics
The total entropy of an isolated system can only increase over time or remain constant.
In ideal cases where the system is in a steady state (equilibrium) or undergoing a reversible process, there is no change in entropy.
The increase in entropy accounts for the irreversibility of natural processes, and the asymmetry between future and past. The arrow of time only goes one way.
Entropy is a measure of the amount of disorder in a system or a counting of the number of possible arrangements of items.

49. Factors affecting Entropy
State of sample: Gases have more entropy by far than liquids, which have more entropy than solids.
A sample with a larger number of particles will have a larger entropy
A sample at a higher temperature will have more entropy

50. Calculating Entropy Change
Significant entropy change happens during phase changes.
When heat is added or removed from a system at a constant temperature (a phase change) the change in entropy is given by:
∆𝑺= 𝑸 𝑻
Symbol for entropy: S
Unit for S is J/K
Entropy is a state function – it has a value that is path independent

51. Change in entropy signs
If heat is added, Q > 0 and S is positive.
If heat is removed, Q < 0 and S is negative.
For the special cases where a process is reversible, S = 0
It is possible to calculate a value for entropy using statistical mechanics with a zero entropy being a single particle at absolute zero as a reference. S= kB log W
Boltzmann’s Grave in Vienna

52. Heat Engines
The purpose of a heat engine is to capture thermal energy and transform it into mechanical energy.
Converting mechanical energy into thermal energy is easy and is usually accomplished with friction.
From the laws of thermodynamics, we know that thermal energy flows from hot objects to cold objects. It is this thermal energy that a heat engine can capture.
Heat engines usually take the form of a gas within a cylinder with a piston. The piston moves in response to state changes of the gas. The piston can do mechanical work, thus completing the transformation to mechanical energy.

53. A Heat Engine Cycle Visualized
Heat is added from a hot heat source (first image) QH
Gas expands doing mechanical work to lift M (second image) W
The gas is cooled isovolumetrically (third image)
Heat is expelled at a lower temperature (fourth image) QC

54. Refrigerators
The second law states heat will not naturally flow from a cold object to a hot object
But this transformation can be accomplished if a refrigerator does work on the system.

55. Versions of the Second Law
Clausius version:
It is impossible for thermal energy to flow from a cold to a hot object without performing work.
Kelvin version:
It is impossible, in a cyclic process, to completely convert heat into mechanical work.
Mechanical work can be completely converted into heat.

56. Engine Efficiency
Engine Efficiency, , is defined as W/QH x100. That is, what percentage of the thermal energy extracted from the heat sink can be transformed into mechanical energy.
Any unconverted energy is vented to a cold sink as QC.
W = QH - QC
The second law requires that there is no such thing a 100% efficient engine.
The most efficient engine possible is called the Carnot engine and is a limit on the efficiency of any engine operating between any two given temperatures.

57. The Carnot Engine Four stages of the cycle:
12 Isothermal compression, Qc at Tc leaves
Heat expelled, Qc< 0
23 adiabatic compression, increase T
34 isothermal expansion QH at TH added
Heat added, QH > 0
41 adiabatic expansion, decrease T
W = QH – QC

58. Carnot Efficiency
S =Q/T for both isothermal steps. S=0 for the adiabatic steps.
Total S is zero because entropy is a state function
0 = QH/TH – QC/TC
QH/TH = QC/TC
QH/QC = TH/TC QC/QH = TC/TH
 = W/QH = (QH – QC)/QH = 1 – QC/QH = 1 – TC/TH
 = 1 – TC/TH Efficiency = 100% only if TC = 0 K or TH = 

59. Other Engines

60. Exit slip and homework
Exit Slip – None
What’s due? (homework for a homework check next class)
Extra Problems
What’s next? (What to read to prepare for the next class)
Start Studying For the Electromagnetism test on Apr 13