Unit 3 Gravitational, Electric and Magnetic Fields Gravitational Fields

Transcription

1 lesson33.notebook September 17, 2013 Unit 3 Gravitational, Electric and Magnetic Fields Gravitational Fields Today's goal: I can explain how gravitational fields influence objects within them and make the correct calculations to determine their quantitative influence. Recall: A force is defined as a push or pull. Gravitational, Electric and Magnetic fields are all examples of action " at a distance" forces. For each of these forces, no "direct contact" is required to achieve an end goal. "Action at a distance" describes the characteristics of these forces but does not explain how the results are achieved. The concept of "fields" was first introduced by Michael Faraday in the early 1 A field is a property of space. An object in that space, or the "source", influences the behaviour of the surrounding space. The field in turn exerts a force on other objects located within the space. The easiest example for us to visualize at this stage is the presence of the Earth and it's effect on the surrounding space. Gravitational fields is the first concept this unit. Of all 3 areas, this topic is the most knowledgeable for us as we have worked with gravity for several units now. Gravitational fields created by objects with mass is defined by the equation: Equation: Graph: Two objects, 200 kg and Discussion:

2 lesson33.notebook September 17, 2013 'Shifting our way of thinking... Up to this point we use the letter 'g' to represent 9.8 m/s 2. We refer to this term as simply "gravity" or "acceleration due to gravity". With knowledge of fields we need to take our understanding to the next level. Equation: Starting with F = mg 'g' is in fact what we call the gravitational field intensity. Gravitational field intensity is the relationship between the force of attraction (due t the mass of the Earth) and the mass of the object. As we move closer to the Earth the intensity of the field increases whereas if we move away the field has less effe on an object. Example: Determine the gravitational field intensity for an object located 1000 km away from Earth.

3 lesson33.notebook September 17, 2013 Gravitational Potential Energy Gravitational potential energy can be defined as the energy required to change an object's distance from the centre of the Earth. To find energy we need to find the area under the curve. This requires us to use math beyond the scope of this course. Below would be the resulting equation for area. A = F 1 F 2 x (r 2 r 1 ) F 1 F 2 r 1 r 2 E g r e Therefore: ΔE g = and E p = is only valid for situations close to the Earth's surface Example: What is the change in GPE of a 64.5 kg astronaut, lifted from the Earth's surface (r E = 6.38 x 10 6 m) to a circular orbit of altitude 4.4 x 10 2 km?

4 lesson33.notebook September 17, 2013 Escaping From a Gravitational Field If we analyze the situation from an energy standpoint: E T = E k + E g, but on the surface of the Earth, E k = 0 J. = E g Therefore, the binding energy for an object on Earth is: The 3 cases that arise for launching projectiles: E r e E r e E r e When we discuss the force to escape the Earth's potential well or achieving circula orbits (overcome GPE) what forces are involved?

5 lesson33.notebook September 17, 2013 Revisiting our equation for Gravitational Potential Energy:

6 lesson33.notebook September 17, 2013 Escaping Earth's gravity well (escape velocity) and Energy of an orbiting satellite: When we discuss the force to escape the Earth's potential well or achieving circula orbits (overcome GPE) what forces are involved?

7 lesson33.notebook September 17, 2013 Example: A 5 x 10 2 kg rocket is to be placed into a circular geosynchronous orbit around the Earth. The period for the rocket will be 24 hours and the mass of the Earth is 5.98 x kg. Answer the following: A) What is the radius of the rocket's orbit? B) What is the GPE of the rocket when it is sitting on the ground before launch? C) What is the total energy of the rocket when it is in orbit? D) How much work must be done on the rocket to place it in orbit? E) How much additional energy would the satellite require to escape from Earth's potential well? Homwork: Handout Page 2 # 3 6, 10 17

8 lesson34.notebook December 04, 2013 Review of Electricity Today's goal: I can explain and apply concepts from electricity (3U) to real world applications. What is electricity? Law of Electric Charges: 1) Opposite charges attract each other. 2) Like charges repel each other. 3) Charged objects attract some neutral objects Electric Charge Transfer The transfer of charge from one object to another is caused by a large difference in the number of unbalanced electrons in the two objects.

9 lesson34.notebook December 04, 2013 Conductor vs Insulator Current Definition: The rate of electron flow. Symbol: Units: Formula: I = Q / s

10 lesson34.notebook December 04, 2013 DC current vs AC current Voltage Definition: The potential difference between two points. How much work is necessary to move 1 coulomb of charge from point A to point B Symbol: Units: J / C

11 lesson34.notebook December 04, 2013 Resistance Definition: The measure of a materials natural opposition to the flow of electrons. Symbol: Units: What does this mean for resistance?

12 lesson34.notebook December 04, 2013 Kirchhoff's Current Law: Kirchhoff's Voltage Law: Ohm's Law

13 lesson34.notebook December 04, 2013 Series vs Parallel Example: What is the total resistance in each circuit? a) b)

14 lesson34.notebook December 04, 2013 Examples: The current in a circuit is 0.1 ma and the resistance is 300 ohms. What is the voltage? Example: An electric belt sander was built for a 120 V outlet. I) If it draws 11 A from the circuit, what is its internal resistance? II) If the internal resistance of the sander is 25 ohms, what is the maximum current that it can use?

15 lesson35.notebook September 17, 2013 Coulomb's Law Today's goal: I can explain how Coulomb developed his law and use the results t solve problems that occur in real world situations. All matter around us contains charged particles and the electric forces between these charged particles determine the strength and properties of substances. Coulomb's Experiment Apparatus: A piece of silver wire attached to a light insulated rod suspended horizontally. At one end of the rod a pith ball was covered in gold foil and the other end has a paper disk to balance the rod. Process: Using a second gold foil covered pith ball and conduction, Coulomb charged both balls with the exact same charge. He then held the second ball stationary and the balls repelled each other. He knew how much force was required to twist the wire so when the balls came to rest, at a constant distance, Coulomb measured the angle of rotation and determine the force between the two balls. Coulomb's Findings: 1 Force is inversely proportional to the square of the distance between charges. ie. 2 Force is directly proportional to the product of the two charges. ie. Combining these two findings results in: F E = k Coulomb's constant (9.0 x 10 9 N m 2 /C 2 )

16 lesson35.notebook September 17, 2013 Putting it all together: 1) Coulomb's Law applies when the charges on the two spheres are very small, and the two spheres are small compared to the distance between them. 2) The forces involved act along a line connecting the two point charges. The charges will repel if alike and attract if unlike. 3) By Newton's third law, the electric force exerted on body A by body B is equal i magnitude and opposite in direction to the force exerted on B by A. A B F F q 1 q 2 r Relationships Stemming from Coulomb's Law: 1) F E = 2) F 1 (r 2 ) 2 F 2 (r 1 ) 2 3) F 1 q 1 q 2 F 2 q 1 'q 2 ' Total Charge The overall charge of an object can be determined by the equation: q = Ne, where q is the amount of charge in Coulomb's (C) N is the total number of electrons in either deficit or excess. e is the charge of an electron. (1.602 x C)

17 lesson35.notebook September 17, 2013 Example: The magnitude of the electrostatic force between two small, essentially point like, charged objects is 5.0 x 10 5 N. Calculate the force for each of the following situations: A) The distance between the charges is doubled, while the size of the charges stays the same. B)The charge on one object is tripled, while the charge on the other is halved. C) Both changes in A and B occur simultaneously.

18 lesson35.notebook September 17, 2013 Example: Determine the magnitude of the force of repulsion between two spheres 1.0 m apart if each has a charge of 1.0 x C. Example: Charged spheres A and B are fixed in position and have charges of 4 x 10 6 C and 2.5 x 10 7 C respectively. Calculate the net force on sphere C whose charge is 6.4 x 10 6 C. A B C cm 10 cm

19 lesson35.notebook September 17, 2013 Example: Identical spheres A, B, C and D, each with a charge of magnitude 5 x 10 6 C, are situated at the corners of a square whose sides are 25 cm long. Two diagonally opposite charges are positive, the other two negative. Calculate the net force acting on each of the four spheres. Homework: page 425, #46 55

20 lesson36.notebook September 17, 2013 Electric Fields Today's goal: I can explain and determine forces acting objects that exist within an electric field. Earlier we discussed that the term "field" simply refers to the space surrounding a source that is under the influence of a "force at a distance". Today we looked at charged particles and how they influence the space around them. Electric charges attract or repel each other without the necessity of coming into contact with each other. The force of attraction is governed by Coulomb's Law. Fo forces, such as electric fields, they cannot be easily viewed by the human eye, however, we must develop a visual system that allows us to understand how they form/look in order to explain the affect they have on other objects within the "field". The basis of field maps are electric field lines. These lines represent the electrostatic force acting on the charges. We generally use a positive charge whe drawing field maps. Rules for Drawing Electric Field Lines 1) The electric field lines always start and point away from any positive charge making up the field. 2) The number of field lines emanating from the charge is proportional to the magnitude of the charge. ***Field line may NOT cross each other*** 3) Fields can be mapped out by first finding equipotential lines. These lines are points where the forces are equal. Field lines are drawn perpendicular to the equipotential lines. 4) Field maps can take on many different shapes but there are 4 basic distribuons. NOTE: The electric force vectors (force of aracon/repulsion) are always tangenal to the drawn field line. It will be demonstrated in the next examples.

21 lesson36.notebook September 17, 2013 Point Charges: There are 4 basic field configurations: Two Equal Point Charges with Opposite Signs (Dipole) Comments:

22 lesson36.notebook September 17, 2013 Two Point Charges with equal magnitude, but opposite sign. Comments: The "Big" picture for point charges (regardless sign): Comments:

23 lesson36.notebook September 17, 2013 Two Unequal opposite point charges: Comments: Some "better" sketches for you notes:

24 lesson36.notebook September 17, 2013 Parallel oppositely charged plates: Comments: 1) The electric field in the region outside the parallel plates is zero (except for slight bulging of the field near the edges "edge effects") 2) The electric field is constant everywhere inside the plates. The field lines are straight, equally spaced and perpendicular to the plates. 3) The magnitude of the electric field at any point between the plates (except at the edges) depends ONLY on the magnitude of the charge on each plate. 4) Electric field strength, ε, is directly proportional to charge on each plate and does not depend on the distance between plates. Single Conductor: Comments:

25 lesson36.notebook September 17, 2013 Electric Field Strength Similar to gravitational fields, electric field strengths can be calculated. Electric Field Strength, ε, at any point is defined as the electric force per unit positive charge and is a vector quantity. ε = F E / q, where: ε Electric field strength F E Force due to electric charge Units: N/C q Charge, C Because this is an analysis of the field strength around one of the charged points we can incorporate it into our equation from yesterday: If F E = then, Example: What is the electric field 0.60 m away from a small sphere with a positiv charge of 1.2 x 10 8 C?

26 lesson36.notebook September 17, 2013 Example: Two charges, one of 3.2 x 10 9 C, the other of 6.4 x 10 9 C, are 42 cm apart. Calculate the net electric field at a point P, 15 cm from the positive charge, on the line connecting the charges. Example: The magnitude of the electric field between the plates of a parallel plate capacitor is 3.2 x 10 2 N/C. How would the field magnitude differ if: A) the charge on each plate were to be doubled? B) the plate separation were to be tripled? Homework: page 426, #56 68

27 lesson37.notebook September 17, 2013 Electric Potential Energy Today's goal: I can explain the concept of electric potential and calculate the associated values that arise in real world applications. From Newton and Gravitational Fields: Now with Electric Fields: Electric Potential Energy (E E ) The energy stored in a system of two charges a distance of 'r' apart: E E = Sign Convention for E E : Similar to Gravitational Potential: E E r Electric Potential We know from our 3U review that Electric Potential is also known as Voltage. With Electric Potential we look at a "positive unit charge, q 2, when it is in the "field" of any other charge, q 1. From this we return to our 3U formula of voltage:

28 lesson37.notebook September 17, 2013 NOTE: It is very important to distinguish between the following: 1) E E the electrical potential energy of a charge at a point. 2) V the electrical potential at the point. The concept of electric potential energy can be extended to include the electric field that results from the distribution of electrical charge rather than just a single point charge. Remember, electric potential is the work necessary to move a unit test charge from one point to another. In this particular case we are dealing with difference in electrical potential between the two points. Electric Potential Difference is the amount of work required per unit charge to move a positive charge from one point (B) to another point (A) in the presence of a electric field, regardless the path. Electric Potential Energy (E E ) Electric Potential (V)

29 lesson37.notebook September 17, 2013 Formulas: Electric Potential Difference By definition: ΔE E = qv 2 qv 1 Difference in Electric Potential For two oppositely charged plates: ε = F E / q F E = εq From our equation of work: E W = FΔd ΔV = V B V A And since E W = ΔE E then, Example: Calculate the electric potential a distance of 0.4 m from a spherical poin charge of +6.4 x 10 6 C.

30 lesson37.notebook September 17, 2013 Example: How much work must be done to increase the potential of a charge of 3.0 x 10 7 C by 120 V? Example: In a uniform electric field, the potential difference between two points 12.0 cm apart is 1.50 x 10 2 V. Calculate the magnitude of the electric field strength.

31 lesson37.notebook September 17, 2013 Example: The electric field strength between two parallel plates is 450 N/C. The plates are connected to a battery with an electric potential difference of 95 V. Wha is the plate separation? Homework: page 427, # page 429, # 83 89

32 lesson38.notebook September 17, 2013 The Motion of Charged Particles in Electrical Fields Today's goal: I can determine the properties of motion of a charged particle in an electric field. Recall Newton's Second Law: If we look at the situation of two similarly charged particles we note that there is an unbalanced force: The unbalanced can be calculated using Coulomb's Law: Rearranging Newton's Second Law and incorporating Coulomb's Law: The problem however is that as distance changes so does the acceleration. This makes it very difficult analyze problems and apply it to applications. If we instead analyze motion from the concept of energy... What we know: E T1 = E E1 = (We are assuming repulsion) As the particles separate the electrical force of repulsion decreases (as discussed on the previous slide), however, the loss in electrical force has been transformed into kinetic energy: E T2 = E E2 + E K2 = Similar to our unit on Energy and Momentum: E T1 = E T2

33 lesson38.notebook September 17, 2013 Example: Two small conducting spheres are placed on insulated air pucks, one anchored and the other is free to move. The mass of each sphere and puck is 0.1 kg and the charge on each sphere is +3.0 x 10 6 C and +5.0 x 10 9 C. If the spheres are originally set 0.25 m apart, how fast will the free sphere be moving when they are 0.65 m apart? Example: The cathode in a typical cathode ray tube (found in older TV's) is heated which makes electrons leave the cathode. The electron is attracted to an oppositely charged anode. If the potential difference between the cathode and the anode is 2.0 x 10 4 V, find the final speed of the electron. (mass of an electron is 9.1 x kg and its charge is x C)

34 lesson38.notebook September 17, 2013 Your typical Inkjet Printer: Homework: page 428 #75, 76, 80, 81

35 lesson39.notebook December 13, 2013 Introduction to Magnetism Magnets naturally occurring magnets are known as lodestone Magnetic Fields Similar to electric fields force of attraction / repulsion caused by dipoles Two defining characteristics Magnetic fields occur naturally in substances that have magnetic character (internal make up) Can effect not only other magnetic substances, but also electric charges Magnetic Character related to the condition of the atoms in the material not fully understood Domain Theory All large magnets are made up of many magnetic regions called domains. The magnetic character of domains comes from the presence of even smaller units called dipoles. Dipoles are called North and South by convention. Dipoles interact with their neighbouring dipoles. If they align with all the poles in one direction, then a larger magnetic domainis produced. Example: A diagram of Domain Theory

36 lesson39.notebook December 13, 2013 Materials that allow for the dipoles to be aligned to create a greater magnetic character are known as ferromagnetic materials. ie nickel, iron, cobalt Law of Magnetic Forces Similar magnetic poles repel each other with a force. Dissimilar magnetic poles attract each other. Magnetic Phenomena Magnetic Induction Ferromagnetic materials can be magnetized. Permanent and Temporary Magnetism Some materials demagnetize easily. Others maintain their magnetic domains. Reverse Magnetization A magnet can have its polarity reversed. Breaking a Large Magnet can create smaller active magnets Maximum Strength There is a limit to how strong a magnet can be.

37 lesson39.notebook December 13, 2013 Sketching Magnetic Fields The number of field lines drawn are proportional the strength of the field. The direction of the field is defined by the direction the north pole of the test magnet would point N S

38 lesson40.notebook December 16, 2013 Artificial Magnetic Fields Electromagnetism Hans Christian Oersted ( ) is credited with the "accidental" discovery o the link between electricity and magnetism. Oersted's Principle: Charge moving through a straight conductor produces a circular magnetic field around the conductor. Electricity passing through a conductor will create a magnetic field without the nee for dipoles. Oersted summarized guidelines for how to determine the resulting magnetic fields as the "right hand rules". Right Hand Rule #1 Grab the conductor with the thumb of the right hand pointing in the direction of conventional flow, or positive (+) current flow. The curved fingers point in the direction of the magnetic field around the conductor.

39 lesson40.notebook December 16, 2013 Example: Right Hand Rule #2 Grab the coiled conductor with the right hand such that the curved fingers point in the direction of conventional flow. The thumb points in the direction of the magnet field within the coil. Outside the coil, the thumb represents the north (N) end of the electromagnet produced by the coil.

40 lesson40.notebook December 16, 2013 Example: A Solenoid Factors that Determine the Strength of an Electromagnet Current in the coil Number of turns in the coil Size of coil Type of material in the coil's centre

41 lesson40.notebook December 16, 2013 Magnetic Permeability (μ) Magnetic permeability is the ratio of the magnetic field strength of the electromagnet with the core present to the field strength of the coil only. Some values: Iron 6100 Steel 2000 Nickel 1000 Oxygen 1.0 Water Homework: page 446 #1 4

42 lesson41.notebook September 17, 2013 The Motor Principle Today's goal: I can explain the Motor principle and can explain factors that effect and calculate magnetic force for real world applications. The Motor Principle states that when two magnetic fields interact, they produce a force. If a current carrying conductor cuts through a uniform magnetic field, it experiences a force directed at 90 degrees to both the direction of charge and the uniform magnetic field. The strength of the force depends on the strength of the uniform field and the field around the conductor. Diagram for the Motor Principle Right Hand Rule #3 Open the right hand so that the fingers point in the direction of the uniform magnetic field, from North (heel) to South (finger tips). Rotate the hand so that the thumb points in the direction of conventional flow. The orientation of the palm indicates the direction of force produced. Diagram

43 lesson41.notebook September 17, 2013 Magnetic Force Equations For the situation described with the right hand rule: F = BILsinθ, where: B The magnetic field strength (T Tesla) I The current in the conductor (A Amps) L The length of the conductor (m metres) θ The angle between the conductor and the magnetic field, in degrees. Calculating Magnetic Field Strength (B) Straight conductor: B = μi/2πr where: μ Magnetic permeability from uctor Flat coiled conductor: B = μni/2r where: N Number of loops in a flat coil Long solenoid: B = μni/l where: L Length of the solenoid

44 lesson41.notebook September 17, 2013 Example: What happens to the strength of magnetic force if the current is halved, the length is tripled and the conductor is rotated 50 degrees from perpendicular to the field? Example: A wire carrying 30 A is suspended 5 m [E] between a shed (power source) and a barn through the Earth's magnetic field (5.0 x 10 5 T). At a point where the wire sags making an angle of 15 degrees with the horizontal, A) What is the magnetic force that acts on the conductor? B) What is the direction of this force in relation to the horizontal wire?

45 lesson41.notebook September 17, 2013 Example: Find the magnetic field strength in air (μ = 4π x 10 7 ) 2.0 cm away from a conductor passing 5 A if: A) It is a straight conductor. B) It is looped 3 times with a radius of 4 cm. Homework: page 476 #22 26, 30 32

46 lesson42.notebook September 17, 2013 Magnetic Induction Today's goal: I can explain magnetic induction and how it applies / can be found in the world around us. Magnetic Force on a Moving Charge Current: The amount of charge passing in a given time. I = Δtnq / Length of conductor: The product of velocity and time for a particle to move from point A to B. Another equation for magnetic force then: Example: A proton of charge +e with a velocity of 2.5 x 10 7 m/s [D] enters a uniform magnetic field and is pushed with a force of 7 x N [R]. Determine the magnitude and direction of the magnetic field.

47 lesson42.notebook September 17, 2013 Centripetal Magnetic Force A particle moving at a constant speed and experiencing a constant magnetic force at 90 degrees will trace a circular path (similar to an orbiting satellite!) The Equation Example: Determine the velocity of a particle (+e) moving in a circular path of radius 8 cm in a plane perpendicular to a 2.0 T uniform magnetic field. (m = 9.1 x kg)

48 lesson42.notebook September 17, 2013 Application of Centripetal Magnetic Force Path of an electron A) v = 0 m/s B) v 0 m/s Electrons in the Earth's Magnetic Field Electricity to Magnetism And Back Again! In 1831 Michael Faraday discovered a concept that complimented Oersted's Principle: Faraday's Law of Electromagnetic Inductionstates that a magnetic field that is moving or changing in intensity in the region around the conductor causes or indu charge to flow in the conductor. Oersted and Faraday The relationship

49 lesson42.notebook September 17, 2013 In 1835 Heinrich Lenz took Faraday one step further: Lenz's Law states that the direction of induced current creates an induced magnetic field that opposes the motion on the inducing magnetic field. Applying Lenz's Law: Finally, in 1864 James Maxwell tied all previous discoveries together: The Four Premises of Maxwell's Equations 1) The distribution of electric charges in space is dictated by the electric field that the charges produce. 2) Magnetic and electric field lines are similar except magnetic lines are continuous, they don't begin or end the way electric field lines do on charges. 3) Electric fields are created by changing magnetic fields. 4) Magnetic fields can be produced by changing electric fields or by moving charges (current). What Maxwell is saying then is:

50 lesson42.notebook September 17, 2013 Finally, The largest impact based on this discovery? Homework: pg 478, #43, 44 pg 475, #7 12