Section 24.5 Magnetic Fields Exert Forces on Moving Charges
Magnetic Fields Sources of Magnetic Fields You already know that a moving charge is the creator of a magnetic field. Effects of Magnetic Fields If a moving charge experiences a magnetic field, it there will be a force on the charge. Sound familiar?!?!
Magnetic Fields Exert Forces on Moving Charges Magnetic fields also exert forces on moving charged particles and on electric currents in wires. There is no magnetic force on a charged particle at rest. There is no magnetic force on a charged particle moving parallel to a magnetic field. Slide 24-3
Magnetic Fields Exert Forces on Moving Charges As the angle α between the velocity and the magnetic field increases, the magnetic force also increases. The force is greatest when the angle is 90. The magnetic force is always perpendicular to the plane containing and. Slide 24-4
Magnetic Force on Particles A magnetic force is exerted on a particle within a magnetic field only if the particle has a charge. the charged particle is moving. with at least a portion of its velocity perpendicular to the magnetic field. Slide 24-5
Magnetic Force on a Charged Particle charge moving F B = magnitude: q: charge in Coulombs v: speed in meters/second B: magnetic field in Tesla Θ: angle between v and B vv qvb sinθ in a magnetic field with an angle not equal to 0 or 180 degrees direction: Right Hand Rule
Magnetic Fields Exert Forces on Moving Charges We determine the correct direction of the force using the right-hand rule for forces. Slide 24-7
The right hand rule to determine magnetic force This right hand rule is a little different than what your book uses, but I think it s easier to remember. You must keep your hand in this configuration and turn your whole wrist!! F Point in the direction of the velocity. Turn your hand so that your middle finger (or three remaining fingers) point in the direction of the field. Your thumb gives you the direction of the force. v B Slide 24-8
Magnetic Fields Exert Forces on Moving Charges Slide 24-9
QuickCheck 24.15 The direction of the magnetic force on the proton is A. To the right. B. To the left. C. Into the screen. D. Out of the screen. E. The magnetic force is zero. Slide 24-10
QuickCheck 24.15 The direction of the magnetic force on the proton is A. To the right. B. To the left. C. Into the screen. D. Out of the screen. E. The magnetic force is zero. Slide 24-11
QuickCheck 24.16 The diagram shows a top view of an electron beam passing between the poles of a magnet. The beam will be deflected A. Toward the north pole of the magnet. B. Toward the south pole of the magnet. C. Out of the plane of the figure D. Into the plane of the figure. Slide 24-12
QuickCheck 24.16 The diagram shows a top view of an electron beam passing between the poles of a magnet. The beam will be deflected A. Toward the north pole of the magnet. B. Toward the south pole of the magnet. C. Out of the plane of the figure D. Into the plane of the figure. Slide 24-13
QuickCheck 24.17 A beam of positively charged particles passes between the poles of a magnet as shown in the figure; the force on the particles is noted in the figure. The magnet s north pole is on the, the south pole on the. A. Left, right B. Right, left C. There s not enough information to tell. Slide 24-14
QuickCheck 24.17 A beam of positively charged particles passes between the poles of a magnet as shown in the figure; the force on the particles is noted in the figure. The magnet s north pole is on the, the south pole on the. A. Left, right B. Right, left C. There s not enough information to tell. Slide 24-15
QuickCheck 24.18 The direction of the magnetic force on the electron is A. Upward. B. Downward. C. Into the screen. D. Out of the screen. E. The magnetic force is zero. Slide 24-16
QuickCheck 24.18 The direction of the magnetic force on the electron is A. Upward. B. Downward. C. Into the screen. D. Out of the screen. E. The magnetic force is zero. Slide 24-17
QuickCheck 24.19 Which magnetic field causes the observed force? Slide 24-18
QuickCheck 24.19 Which magnetic field causes the observed force? C. Slide 24-19
Sample Problem What is the magnetic force exerted on a 3.0 µc charge moving north at 300,000 m/s in a magnetic field of 200 mt if the field is directed a) North. b) South. c) East. d) West. Slide 24-20
Sample Problem What is the magnetic force exerted on a 3.0 µc charge moving north at 300,000 m/s in a magnetic field of 200 mt if the field is directed a) North. b) South. c) East. d) West. F F F F F F B B B B B B = qvbsinθ = 6 3 ( 3 10 C)( 300,000 m/s)( 200 10 T) sin( 0 ) = 0 = qvbsinθ = = F F F B B B = qvbsinθ = 6 3 ( 3 10 C)( 300,000 m/s)( 200 10 T) sin( 180 ) = 0 6 3 ( 3 10 C)( 300,000 m/s)( 200 10 T) sin( 90 ) 0.18 N, down (toward Earth) F F F B B B = qvbsinθ = = 6 3 ( 3 10 C)( 300,000 m/s)( 200 10 T) sin( 90 ) 0.18 N, up (toward sky)
Sample Problem Calculate the magnitude and direction of the magnetic force in the situation below. F F F B B B = qvbsinθ = 6 3 ( 3 10 C)( 300,000 m/s)( 200 10 T) sin( 34 ) = 0.101 N, out of the page
Conceptual Example 24.6 Determining the force on a moving electron An electron is moving to the right in a magnetic field that points upward, as in FIGURE 24.26. What is the direction of the magnetic force? Slide 24-23
Magnetic forces are always orthogonal (at right angles) to the plane established by the velocity and magnetic field vectors. can accelerate charged particles by changing their direction. can cause charged particles to move in circular or helical paths. Slide 24-24
Magnetic forces cannot do work on charged particles. Why? The force is always perpendicular to the motion. What are the implications of this? They cannot change the speed or kinetic energy of charged particles. Slide 24-25
This means magnetic forces are centripetal! Remember that centripetal acceleration is Therefore, centripetal force is v 2 a = c r mv 2 ΣF c = ma c = r Slide 24-26
Paths of Charged Particles in Magnetic Fields When we studied the motion of objects subject to a force that was always perpendicular to the velocity, the result was circular motion at a constant speed. For example, a ball moved at the end of a string moved in a circle due to the perpendicular force of tension in the string. For a charged particle moving in a magnetic field, the magnetic force is always perpendicular to and so it causes the particle to move in a circle. Slide 24-27
Paths of Charged Particles in Magnetic Fields A particle moving perpendicular to a uniform magnetic field undergoes uniform circular motion at constant speed. Slide 24-28
Paths of Charged Particles in Magnetic Fields Derive an equation for the radius of orbit for a charged particle in a magnetic field. Slide 24-29
Sample Problem What is the orbital radius of a proton moving at 20,000 m/s perpendicular to a 40 T magnetic field? Slide 24-30
Sample Problem What is the orbital radius of a proton moving at 20,000 m/s perpendicular to a 40 T magnetic field? ma c = ΣF 2 v m = FB r 2 v m = qvbsinθ r v m = qbsinθ r mv r = = qbsinθ c 27 ( 1.67 10 kg)( 20,000 m/s) 19 ( 1.6 10 C)( 40 T) sin( 90 ) = 5.22 10 6 m
Paths of Charged Particles in Magnetic Fields The motion of a charged particle when its velocity is neither parallel nor perpendicular to the magnetic field: Slide 24-32
Paths of Charged Particles in Magnetic Fields Slide 24-33
Paths of Charged Particles in Magnetic Fields Slide 24-34
Paths of Charged Particles in Magnetic Fields High-energy particles stream out from the sun in the solar wind, some of which becomes trapped in the earth s magnetic field. The particles spiral in helical trajectories along the earth s magnetic field lines. When they enter the atmosphere at the poles, they ionize gas, creating the aurora. Slide 24-35
Sample Problem An electric field of 2000 N/C is directed to the south. A proton is traveling at 300,000 m/s to the west. What is the magnitude and direction of the force on the proton? Describe the path of the proton. Ignore gravitational effects. v F F F = qe = ( 19 )( N 1.6 10 C 2000 ) = 3.2 10 16 N C E The force is south. (Since the charge is positive, it will experience a force in the direction of the field.) Since the horizontal velocity is unchanged, the proton will follow a parabolic path downward. Slide 24-36
Sample Problem A magnetic field of 2000 mt is directed to the south. A proton is traveling at 300,000 m/s to the west. What is the magnitude and direction of the force on the proton? Describe the path of the proton. Ignore gravitational effects. v F F F = qvbsinθ = 19 m -3 ( 1.6 10 C)( 300,000 )( 2000 10 T) sin( 90 ) = 9.6 10 14 N s The force is up (out of the page). (Since the charge is positive, use the right hand rule.) It will move in a circular path. B Slide 24-37
Sample Problem How would you arrange a magnetic field and an electric field so that a charged particle of velocity v would pass straight through without deflection? Slide 24-40
QuickCheck 24.20 Which magnetic field (if it s the correct strength) allows the electron to pass through the charged electrodes without being deflected? Slide 24-41
QuickCheck 24.20 Which magnetic field (if it s the correct strength) allows the electron to pass through the charged electrodes without being deflected? E. Slide 24-42
Electric and Magnetic Fields Together B e- E This electron will experience an upward force from the electric field (opposite the direction of the field) and a downward force from the magnetic field (left hand rule). Slide 24-43
Sample Problem It is found that protons traveling at 20,000 m/s pass undeflected through the velocity filter below. What is the magnitude and direction of the magnetic field between the plates? e 20,000 m/s 0.02 m 400 V Solution on next page. Slide 24-44
Solution V = Ed Undeflected means that ΣF = 0. ΣF F e = qe = qvsinθ E = B vsinθ B = = 0 = F F B B F qe = qvbsinθ qvbsinθ qvsinθ N ( 20,000 C ) m ( 20,000 ) s e sin 90 = 1T, into the page E = V d = 400 V 0.02 m = 20,000 V m Slide 24-45
Electromagnetic Flowmeters An electromagnetic flowmeter is a device that can be used to measure the blood flow in an artery. It applies a magnetic field across the artery, which separates the positive and negative ions in the blood. The flowmeter measures the potential difference due to the separation of the ions. The faster the blood s ions are moving, the greater the forces separating the ions become, therefore generating a higher voltage. Therefore, the measured voltage is proportional to the velocity of the blood. Slide 24-46
Electromagnetic Flowmeters Slide 24-47
Electromagnetic Flowmeters Slide 24-48
Don t Try It Yourself: Magnets and Television Screens The image on a cathode-ray tube television screen is drawn by an electron beam that is steered by magnetic fields from coils of wire. Other magnetic fields can also exert forces on the moving electrons. If you place a strong magnet near the TV screen, the electrons will be forced along altered trajectories and will strike different places on the screen than they are supposed to, producing an array of bright colors. (The magnet can magnetize internal components and permanently alter the image, so do not do this to your television!) Slide 24-57
Section 24.6 Magnetic Fields Exert Forces on Currents
The Form of the Magnetic Force on a Current We learned that the magnetic field exerts no force on a charged particle moving parallel to a magnetic field. If a current-carrying wire is parallel to a magnetic field, we also find that the force on it is zero. There is a force on a currentcarrying wire that is perpendicular to a magnetic field. Slide 24-59
The right hand rule to determine magnetic force This right hand rule is a little different than what your book uses, but I think it s easier to remember. You must keep your hand in this configuration and turn your whole wrist!! F Point in the direction of the CURRENT (velocity of the charges). Turn your hand so that your middle finger (or three remaining fingers) point in the direction of the field. Your thumb gives you the direction of the force. v B Slide 24-60
Calculating Magnetic Force on a Current Derive an equation that can be used to calculate magnetic force on a current-carrying wire. Slide 24-61
Review What is the equation for a magnetic force on a moving charge? F = qvbsinθ What if there are many moving charges, like a current in a wire? F F = = x q Bsinθ t q xbsinθ = t q t xbsinθ = x represents the length of the wire (the distance the charges move) I xbsinθ Slide 24-62
Magnetic Force on Current- Carrying Wire F B I: current in Amps l: length in meters B: magnetic field in Tesla θ: angle between current and field = IlB sinθ Slide 24-63
Sample Problem What is the force on a 100 m long wire bearing a 30 A current flowing north if the wire is in a downwarddirected magnetic field of 400 mt? F F F B B B = IlB sinθ = ( )( )( 3 30 100 m 400 10 T) C s = 1200 N sin90 The force is west (use the right hand rule remember we assume positive charges are flowing in the wire). Slide 24-65
Sample Problem A wire is in a magnetic field that is directed out of the page. What is the magnetic field strength if the current in the wire is 15 A and the force is downward and has a magnitude of 40 N/m? What is the direction of the current? ANS: 2.67 T, right Slide 24-66
Which way will this loop of wire rotate a) if the current is clockwise? b) if the current is counterclockwise? B a) The right side of the loop will rotate out of the page. b) The right side of the loop will rotate into the page. Slide 24-67
QuickCheck 24.23 The horizontal wire can be levitated held up against the force of gravity if the current in the wire is A. Right to left. B. Left to right. C. It can t be done with this magnetic field. Slide 24-70
QuickCheck 24.23 The horizontal wire can be levitated held up against the force of gravity if the current in the wire is A. Right to left. B. Left to right. C. It can t be done with this magnetic field. Slide 24-71
Example 24.11 Magnetic force on a power line A DC power line near the equator runs east-west. At this location, the earth s magnetic field is parallel to the ground, points north, and has magnitude 50 µt. A 400 m length of the heavy cable that spans the distance between two towers has a mass of 1000 kg. What direction and magnitude of current would be necessary to offset the force of gravity and levitate the wire? (The power line will actually carry a current that is much less than this; 850 A is a typical value.) Slide 24-72
Example 24.11 Magnetic force on a power line (cont.) PREPARE First, we sketch a top view of the situation, as in FIGURE 24.38. The magnetic force on the wire must be opposite that of gravity. An application of the right-hand rule for forces shows that a current to the east will result in an upward force out of the page. Slide 24-73
Example 24.11 Magnetic force on a power line (cont.) SOLVE The magnetic field is perpendicular to the current, so the magnitude of the magnetic force is given by Equation 24.10. To levitate the wire, this force must be opposite to the weight force but equal in magnitude, so we can write mg = ILB where m and L are the mass and length of the wire and B is the magnitude of the earth s field. Solving for the current, we find directed to the east. Slide 24-74
Example 24.11 Magnetic force on a power line (cont.) ASSESS The current is much larger than a typical current, as we expected. Slide 24-75
Forces Between Currents Because a current produces a magnetic field, and a magnetic field exerts a force on a current, it follows that two currentcarrying wires will exert forces on each other. A wire carrying a current I 1 will create a magnetic field 1. Slide 24-76
Forces Between Currents A second wire with current I 2 will experience the magnetic force due to the wire with current I 1. Using the right-hand rule for forces, we can see that when I 2 is in the same direction as I 1, the second wire is attracted to the first wire. If they were in opposite directions, the second wire would be repelled. Slide 24-77
Forces Between Currents The magnetic field created by the wire with current I 2 will also exert an attractive force on the wire with current I 1. The forces on the two wires form a Newton s third law action/reaction pair. The forces due to the magnetic fields of the wires are directed in opposite directions and must have the same magnitude. Slide 24-78
Forces Between Currents Slide 24-79
Example 24.12 Finding the force between wires in jumper cables You may have used a set of jumper cables connected to a running vehicle to start a car with a dead battery. Jumper cables are a matched pair of wires, red and black, joined together along their length. Suppose we have a set of jumper cables in which the two wires are separated by 1.2 cm along their 3.7 m (12 ft) length. While starting a car, the wires each carry a current of 150 A, in opposite directions. What is the force between the two wires? Slide 24-80
Example 24.12 Finding the force between wires in jumper cables (cont.) PREPARE Our first step is to sketch the situation, noting distances and currents, as shown in FIGURE 24.41. Let s find the force on the red wire; from the discussion above, the force on the black wire has the same magnitude but is in the opposite direction. Slide 24-81
Example 24.12 Finding the force between wires in jumper cables (cont.) The force on the red wire is found using a two-step process. First, we find the magnetic field due to the current in the black wire at the position of the red wire. Then, we find the force on the current in the red wire due to this magnetic field. Slide 24-82
Example 24.12 Finding the force between wires in jumper cables (cont.) SOLVE The magnetic field at the position of the red wire, due to the current in the black wire, is According to the right-hand rule for fields, this magnetic field is directed into the page. The magnitude of the force on the red wire is then Slide 24-83
Example 24.12 Finding the force between wires in jumper cables (cont.) The direction of the force can be found using the right-hand rule for forces. The magnetic field at the position of the red wire is into the page, while the current is to the right. This means that the force on the red wire is in the plane of the page, directed away from the black wire. Thus the force between the two wires is repulsive, as we expect when their currents are directed oppositely. Slide 24-84
Example 24.12 Finding the force between wires in jumper cables (cont.) ASSESS These wires are long, close together, and carry very large currents. But the force between them is quite small much less than the weight of the wires. In practice, the forces between currents are not an important consideration unless there are many coils of wire, leading to a large total force. This is the case in an MRI solenoid. Slide 24-85
Forces Between Current Loops Just as there is an attractive force between parallel wires that have currents in the same direction, there is an attractive force between parallel loops with currents in the same direction. There is a repulsive force between parallel loops with currents in opposite directions. Slide 24-86
Forces Between Current Loops The field of a current loop is very similar to that of a bar magnet. A current loop, like a bar magnet, is a magnetic dipole with a north and a south pole. Slide 24-87
Forces Between Current Loops Slide 24-88
QuickCheck 24.24 The diagram below shows slices through two adjacent current loops. Think about the force exerted on the loop on the right due to the loop on the left. The force on the right loop is directed A. To the left. B. Up. C. To the right. D. Down. Slide 24-89
QuickCheck 24.24 The diagram below shows slices through two adjacent current loops. Think about the force exerted on the loop on the right due to the loop on the left. The force on the right loop is directed A. To the left. B. Up. C. To the right. D. Down. Slide 24-90
Example Problem A 10 cm length of wire carries a current of 3.0 A. The wire is in a uniform field with a strength of 5E-3 Tesla as in the following diagram. What are the magnitude and direction of the force on this segment of wire? Slide 24-91
Summary: General Principles Text: p. 794 Slide 24-92
Summary: Important Concepts Text: p. 794 Slide 24-93
Summary: Applications Text: p. 794 Slide 24-94