10/24/2012 PHY 102. (FAWOLE O.G.) Good day. Here we go..

Size: px

Start display at page:

Download "10/24/2012 PHY 102. (FAWOLE O.G.) Good day. Here we go.."

Warren Moore
5 years ago
Views:

1 Good day. Here we go.. 1

2 PHY102- GENERAL PHYSICS II Text Book: Fundamentals of Physics Authors: Halliday, Resnick & Walker Edition: 8 th Extended Lecture Schedule TOPICS: Dates Ch. 28 Magnetic Fields 12 th & 15 th October Ch. 29 Magnetic Fields due to Currents 17 th & 19th October Date for 1 st Test: 10 November, 2012 (Tentative) 2 MR. O.G. FAWOLE Rm. G14 F {Dept. of Physics.} 10/24/201 12

3 CHAPTER 28. MAGNETIC FIELD What is Physics? What Produces a Magnetic Field? The Definition of Magnetic Field Crossed Fields: Discovery of the Electron Crossed Fields: The Hall Effect A Circulating Charged Particle Cyclotrons and Synchrotrons Magnetic Force on a Current-Carrying wire Torque on a Current Loop The Magnetic Dipole Moment 3

4 1. MAGNETIC FIELDS 4

5 SUB-SECTIONS SECTIONS Magnetic Fields. You must understand the similarities and differences between electric fields and field lines, and magnetic fields and field lines. Magnetic Force on Moving Charged Particles. You must be able to calculate the magnetic force on moving charged particle. Magnetic Flux and Gauss Law for Magnetism. You must be able to calculate magnetic flux and recognize the consequences of Gauss Law for Magnetism. 5 10/24/ Motion of a Charged Particle in a Uniform Magnetic Field. You must be able to calculate the trajectory and energy of a charged particle moving in a uniform magnetic field.

6 MAGNETISM Recall how there are two kinds of electrical charge (+ and -), and likes repel opposites attract. Similarly, there are two kinds of magnetic poles (north and south), and like poles repel, opposites attract. S N S N Attract S N Repel N S S N S N Repel Attract S N N S 6

7 THE MAGNETIC FIELD A magnetic field exists in the region around a magnet. The magnetic field is a vector that has both magnitude and direction. The direction of the magnetic field at any point in space is the direction indicated by the north pole of a small compass needle placed at that point. 7

8 MAGNETIC FIELD. The earth has associated with it a magnetic field, with poles near the geographic poles. The pole of a magnet attracted to the earth s north geographic pole is the magnet s north pole. The pole of a magnet attracted to the earth s south geographic pole is the magnet s south pole. There are two ways to produce magnetic field namely: Using moving charged particles, such as a current in a wire, to make an electromagnet Using elementary particles such as electron, because they have intrinsic magnetic fields around them Just as we used the electric field to help us explain and visualize electric forces in space, we use the magnetic field to help us explain and visualize magnetic forces in space. Magnetic field lines point in the same direction that the north pole of a compass would point. N PHY S 102. (FAWOLE O.G.) 8

The magnetic field at any point is tangential to the magnetic field line at that point.

9 THE MAGNETIC FIELD LINE The lines originate from the north pole and end on the south pole; they do not start or stop in mid-space. The magnetic field at any point is tangential to the magnetic field line at that point. The strength of the field is proportional to the number of lines per unit area that passes through a surface oriented perpendicular to the lines. 9 The magnetic field lines will never come to cross each other.

The magnetic fields of the electrons in certain materials add together to give a net

10 WHAT PRODUCES A MAGNETIC FIELD? Moving electrically charged particles, such as a current, produce a magnetic field Permanent magnet. Elementary particles such as electrons have an intrinsic magnetic field around them. The magnetic fields of the electrons in certain materials add together to give a net magnetic field around the material. Such addition is the reason why a permanent magnet has a permanent magnetic field. In other materials, the magnetic fields of 10 the electrons cancel out, giving no net magnetic field surrounding the material

11 THE DEFINITION OF MAGNETIC FIELD, B We use the symbol B for magnetic field. Remember: magnetic field lines point away from north poles, and towards south poles. The SI unit for magnetic field is the Tesla. S N Remember, units of field are force per something. The earth s magnetic field has a magnitude of roughly 0.5 G, or T. A powerful permanent magnet, like the kind you might find in headphones, might produce a magnetic field of 1000 G, or 0.1 T. Superconducting magnets can produce a field of over 10 T. Where G (gauss) is an old unit of magnetic field. 1 tesla(t) = 10 4 gauss(g). 11

MAGNETIC FIELD, B. To determine the electric field E at a point, a test particle of charge, q at rest is placed at the point. The electric force, F E acting on the particle can then be measured.

12 MAGNETIC FIELD, B. To determine the electric field E at a point, a test particle of charge, q at rest is placed at the point. The electric force, F E acting on the particle can then be measured. E is then given by: We cannot define magnetic field, like this because a magnetic monopole is not available. Hence, we have to define, in another way, in terms of the magnetic force, F B exerted on a moving electrically charged test particle. Magnetic field, B can be said to be a vector quantity that is directed along the zero-force axis. When the particle velocity, v, is directed perpendicular to the direction of the zero-force axis, the magnetic field is given as: (1) where q is the charge of the particle. (2) A charged particle moving in a magnetic field experiences a force. The magnetic force equation predicts the effect of a magnetic field on a moving charged particle. 12

13 MAGNETIC FORCE ON A CHARGED PARTICLE When a charge is placed in a magnetic field, it experiences a magnetic force if two conditions are met: 1. The charge must be moving. No magnetic force acts on a stationary charge. 2. The velocity of the moving charge must have a component that is perpendicular to the direction of the field. 13

14 F=qv B force on particle velocity of charged particle magnetic field vector What is the magnetic force if the charged particle is at rest? What is the magnetic force if v is (anti-)parallel to B? The force F B on the particle is equal to the charge q times the cross product of its velocity v, and the magnetic field B (all measured in the same frame of reference). RECALL: A X B = ABsinθ. Hence; F = q vb sin = q v B = q vb θ (3) where θ is the angle between the directions of velocity, v and magnetic field, B 14

MAGNETIC FORCE ON A CHARGED PARTICLE Right-Hand Rule The

through a magnetic field is always perpendicular to and.

15 MAGNETIC FORCE ON A CHARGED PARTICLE Right-Hand Rule The force acting on a charged particle moving with velocity through a magnetic field is always perpendicular to and. 15 Fingers out in direction of velocity, thumb perpendicular to them. rotate your hand until your palm points in the direction of magnetic field. Thumb points in direction of magnetic force on + charge.

16 Foolproof technique for calculating both magnitude and direction of magnetic force. hints for solving problems: F = qv B ˆi ˆj kˆ F = q d e t v v v B B B F = q vb sinθ x y z x y z A charged particle moving along or opposite to the direction of a magnetic field will experience no magnetic force. Conversely, the fact that there is no magnetic force along some direction does not mean there is no magnetic field along or opposite to that direction. It s OK to use F = q vb if you know that v and B are perpendicular, or you are calculating a minimum field to produce a given force 16 (understand why).

17 MODEL PROBLEMS. 1. An electron that has a velocity, = (2.0 x 10 6 m/s) i + (3.0 x 10 6 m/s) j moves through the uniform magnetic field, = (0.030 T)i (0.15 T)j. (a). Find the force on the electron due to the magnetic field. (b). Repeat your calculation for a proton having the same velocity. SOLUTION. 1(a). The force on the electron is = (-1.6 x C){(2.0 x 10 6 m/s)(-0.15 T) (3.0 x 10 6 m/s)(0.030 T)} = (6.2 x N) The magnitude of is 6.2 x N and it points in the direction of the positive z direction (b) This implies repeating the calculations in (a) with a change in the sign of the charge. Hence, has the same magnitude but points in the negative z direction = - (6.2 x N) 17

18 2. A proton travelling at 23.0º with respect to the direction of a magnetic field of strength 2.60 mt experiences a magnetic force of 6.50 x N. Calculate (a) the proton speed and (b) its kinetic energy in ev. SOLUTION. 1 (a). From eqn. (3) (b). The kinetic energy of the proton is K.E= PRACTICE Since 1.6 x J = 1 ev 1.34 x J = 835 ev 18

19 2 (a). A proton is moving with a velocity v = v 0 j in a region of uniform magnetic field. The resulting force is F = F 0 i. What is the magnetic field (magnitude and direction)? (b) In the above example, what is the minimum magnetic field that can be present (magnitude and direction)? USEFUL HINTS F = q vb sinθ A charged particle moving along or opposite to the direction of a magnetic field will experience no magnetic force. Conversely, the fact that there is no magnetic force along some direction does not mean there is no magnetic field along or opposite to that direction. It s OK to use F = q vb if you know that v and B are perpendicular, or you are calculating a minimum field to produce a given force (understand why). 19

always parallel (or anti-parallel) to the electric field direction.

20 DIFFERENCES OF ELECTRIC AND MAGNETIC FIELDS 1. Direction of forces The electric force on a charged particle (both moving and stationary) is always parallel (or anti-parallel) to the electric field direction. The magnetic force on a moving charged particle is always perpendicular to both magnetic field and velocity of the particle. No magnetic force on a stationary charged particle. 20

21 2. THE WORK DONE ON A CHARGED PARTICLE: The electric force can do work on the particle. The magnetic force cannot do work and change the kinetic energy of the charged particle. 21

22 CROSSED FIELDS: DISCOVERY OF THE ELECTRON Both an electric field, E and magnetic field, B can produce a force on a charged particle. When the two fields are perpendicular to each other, they are said to be cross fields. Studying the crossed field using J.J. Thomson s expt. that led to the discovery of electrons. 22 Cathode ray tube

23 CROSSED FIELDS CONTD. In the arrangement the figure above, electron are forced up the page by the electric field and down the page by magnetic field meaning the forces are in opposition. Thomson s procedure was equivalent to the following series of steps. 1. Set and to zero and note the position of the spot on screen S due to the undeflected beam. 2. Turn on and measure the resulting beam deflection y. 3. Maintaining, now turn on and adjust its value until the beam returns to the undeflected position. (With the forces in opposition, they can be made to cancel.) 23

CROSSED FIELDS CONTD. The deflection of the particle at the far end of the plate is: Where v is the particle s speed, m its mass and q its charge, L is the length of the plates.

24 CROSSED FIELDS CONTD. The deflection of the particle at the far end of the plate is: Where v is the particle s speed, m its mass and q its charge, L is the length of the plates. When the two fields are adjusted so that the two deflecting forces cancel (step in 3), then eqn (1) = eqn (3) qe = qvb sin90 = qvb v = Substituting for v in eq. (4): (4) (5) 24

CROSSED FIELDS: THE HALL EFFECT Can the drifting conduction

field? In 1879, E.H Edwin showed they can.

carriers in a conductor are positively or negatively charged.

25 CROSSED FIELDS: THE HALL EFFECT Can the drifting conduction electrons in a copper wire also be deflected by a magnetic field? In 1879, E.H Edwin showed they can. The Hall effect allows us to find out whether the charge carriers in a conductor are positively or negatively charged. The number of such carriers per unit volume of the conductor can also be determined. 25

26 CROSSED FIELDS: THE HALL EFFECT Can the drifting conduction electrons in a copper wire also be deflected by a magnetic field? A Hall potential difference V of magnitude V = Ed is associated with the electric field across strip width d When the electric and magnetic forces are in balance The drift speed, V d = i neld B E A = = v B ld d Where J is the charge density in the strip, A is the cross-sectional area of the strip and n is the number density of charge carriers i.e. number per unit volume, =A/d is the thickness of the strip. 26

EXAMPLE. A metal strip 6.50 cm long, 0.850 cm wide and 0.

20 mt directed perpendicular to the strip as shown in the figure. A potential difference of 3.

27 EXAMPLE. A metal strip 6.50 cm long, cm wide and mm thick moves with constant velocity, v through a uniform magnetic field B = 1.20 mt directed perpendicular to the strip as shown in the figure. A potential difference of 3.90 nv is measured between point x and y across the strip. Calculate the speed v. Solution For a free charge q inside the metal strip with velocity v, we have We set this force equal to zero and use the relation between (uniform) electric field and potential difference. Thus, = 3.82 x 10-4 m/s. 27

THE MOTION OF A CHARGED PARTICLE IN A CONSTANT MAGNETIC FIELD If a particle moves in a circular path at a constant speed, it is sure that the net force acting on the particle is constant

28 THE MOTION OF A CHARGED PARTICLE IN A CONSTANT MAGNETIC FIELD If a particle moves in a circular path at a constant speed, it is sure that the net force acting on the particle is constant in magnitude and direction, pointing towards the centre of the circle, always perpendicular to the particle s velocity. Applying Newton s 2 nd law to uniform circular motion, we have: (6) 28

29 The force acting on the particle has magnitude of q VB, hence (7) (8) Frequency f = The angular frequency, Note that T, f and ω do not depend on the speed of the particle (provided the speed is less than the speed of light). HELICAL PATHS (9) (10) When the velocity of the charged particle has a component parallel to the magnetic field, the particle will move in a helical path about the direction of the field vector. 29 Read more about motion of charged particle in helical paths.

30 PRACTICE QUESTION Fig. 28a 1. A proton is released from rest at point A, which is located next to the positive plate of a parallel plate capacitor (see Figure 28a). The proton then accelerates toward the negative plate, leaving the capacitor at point B through a small hole in the plate. The electric potential of the positive plate is 2100 V greater than that of the negative plate, so V A V B = 2100 V. Once outside the capacitor, the proton travels at a constant velocity until it enters a region of constant magnetic field of magnitude 0.10 T. The velocity is perpendicular to the magnetic field, which is directed out of the page in Figure 28a. Find (a) the speed v B of the proton when it leaves the negative plate of the capacitor, and (b) the radius r of the circular path on which the proton moves in the magnetic field. 30

31 CYCLOTRON AND SYNCHROTRON The number of revolutions per unit time is f = ω/2π. This frequency f is independent of the radius R of the path. It is called the cyclotron frequency; in a particle accelerator called a cyclotron, particles moving in nearly circular paths are given a boost twice each revolution, increasing their energy and their orbital radii but not their angular speed or frequency. For resonance condition: (12) (11) 31

MAGNETIC FORCE ON A CURRENT-CARRYING WIRE Fig. 28b Fig. 28c explains what happens in Fig. 28b. A conduction electron drifting downwards with an assumed drift speed.

32 MAGNETIC FORCE ON A CURRENT-CARRYING WIRE Fig. 28b Fig. 28c explains what happens in Fig. 28b. A conduction electron drifting downwards with an assumed drift speed. A force of magnitude must act on such electron. This force must be directed to the RIGHT. Consider a length L of the wire in fig. 28c. All the conduction electrons in this section of the wire will drift past the xx plane in a time t = Fig. 28c (13) eq. (13) gives the magnetic force acting on a length L of straight wire carrying a current i and immersed in a magnetic field B that is perpendicular to the wire. 32

33 MAGNETIC FORCE ON A CURRENT-CARRYING WIRE CONTD. Fig. 28d If the magnetic field is not perpendicular to the wire, as in Fig. 28d, the magnetic force is given by (14) Define as a length vector that has magnitude L and is directed along the wire segment in the direction of the (conventional) current. F B is always perpendicular to the plane defined by vectors L and B as shown in fig. 28d. If a wire is not straight or the field is not uniform. We can imagine the wire broken up into small straight segments and then apply: (15) In the differential limit we can write: (16) 33

34 SAMPLE PROBLEM A straight, horizontal length of copper wire has a current i=28 A flowing through it. What are the magnitude and direction of the minimum magnetic field needed to suspend the wire that is, to balance the gravitational force on it? The linear density (mass per unit length) of the wire is 46.6 g/m. 34

35 TORQUE ON A CURRENT LOOP F = net 0 The force F 4 has the same magnitude as force F 2 but they act in opposite directions, hence they cancel out exactly. The vector, which is normal to the plane of the coil with direction determined by right hand rule shown in fig. (b). 35

36 τ = iab sin θ A(=ab) is area of loop (15) When a current-carrying loop is placed in a magnetic field, the loop tends to rotate such that its normal becomes aligned with the magnetic field 36

37 TORQUE ON A CURRENT COIL A coil containing N loops, each of area A, the force on each side is N times larger, and the torque on the coil becomes: (16) Where NiA are properties of the coil: its number of turns, its area and the current it carries 37

38 THE MAGNETIC DIPOLE MOMENT, OF A COIL. To account for the torque on the coil due to the magnetic field, a magnetic dipole moment is assigned. The magnitude of is given by: (17) Recall: For electric field, torque exerted by an electric field on an electric dipole is given by: The magnetic potential energy (18) If an applied torque rotates a magnetic dipole from an initial orientation to another orientation, then the work W a done on the dipole by the applied torque is (19) 38

39 Model Question A circular wire loop of radius 15.0 m carries a current of 2.0 A. It is placed so that the normal to its plane makes an angle of 41.0 o with a uniform magnetic field of magnitude 12.0 T. (a) Calculate the magnitude of the magnetic dipole moment of the loop. (b) What is the magnitude of the torque acting on the loop? Solution (a) = A.m 2 (b) Torque τ = µ x B = µbsinθ = (0.184Am 2 )(12.0 T) sin 41.0 o = 1.45 Nm 39

40 CHAPTER 29. MAGNETIC FIELD DUE TO CURRENTS What is Physics? Calculating the Magnetic Field Due to a Current Force Between Two Parallel Currents Ampere's Law Solenoids and Toroids A Current-Carrying Coil as a Magnetic Dipole 40

41 WHAT IS PHYSICS? A moving charged particle produces a magnetic field around itself Activation of regions in the human brain by certain activities such as reading produces magnetic field that can be detected by several devices. 41

MAGNETIC FIELD DUE TO A CURRENT Jean-Baptiste Biot

magnetic field from any electric current Choose a

carrying a current i The field db from this element

Biot-Savart Law in scalar form: µ 0 =4πx10 7 T m/a=1.

42 MAGNETIC FIELD DUE TO A CURRENT Jean-Baptiste Biot ( ) Quantitative rule for computing the magnetic field from any electric current Choose a differential element of wire of length ds and carrying a current i The field db from this element at a point located by the vector r is given by the Biot-Savart Law in scalar form: µ 0 =4πx10 7 T m/a=1.26 x 10-6 T.m/A (permeability constant) Felix Savart ( ) 42

43 BIOT-SAVART LAW An infinitely long straight wire carries a current i. Determine the magnetic field generated at a point located at a perpendicular distance R from the wire. Choose an element ds as shown Biot-Savart Law: db points INTO the page. This is the vector form of the Biot- Savart law. Integrate over all such elements This law can be used to calculate thenet magnetic field B produced at a point by various 43 Where θ is the angle between the directions of ds and r, a unit vector that points from ds towards P.

44 MAGNETIC FIELD DUE TO A CURRENT IN A LONG STRAIGHT WIRE db = µ 0 i sinθids 2 4π r 44

45 RIGHT-HAND RULE Grasp the element in your right hand with your extended thumb pointing in the direction of the current. Your fingers will then naturally curl around in the direction of the magnetic field lines due to that element. 45

46 MAGNETIC FIELD DUE TO A CURRENT IN A CIRCULAR ARC OF WIRE φ = 2 π 46

47 A CURRENT-CARRYING COIL AS A MAGNETIC DIPOLE For a loop, ϕ=2π, at the center of the loop B c = µ i 0 2R B( z) = µ 0 µ 2 π ( R z ) + If 2 2 3/2 47

48 SAMPLE PROBLEM The wire in Fig. 29-8a carries a current i and consists of a circular arc of radius R and central angle rad, and two straight sections whose extensions intersect the center C of the arc. What magnetic field does the current produce at C? 48

49 TWO CURRENT-CARRYING WIRES EXERT MAGNETIC FORCES ON ONE ANOTHER Two long parallel wire carrying currents exert forces on each other. To find the force on a current-carrying wire due to a second current-carrying wire, first find the field due to the second wire at the site of the first wire. Then find the force on the first wire due to that field. 49 Parallel currents attract each other, and anti-parallel currents repel each other.

To find the force on wire b due to the current in wire a. That current produces a magnetic field B a and it is this magnetic field that produces the force on wire b.

50 To find the force on wire b due to the current in wire a. That current produces a magnetic field B a and it is this magnetic field that produces the force on wire b. (1) The force, produced on wire b due to B a : (2) The vectors L and B are perpendicular to each other, hence: The direction of (3) is the direction of the cross product The Ampere, A, is that constant current which, if maintained in two straight, parallel conductors of infinite length, of negligible circular cross section, and placed 1 m apart in a vacuum, would produce on each of these conductors a force of magnitude of wire length. 50

51 AMPERE'S LAW The loop on the integral sign means that is to be integrated around a closed loop, called an Amperian loop. The current i enc is the net current encircled by that closed loop. Curl your right hand around the Amperian loop, with the fingers pointing in the direction of integration. A current through the loop in the general direction of your outstretched thumb is assigned a plus sign, and a current generally in the opposite direction is assigned a minus sign.

52 Ampère s Law B ds = µ I 0 Using arguments similar to the ones use for Gauss s Law, we can show that this result Is independent of the shape of the curve around the current. Is independent of where the current passes through the curve. Depends only on the total amount of current through the area enclosed by the integration path. 52

53 Magnetic Field inside a current-carrying wire 2 Ithrough JAcircle r J = = π Assume a uniform current density, with I the total current passing through wire of radius R. J 2 I 2 r = I 2 through = π r J = I 2 π R R 2 µ 0r B ds = µ 2 0Ithrough = I 2 µ 0r R 2π rb = I 2 R B ds = Bl = 2π rb µ 0I µ 0I B = r Recall: B = outside. 2 2π R 2π r 53

54 MAGNETIC FIELD OF A SOLENOID For a long ideal solenoid h is the (arbitrary) length of the segment from a to b in the figure. 54 where n is the number of turns per unit length of the solenoid

55 MAGNETIC FIELD OF A TOROID 55

56 EXAMPLE 1. A solenoid is 0.50 m long, has three layers of windings of 750 turns each, and carries a current of 4.0 A. What is the magnetic field at the center of the solenoid? Solution: 56

Key Contents. Magnetic fields and the Lorentz force. Magnetic force on current. Ampere s law. The Hall effect

Key Contents. Magnetic fields and the Lorentz force. Magnetic force on current. Ampere s law. The Hall effect Magnetic Fields Key Contents Magnetic fields and the Lorentz force The Hall effect Magnetic force on current The magnetic dipole moment Biot-Savart law Ampere s law The magnetic dipole field What is a