Chapter 30. Sources of the Magnetic Field Amperes and Biot-Savart Laws

Transcription

1 Chapter 30 Sources of the Magnetic Field Amperes and Biot-Savart Laws

2 F B on a Charge Moving in a Magnetic Field Magnitude proportional to charge and speed of the particle Direction depends on the velocity of the particle and the direction of the magnetic field It is perpendicular to both F B = q v x B F B is the magnetic force q is the charge v is the velocity of the moving charge B is the magnetic field

3 Direction given by Right-Hand Rule Fingers point in the direction of v (for positive charge; opposite direction if negative) Curl fingers in the direction of B Then thumb points in the direction of v x B; i.e. the direction of F B

4 Hall Effect When the charge carriers are negative, the upper edge of the conductor becomes negatively charged When the charge carriers are positive, the upper edge becomes positively charged Sign of Hall voltage, V H, gives the sign of the charges

5 Biot-Savart Law Introduction Biot and Savart conducted experiments on the force exerted by an electric current on a nearby magnet They arrived at a mathematical expression that gives the magnetic field at some point in space due to a current

6 Biot-Savart Law Set-Up Measured magnetic field db at some point P, distance r from wire Wire carrying a steady current of I Length element is ds Biot & Savart found that db is proportional to current, I, length, ds and sin( ) is inversely proportional to r 2

7 Biot-Savart Law Equation Observations are summarized in the Biot-Savart law: d B 0 4 I d s ˆ r r 2 Constant o called the permeability of free space o = 4 x 10-7 T. m / A The magnetic field is due to the currentcarrying conductor

8 Total Magnetic Field db is the field created by the current in the length segment ds To find total field, B, sum up contributions from all current elements I.ds B 0 4 I d s ˆ r The integral is over the entire current distribution B-S Law is also valid for a current consisting of charges flowing through space r 2

9 Quick Quiz 30.1 Consider the current in the length of wire shown in the figure below. Rank the points A, B, and C in terms of magnitude of the magnetic field due to the current in the length element shown, from greatest to least. (a) A, B, C (b) B, C, A (c) C, B, A (d) C, A, B (e) An equal field applies at all these points.

10 Quick Quiz 30.1 Answer: (b). Point B is closest to the current element. Point C is farther away and the field is further reduced by the sin θ factor in the cross product because θ = 0. d s x r ˆ. The field at A is zero

11 B Compared to E, 1 Distance Magnitude of magnetic field varies as inverse square of distance from source Magnitude of electric field due to point charge also varies as inverse square of distance from the charge

12 B Compared to E, 2 Direction Electric field created by a point charge is radial in direction Magnetic field created by a current element is perpendicular to both the length element ds and the unit vector rˆ

13 B Compared to E, 3 Source Electric field is established by an isolated electric charge Current element producing a magnetic field must be part of an extended current distribution Need to integrate over entire current distribution

14 B Compared to E, 4 Ends of Field Lines Magnetic field lines have no beginning and no end They form continuous circles Electric field lines begin on positive charges and end on negative charges

15 B for Long, Straight Conductor Thin, straight wire carrying constant current I d s r ˆ dx sin( ) k ˆ Integrating over all the current elements gives (see ex 30.1) B 0 I 4 a 2 1 sin( ) d 0 I 4 a cos( 1 ) cos( 2 )

16 B for a Long, Straight Conductor, Special Case For infinitely long, straight wire 1 = 0 and 2 = 1 cos( 1 1 cos( 2 1 The field becomes B 0 I 2 a

17 B for a Long, Straight Conductor, Direction Magnetic field lines are circles concentric with the wire Field lines lie in planes perpendicular to to wire Magnitude of B is constant on any circle of radius a The right-hand rule for determining the direction of B is shown Grasp wire with thumb in direction of current. Fingers wrap in direction of B.

18 B for a Curved Wire Segment What is the field at point O due to the wire segment? I and R are constants ds is parallel to r along AA and CC so we only need to calculate the field from AC. We find that B 0 I 4 R in radians See Example 30.2

19 B for a Circular Loop of Wire Put = 2 in previous result B I 0 2 I 0 4 R 2 R This is the field at the centre of the loop

20 Demo Ec10: Forces between parallel wires Wires can be made to attract and repel one another depending on the direction of the current flow through each wire.

21 Magnetic Force Between Two Parallel Conductors Two parallel wires carrying steady currents I 1 and I 2 Field B 2 due to the current in wire 2 exerts a force on wire 1 of F 1 = B 2 I 1 l However B 2 = 0 I 2 / 2 a, so F 1 0 I 1 I 2 2 a l

22 Magnetic Force Between Two Parallel Conductors, cont. F 1 0 I 1 I 2 2 a l Parallel conductors carrying currents in the same direction attract each other Parallel conductors carrying current in opposite directions repel each other

23 Quick Quiz 30.2 For I 1 = 2 A and I 2 = 6 A in the figure below, which is true: (a) F 1 = 3F 2 (b) F 1 = F 2 /3 (c) F 1 = F 2

24 Quick Quiz 30.2 Answer: (c). F 1 = F 2 as required by Newton s third law. Another way to arrive at this answer is to realize that F 1 0 I 1 I 2 2 a gives the same result whether the multiplication of currents is (2 A)(6 A) or (6 A)(2 A). l

25 Quick Quiz 30.3 A loose spiral spring is hung from the ceiling, and a large current is sent through it. The coils move (a) closer together (b) farther apart (c) they do not move at all

26 Quick Quiz 30.3 Answer: (a). The coils act like wires carrying parallel currents in the same direction and hence attract one another.

27 Definition of the Ampere The force between two parallel wires can be used to define the ampere F 1 l 0 I 1 I 2 2 a with T m A -1 When the magnitude of the force per unit length between two long parallel wires that carry identical currents and are separated by 1 m is 2 x 10-7 N/m, the current in each wire is defined to be 1 A

28 Definition of the Coulomb The SI unit of charge, the coulomb, is defined in terms of the ampere When a conductor carries a steady current of 1 A, the quantity of charge that flows through a cross section of the conductor in 1 s is 1 C

29 Magnetic Field of a Wire Compass can be used to detect the magnetic field When there is no current in the wire, there is no field due to the current Needle points towards the Earth s north pole

30 Magnetic Field of a Wire, 2 When wire carries a current a compass needle deflects in a direction tangent to the circle Shows the direction of the magnetic field produced by the wire

31 Ampere s Law The line integral of B. ds around any closed path equals µ o I, where I is the total steady current passing through any surface bounded by the closed path. B ds 0 I

32 Ampere s Law, cont. Ampere s law describes the creation of magnetic fields by all continuous current configurations Direction: put thumb of right hand in the direction of the current through loop. Your fingers curl in the direction you integrate around the loop.

33 Quick Quiz 30.4 Rank the magnitudes of B.d s for the closed paths in the figure below, from least to greatest. (a) a, b, c, d (b) b, d, a, c (c) c, d, b, a (d) c, b, a, d (e) d, c, a, b

34 Quick Quiz 30.4 B ds 0 I Answer: (b). Ampere s law (above) indicates that the value of the line integral depends only on the net current through each closed path. Path b encloses 1 A, path d encloses 3 A, path a encloses 4 A, and path c encloses 6 A.

35 Quick Quiz 30.5 Rank the magnitudes of B.d s for the closed paths in the figure below, from least to greatest. (a) a, b, c, d (b) b, c, d, a (c) b, d, a, c (d) d, c, a, b (e) The criteria are badly chosen in this case.

36 Quick Quiz 30.5 Answer: (e). Ranked from least to greatest, first comes b, then a = c = d. Paths a, c and d all give the same nonzero value μ 0 I because the size and shape of the paths do not matter. Path b does not enclose the current, and hence its line integral is zero.

37 MFM04AN1: Ampere s law for magnetic field near a currentcarrying wire

38 Field Due to a Long Straight Wire from Ampere s Law Calculate the magnetic field at a distance r from the centre of a wire carrying a steady current I The current is uniformly distributed through the cross section of the wire

39 Field Due to Long Straight Wire 1. Outside of the wire, r > R B ds B 2 r 0 I B 0 I 2 r 2. Inside the wire, current I' at radius r: I r 2 R 2 I, so that r 2 B ds B 2 r 0 I 0 R 2 I B 0 I 2 R 2 r

40 Field Due to a Long Straight Wire Results Summary Field proportional to r inside the wire Field varies as 1/r outside the wire Both equations are equal at r = R Note: much easier to calculate using Ampere s rather than Biot-Savart law!

41 Magnetic Field of a Toroid Find field at distance r from the centre of the toroid Toroid has N turns of wire B ds B 2 r 0 NI B 0 NI 2 r

42 Magnetic Field of a Solenoid A solenoid is a long wire wound in the form of a helix Reasonably uniform magnetic field can be produced in its interior. Here the field lines are approximately parallel uniformly distributed close together

43 Magnetic Field of a Tightly Wound Solenoid Field distribution is similar to a bar magnet As the length of the solenoid increases the interior field becomes more uniform the exterior field becomes weaker i.e. an ideal solenoid

44 Ideal Solenoid Characteristics An ideal solenoid is approached when: turns closely spaced length much greater than radius of turns

45 Field in interior of a Solenoid Apply Ampere s law Consider rectangle with side l parallel to the interior field and side w perpendicular to the field The side of length l inside the solenoid contributes to the field This is path 1 in the diagram B d s B d s B path 1 path 1 ds B

46 Magnetic Field of Solenoid Total current through the rectangular path equals the current through each turn multiplied by the number of turns B d s B Ampere s law then gives o NI B 0 N l I 0 ni n = N / l is the number of turns per unit length valid only at points near centre of long solenoid

47 Quick Quiz 30.6 Consider a solenoid that is very long compared to the radius. Of the following choices, the most effective way to increase the magnetic field in the interior of the solenoid is: (a) double its length, keeping the number of turns per unit length constant (b) reduce its radius by half, keeping the number of turns per unit length constant (c) overwrapping the entire solenoid with an additional layer of current-carrying wire

48 Quick Quiz 30.6 Answer: (c). The magnetic field in a very long solenoid is independent of its length or radius. Overwrapping with an additional layer of wire increases the number of turns per unit length.

49 Ampere s vs. Gauss s Law B ds 0 I E da q 0 Integrals around closed path vs. closed surface. i.e. 2D vs. 3D geometrical figures Integrals related to fundamental constant x source of the field. Concept of Flux the flow of field lines through a surface.

50 MFM05AN1: Magnetic Flux

51 Magnetic Flux Magnetic flux defined in a similar way to electric flux Consider an area element da on an arbitrarily shaped surface, and magnetic field B The magnetic flux F B is F B B.d A Units T. m 2 = Wb Wb is a weber

52 Magnetic Flux through a Plane Suppose a plane, of area A, makes an angle with da Magnetic flux F B = BA cos a) When field parallel to plane, and F B = 0 b) When field perpendicular to plane, and F B = BA maximum value

53 Gauss Law in Magnetism Magnetic fields do not begin or end at any point i.e. they form closed loops, with the number of lines entering a surface equaling the number of lines leaving that surface Gauss law in magnetism says: F B.d A 0 B Would not be true if magnetic monopoles were found!

54 MFA05AN2: Gauss s law for magnetism, F B =0

55 Earth s Magnetic Field The Earth s magnetic field resembles that of a huge bar magnet buried deep in its interior South magnetic pole is located near the north geographic pole North magnetic pole is located near the south geographic pole

56 Dip Angle of Earth s Magnetic Field A compass needle free to rotate vertically as well as horizontally will point towards the Earth s surface Angle between horizontal and direction of the magnetic field is called the dip angle The closer the compass is moved to the magnetic poles, the farther from horizontal its needle will be Needle horizontal at equator, with dip angle = 0 Needle points straight down at south magnetic pole, with dip angle = 90

57 Earth s Magnetic Poles Dip angle of 90 near Hudson Bay in Canada Earth s south magnetic pole Dip angle of -90 near Dumont Durville, Antarctica Earth s north magnetic pole The magnetic and geographic poles are not at the same location The difference between true north and magnetic north is called the magnetic declination Declination varies by location on the Earth s surface

58 Source of the Earth s Magnetic Field There cannot be large masses of permanently magnetized materials in Earth s core since the high temperatures there prevent materials from retaining permanent magnetization Most likely source of the Earth s magnetic field is believed to be convection currents in the liquid part of the core It is also likely related to rate of the Earth s rotation: dynamo action QuickTime and a TIFF (Uncompressed) decompressor are needed to see this picture.

59 Reversals of the Earth s Magnetic Field The direction of the Earth s magnetic field reverses every few million years Evidence of these reversals are found in basalts resulting from volcanic activity The origin of the reversals is not understood

60 Demo Ec19: Magnetic Flux An Earth inductor compass. Coil is rotated at constant speed, with its axis orientated either parallel or anti-parallel to the Earth s magnetic field. The galvanometer will show a different deflection, depending on the direction the axis is pointing. Puzzle: try to work out the direction of the Earth s magnetic field?

61 End of Chapter

62 Quick Quiz 30.7 In an RC circuit, the capacitor begins to discharge. During the discharge, in the region of space between the plates of the capacitor, there is (a) conduction current but no displacement current (b) displacement current but no conduction current (c) both conduction and displacement current (d) no current of any type

63 Quick Quiz 30.7 Answer: (b). There can be no conduction current because there is no conductor between the plates. There is a timevarying electric field because of the decreasing charge on the plates, and the time-varying electric flux represents a displacement current.

64 Quick Quiz 30.8 The capacitor in an RC circuit begins to discharge. During the discharge, in the region of space between the plates of the capacitor, there is (a) an electric field but no magnetic field (b) a magnetic field but no electric field (c) both electric and magnetic fields (d) no fields of any type

65 Quick Quiz 30.8 Answer: (c). There is a time-varying electric field because of the decreasing charge on the plates. This time-varying electric field produces a magnetic field.

66 Quick Quiz 30.9 Which material would make a better permanent magnet? -one whose hysteresis loop looks like Figure (a) -one whose hysteresis loop looks like Figure (b)

67 Quick Quiz 30.9 Answer: (a). The loop that looks like Figure 30.32a is better because the remnant magnetization at the point corresponding to point b in Figure is greater.

68 Quick Quiz If we wanted to cancel the Earth s magnetic field by running an enormous current loop around the equator, the current loop would be directed: (a) east to west (b) west to east

69 Quick Quiz Answer: (b). The lines of the Earth s magnetic field enter the planet in Hudson Bay and emerge from Antarctica; thus, the field lines resulting from the current would have to go in the opposite direction. Compare Figure 30.7a with Figure