1. 2 Today’s agendum: Magnetic Fields. You must understand the similarities and differences between electric fields and field lines, and magnetic fields.

2 Today’s agendum: 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 particles. Magnetic Flux and Gauss’ “Law” for Magnetism. You must be able to calculate magnetic flux and recognize the consequences of Gauss’ “Law” for Magnetism. 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.

3 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 S N S N Repel Attract S N S N S N RepelAttract S N

4 There is one difference between magnetism and electricity: it is possible to have isolated + or – electric charges, but isolated N and S poles have never been observed. +- S N I.E., every magnet has BOTH a N and a S pole, no matter how many times you “chop it up.” S N S N S N

5 The earth has associated with it a magnetic field, with poles near the geographic poles. http://hyperphysics.phy- astr.gsu.edu/hbase/magnetic/ magearth.html 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. N S 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 Fields

6 Magnetic field lines point in the same direction that the north pole of a compass would point. Magnetic field lines are tangent to the magnetic field. The more magnetic field lines in a region in space, the stronger the magnetic field. Outside the magnet, magnetic field lines point away from N poles (*why?). *The N pole of a compass would “want to get to” the S pole of the magnet. Later I’ll give a better definition for magnetic field direction.

7 Here’s a “picture” of the magnetic field of a bar magnet, using iron filings to map out the field. The magnetic field ought to “remind” you of the earth’s field.

8 We use the symbol B for magnetic field. Magnetic field lines point away from north poles, and towards south poles. S N The SI unit* for magnetic field is the Tesla. In a bit, we’ll see how the units are related to other quantities we know about, and later in the course we’ll see an “official” definition of the units for the magnetic field. *Old unit, still sometimes used: 1 Gauss = 10 -4 Tesla. These units come from the magnetic force equation, which appears two slides from now.

9 The earth’s magnetic field has a magnitude of roughly 0.5 G, or 0.00005 T. A powerful perm- anent magnet, like the kind you might find in headphones, might produce a magnetic field of 1000 G, or 0.1 T. An electromagnet like this one can produce a field of 26000 G = 26 kG = 2.6 T. Superconducting magnets can produce a field of over 10 T. Never get near an operating superconducting magnet while wearing a watch or belt buckle with iron in it! http://liftoff.msfc.nasa.gov/academy/space/mag_field.html

11 A charged particle moving in a magnetic field experiences a force. Magnetic Fields and Moving Charges The magnetic force equation predicts the effect of a magnetic field on a moving charged particle. force on particle magnetic field vector velocity of charged particle What is the force if the charged particle is at rest?

12  Vector notation conventions:  is a vector pointing out of the paper/board/screen (looks like an arrow coming straight for your eye).  is a vector pointing into the paper/board/screen (looks like the feathers of an arrow going away from eye). 

13 Magnitude of magnetic force (and the meaning of v  and B  ): +  B v +  B v vv BB Cross product as presented in Physics 103.

14 Use right hand rule: fingers out, thumb perpendicular to them (curl fingers through smallest angle from v to B) thumb points in direction of F Direction of magnetic force--- Your text presents two alternative variations (curl your fingers, imagine turning a right-handed screw). There is one other variation on the right hand rule. I’ll demonstrate all variations in lecture sooner or later. still more variations: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html rotate your hand until your palm points in the direction of B

15 Direction of magnetic force: +  F B v  F B v -  v B - F?

16 “Foolproof” technique for calculating both magnitude and direction of magnetic force.

17 Alternative* view of magnetic field units. *“Official” definition of units coming later. Remember, units of field are force per “something.”

19 Magnetic Flux and Gauss’ “Law” for Magnetism Magnetic Flux We have used magnetic field lines to visualize magnetic fields and indicate their strength. We are now going to count the number of magnetic field lines passing through a surface, and use this count to determine the magnetic field. B

20 The magnetic flux passing through a surface is the number of magnetic field lines that pass through it. Because magnetic field lines are drawn arbitrarily, we quantify magnetic flux like this:  M =BA. If the surface is tilted, fewer lines cut the surface. B A If these slides look familiar, refer back to lecture 4! B 

21 B   A The “amount of surface” perpendicular to the magnetic field is A cos . A  = A cos  so  M = BA  = BA cos . We define A to be a vector having a magnitude equal to the area of the surface, in a direction normal to the surface. Because A is perpendicular to the surface, the amount of A parallel to the electric field is A cos . Remember the dot product from Physics 103?

22 If the magnetic field is not uniform, or the surface is not flat… divide the surface into infinitesimal surface elements and add the flux through each… dA B your starting equation sheet has…

23 If the surface is closed (completely encloses a volume)… B …we count lines going out as positive and lines going in as negative… dA a surface integral, therefore a double integral But there are no magnetic monopoles in nature (experimental fact). If there were more flux lines going out of than into the volume, there would be a magnetic monopole inside.

24 B Therefore dA Gauss’ “Law” for Magnetism! Gauss’ “Law” for magnetism is not very useful in this course. The concept of magnetic flux is extremely useful, and will be used later!

25 You have now learned Gauss’s “Law” for both electricity and magnetism. These equations can also be written in differential form:

27 Example: an electron travels at 2x10 7 m/s in a plane perpendicular to a 0.01 T magnetic field. Describe its path. Motion of a charged particle in a uniform magnetic field

28 Example: an electron travels at 2x10 7 m/s in a plane perpendicular to a 0.01 T magnetic field. Describe its path. The above paragraph is a description of uniform circular motion. The electron will move in a circular path with an acceleration equal to v 2 /r, where r is the radius of the circle. The force on the electron (remember, its charge is -) is always perpendicular to the velocity. If v and B are constant, then F remains constant (in magnitude). -                                                         B v F v F - -

29 Motion of a proton in a uniform magnetic field +                                        B out v FBFB + + v v FBFB FBFB r The force is always in the radial direction and has a magnitude qvB. For circular motion, a = v 2 /r so The rotational frequency f is called the cyclotron frequency The period T is Thanks to Dr. Waddill for the use of the picture and following examples.

30 Helical motion in a uniform magnetic field If v and B are perpendicular, a charged particle travels in a circular path. v remains constant but the direction of v constantly changes. If v has a component parallel to B, then v  remains constant, and the charged particle moves in a helical path. v  vv B + There won’t be any test problems on helical motion.

31 Lorentz Force “Law” If both electric and magnetic fields are present, Applications The applications in the remaining slides are in your textbook sections for the next lecture. If I have time, I will show them today. The energy calculation in the mass spectrometer example is often useful in homework.

32 Velocity Selector  E - - - - - - - -                                 v B out + + q When the electric and magnetic forces balance then the charge will pass straight through. This occurs when F E = F B or Thanks to Dr. Waddill for the use of the picture.

33       B  S VV +q+q x r Mass Spectrometers Mass spectrometers separate charges of different mass. When ions of fixed energy enter a region of constant magnetic field, they follow a circular path. The radius of the path depends on the mass/charge ratio and speed of the ion, and the magnitude of the magnetic field. Thanks to Dr. Waddill for the use of the picture.

34       B  S VV +q+q x r Example: Ions from source S enter a region of constant magnetic field B that is perpendicular to the ions path. The ions follow a semicircle and strike the detector plate at x = 1.7558 m from the point where they entered the field. If the ions have a charge of 1.6022 x 10 -19 C, the magnetic field has a magnitude of B = 80.0 mT, and the accelerating potential is  V = 1000.0 V, what is the mass of the ion? Radius of ion path: Unknowns are m and v.

35       B  S VV +q+q x r Conservation of energy gives speed of ion. The ions leave the source with approximately zero kinetic energy 0 A proton accelerates when it goes from a more positive V to a less positive V; i.e., when  V is negative. That’s what this minus sign means. Caution! V is potential, v (lowercase) is speed.

36       B  S VV +q+q x r

37       B  S VV +q+q x r 1 atomic mass unit (amu) equals 1.66x10 -27 kg, so uranium-238!

38 Today’s agendum: Review and some interesting consequences of F=qvxB. You must understand the similarities and differences between electric forces and magnetic forces on charged particles. Magnetic forces on currents and current-carrying wires. You must be able to calculate the magnetic force on currents. Magnetic forces and torques on current loops. You must be able to calculate the torque and magnetic moment for a current-carrying wire in a uniform magnetic field. Applications: galvanometers, electric motors, rail guns. You must be able to use your understanding of magnetic forces and magnetic fields to describe how electromagnetic devices operate.

39 Reminder: signs Include the sign on q, properly account for the directions of any two of the vectors, and the direction of the third vector is calculated “automatically.” If you determine the direction “by hand,” use the magnitude of the charge. Everything in this equation is a magnitude. The sign of r had better be +!

40 The electric force acts in the direction of the electric field. Magnetic and Electric Forces The magnetic force acts perpendicular to the magnetic field. The electric force is nonzero even if v=0. The magnetic force is zero if v=0.

41 The electric force does work in displacing a charged particle. Magnetic and Electric Forces The magnetic force does no work in displacing a charged particle! E F + D + v  B F ds Amazing!

43 So far, I’ve lectured about magnetic forces on moving charged particles. Magnetic Forces on Currents Actually, magnetic forces were observed on current-carrying wires long before we discovered what the fundamental charged particles are. Experiment, followed by theoretical understanding, gives If you know about charged particles, you can derive this from the equation for the force on a moving charged particle. It is valid for a straight wire in a uniform magnetic field. For reading clarity, I’ll use L instead of the l your text uses.

44 Here is a picture to help you visualize. It came from http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html

45                                 L I F B Valid for straight wire, length L inside region of magnetic field constant magnetic field, constant current I, direction of L is direction of conventional current I. You could apply this equation to a beam of charged particles moving through space, even if the charged particles are not confined to a wire.

46 What if the wire is not straight?                                           I B ds dF Integrate over the part of the wire that is in the magnetic field region.

47 Example: a wire carrying current I consists of a semicircle of radius R and two horizontal straight portions each of length L. It is in a region of constant magnetic field as shown. What is the net magnetic force on the wire?                                 B  LL R I x y There is no magnetic force on the portions of the wire outside the magnetic field region.

48                                 B  LL R I x y First look at the two straight sections. L  B, so F1F1 F2F2

49                                 B  LL R I x y Next look at the semicircular section. ds  B, so F1F1 F2F2 Calculate the incremental force dF on an incremental ds of current-carrying wire. dF  dd ds ds subtends the angle from  to  +d . The incremental force is Arc length Finally,

50                       B  LL R I x y Calculate the y- component of F. F1F1 F2F2 dF  dd ds dF y Interesting—just the force on a straight horizontal wire of length 2R.

51                       B  LL R I x y F1F1 F2F2 dF  dd ds Or, you can calculate the x component of F. Does symmetry give you F x immediately? dF x

52                                 B  LL R I x y Total force: F1F1 F2F2 dF  ds FyFy

53 Example: a semicircular closed loop of radius R carries current I. It is in a region of constant magnetic field as shown. What is the net magnetic force on the loop of wire?                                 B  R I x y We calculated the force on the semicircular part in the previous example (current is flowing in the same direction there as before). FCFC

54                                 B   R I x y FCFC Next look at the straight section. L  B, and L=2R so F s is directed in the –y direction (right hand rule). FSFS The net force on the closed loop is zero! This is true in general for closed loops in a uniform magnetic field.

56 Magnetic Forces and Torques on Current Loops We showed (not in general, but illustrated the technique) that the net force on a current loop in a uniform magnetic field is zero. No net force means no motion. No net force means no motion.NOT. Example: a rectangular current loop of area A is placed in a uniform magnetic field. Calculate the torque on the loop. B I Let the loop carry a counterclockwise current I and have length L and width W. W L The drawing is not meant to imply that the top and bottom parts are outside the magnetic field region.

57 B I W L There is no force on the “horizontal” segments because the current and magnetic field are in the same direction. The vertical segment on the left “feels” a force “out of the page.” The vertical segment on the right “feels” a force “into the page.” The two forces have the same magnitude: F L = F R = I L B. Because F L and F R are in opposite directions, there is no net force on the current loop, but there is a net torque.   FLFL FRFR

58 Top view of current loop, looking “down,” at the instant the magnetic field is parallel to the plane of the loop. FLFL FRFR B  ILIL  IRIR In general, torque is area of loop = WL

59 When the magnetic field is not parallel to the plane of the loop… FLFL FRFR B  ILIL  IRIR A   Define A to be a vector whose magnitude is the area of the loop and whose direction is given by the right hand rule (cross A into B to get  ). Then

60 FLFL FRFR B  ILIL  IRIR A   Magnetic Moment of a Current Loop Alternative way to get direction of A: curl your fingers (right hand) in direction of current; thumb points in direction of A. IA is defined to be the magnetic moment of the current loop. Your starting equation sheet has:

61 Energy of a magnetic dipole in a magnetic field You don’t realize it yet, but we have been talking about magnetic dipoles for the last 5 slides. A current loop, or any other body that experiences a magnetic torque as given above, is called a magnetic dipole. Energy of a magnetic dipole? You already know this:Today: Electric DipoleMagnetic Dipole

63 The Galvanometer Now you can understand how a galvanometer works… http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/galvan.html#c1 When a current is passed through a coil connected to a needle, the coil experiences a torque and deflects. See the link below for more details.

64 Electric Motors http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/mothow.html#c1

65 Hyperphysics has nice interactive graphics showing how dc and ac motors work.dc ac

66 Today’s agendum: Magnetic Fields Due To A Current. You must be able to calculate the magnetic field due to a moving charged particle. Biot-Savart “Law.” You must be able to use the Biot-Savart “Law” to calculate the magnetic field of a current- carrying conductor (for example: a long straight wire). We’ve been working with the effects of magnetic fields without considering where they come from. Today we learn about sources of magnetic fields.

67 But first, a note on the right hand rule. I personally find the three-fingered axis system often (but not always) to be the most useful.     No, as long as you keep the right order. All three of these will work:          “In and does it matter which finger I use for what?” No, as long as you keep the right order. 

68    This works:    This doesn’t: Switching only two is wrong! “The right-hand rule is unfair! Physics is discriminating against left-handers!” No, you can get the same results with left-hand axes and left hand rules. See this web page.this

69 Today’s agendum: Magnetic Fields Due To A Current. You must be able to calculate the magnetic field due to a moving charged particle. Biot-Savart “Law.” You must be able to use the Biot-Savart “Law” to calculate the magnetic field of a current- carrying conductor (for example, a long straight wire).

70 Biot-Savart “Law”: magnetic field of a current element + B r v It is experimentally observed that a moving point charge q gives rise to a magnetic field  0 is a constant, and its value is  0 =4  x10 -7 T·m/A Let’s start with the magnetic field of a moving charged particle. Remember: the direction of r is always from the source point (the thing that causes the field) to the field point (the location where the field is being measured. 

71 This is example 28.1 in your text. Example: proton 1 has a speed v 0 (v 0 <<c) and is moving along the x-axis in the +x direction. Proton 2 has the same speed and is moving parallel to the x-axis in the –x direction, at a distance r directly above the x-axis. Determine the electric and magnetic forces on proton 2 at the instant the protons pass closest to each other. 1 r x y z 2 v0v0 v0v0 The electric force is E FEFE

72 1 r x y z 2 FEFE At the position of proton 2 there is a magnetic field due to proton 1. B1B1 v0v0 v0v0

73 1 r x y z 2 FEFE B1B1 Proton 2 “feels” a magnetic force due to the magnetic field of proton 1. FBFB v0v0 v0v0 What would proton 1 “feel?” Caution! Relativity overrules Newtonian mechanics! However, in this case, the force is “equal & opposite.”

74 1 r x y z 2 FEFE B1B1 Both forces are in the +y direction. The ratio of their magnitudes is FBFB v0v0 v0v0 Later we will find that

75 1 r x y z 2 FEFE B1B1 Thus FBFB v0v0 v0v0 If v 0 =10 6 m/s, then Don’t you feel sorry for the poor, weak magnetic force?

76 Today’s agendum: Magnetic Fields Due To A Current. You must be able to calculate the magnetic field due to a moving charged particle. Biot-Savart “Law.” You must be able to use the Biot-Savart “Law” to calculate the magnetic field of a current- carrying conductor (for example, a long straight wire).

77 From the equation for the magnetic field of a moving charged particle, it is “easy” to show that a current I in a little length d l of wire gives rise to a little bit of magnetic field. dB r dldl The Biot-Savart “Law” I You may see the equation written using If you like to be more precise in your language, substitute “an infinitesimal” for “a little length” and “a little bit of.” I often use ds instead of d l because the script l does not display very well.

78 Applying the Biot-Savart “Law”  I ds r  dB

79 Example: calculate the magnetic field at point P due to a thin straight wire of length L carrying a current I. (P is on the perpendicular bisector of the wire at distance a.) ds is an infinitesimal quantity in the direction of dx, so I y r x  dB P ds  x z  a L

80 I y r x  dB P ds  x z  a L

81 I y r x  dB P ds  x z  a look integral up in tables, use the web,or use trig substitutions web L

82 I y r x  dB P ds  x z  a

83 I y r x  dB P ds  x z  a When L , or

84 I B r Magnetic Field of a Long Straight Wire We’ve just derived the equation for the magnetic field around a long, straight wire* with a direction given by a “new” right-hand rule. *Don’t use this equation unless you have a long, straight wire!

85 Looking “down” along the wire:  I B The magnetic field is not constant. At a fixed distance r from the wire, the magnitude of the magnetic field is constant. The magnetic field direction is always tangent to the imaginary circles drawn around the wire, and perpendicular to the radius “connecting” the wire and the point where the field is being calculated.

86 Today’s agendum: Force Between Current-Carrying Conductors. You must be able to begin with starting equations and calculate forces between current- carrying conductors. Magnetic Fields Due To A Current Loop. You must be able to apply the Biot-Savart “Law” to calculate the magnetic field of a current loop.

87 Magnetic Field of a Current-Carrying Wire “Knowledge advances when Theory does battle with Real Data.”—Ian Redmount It is experimentally observed that parallel wires exert forces on each other when current flows. I1I1 I2I2  F 12 F 21 I1I1 I2I2  F 12 F 21

88 The magnitude of the force depends on the two currents, the length of the wires, and the distance between them. I1I1 I2I2  F 12 F 21 d L The wires are electrically neutral, so this is not a Coulomb force. We showed that a long straight wire carrying a current I gives rise to a magnetic field B at a distance r from the wire given by I B r Remember, the direction of the field is given by another (different) right-hand rule: grasp the wire and point the thumb of your right hand in the direction of I; your fingers curl around the wire and show the direction of the magnetic field. The magnetic field of one wire exerts a force on a nearby current-carrying wire.

89 Example: use the expression for B due to a current-carrying wire to calculate the force between two current-carrying wires. I1I1 I2I2  F 12 d L L2L2 L1L1 B2B2  x y The force per unit length of wire is

90 I1I1 I2I2  F 12 F 21 d L L2L2 L1L1 B1B1  x y The force per unit length of wire is

91 If the currents in the wires are in the opposite direction, the force is repulsive. I1I1 I2I2  F 12 F 21 d L L2L2 L1L1 y

92 I1I1 I2I2  F 12 F 21 d L2L2 L1L1 The official definition of the Ampere: 1 A is the current that produces a force of 2x10 -7 N force per meter of length between two long parallel wires placed 1 meter apart in empty space.

93 Today’s agendum: Force Between Current-Carrying Conductors. You must be able to begin with starting equations and calculate forces between current- carrying conductors. Magnetic Fields Due To A Current Loop. You must be able to apply the Biot-Savart “Law” to calculate the magnetic field of a current loop.

94 Magnetic Field of a Current Loop P a x dB x  A circular ring of radius a carries a current I as shown. Calculate the magnetic field at a point P along the axis of the ring at a distance x from its center. dldl 90-  I y dB y dB x  90-  z Complicated diagram! You are supposed to visualize the ring lying in the yz plane. d l is in the yz plane. r is in the xy plane and is perpendicular to d l. Thus Also, dB must lie in the xy plane (perpendicular to d l ) and is also perpendicular to r.

95 P a x dB x  dldl 90-  I y dB y dB x  90-  z By symmetry, B y will be 0. Do you see why?

96 P x dB z x dldl I y dB y dB x z When d l is not centered at z=0, there will be a z-component to the magnetic field, but by symmetry B z will still be zero.

97 I, x, and a are constant as you integrate around the ring! P a x dB x  dldl 90-  I y dB y dB x  90-  z

98 At the center of the ring, x=0. P a x dB x  dldl 90-  I y dB y dB x  90-  z For N tightly packed concentric rings (a coil)…

99 Magnetic Field at the center of a Current Loop A circular ring of radius a lies in the xy plane and carries a current I as shown. Calculate the magnetic field at the center of the loop. a z dldl I x y One-page derivation, so you won’t have to go through all the above steps (to be worked at the blackboard)! The direction of the magnetic field will be different if the plane of the loop is not in the xy plane. I’ll explain the “funny” orientation of the axes in lecture.

100 Today’s agendum: Ampere’s “Law.” You must be able to use Ampere’s “Law” to calculate the magnetic field for high-symmetry current configurations. Solenoids. You must be able to use Ampere’s “Law” to calculate the magnetic field of solenoids and toroids. You must be able to use the magnetic field equations derived with Ampere’s “Law” to make numerical magnetic field calculations for solenoids and toroids.

101 I B r Ampere’s “Law” Just for kicks, let’s evaluate the line integral along the direction of B over a closed circular path around a current-carrying wire. ds The above calculation is only for the special case of a long straight wire, but you can show that the result is valid in general.

102 Ampere’s “Law” I is the total current that passes through a surface bounded by the closed (and not necessarily circular) path of integration. Ampere’s “Law” is useful for calculating the magnetic field due to current configurations that have high symmetry. I B r ds The current I passing through a loop is positive if the direction of integration is the same as the direction of B from the right hand rule. I B r ds positive I negative I

103 Your text writes General form of Ampere’s “Law”: I1I1 ds If your path includes more than one source of current, add all the currents (with correct sign). I2I2 because the current that you use is the current “enclosed” by the closed path over which you integrate. The reason for the 2 nd term on the right will become apparent later. Ignore it for now.

104 Example: a cylindrical wire of radius R carries a current I that is uniformly distributed over the wire’s cross section. Calculate the magnetic field inside and outside the wire. I R Cross-section of the wire: R r  direction of I B

105 R r  direction of I B Over the closed circular path r: Solve for B: B is linear in r.

106 R r  direction of I B Outside the wire: (as expected). B rR Plot: A lot easier than using the Biot-Savart “Law”!

107 Calculating Electric and Magnetic Fields Electric Field in general: Coulomb’s “Law” for high symmetry configurations: Gauss’ “Law” Magnetic Field in general: Biot-Savart “Law” for high symmetry configurations: Ampere’s “Law” This analogy is rather flawed because Ampere’s “Law” is not really the “Gauss’ “Law” of magnetism.”

108 Today’s agendum: Ampere’s “Law.” You must be able to use Ampere’s “Law” to calculate the magnetic field for high-symmetry current configurations. Solenoids. You must be able to use Ampere’s Law to calculate the magnetic field of solenoids and toroids. You must be able to use the magnetic field equations derived with Ampere’s “Law” to make numerical magnetic field calculations for solenoids and toroids.

109 Magnetic Field of a Solenoid A solenoid is made of many loops of wire, packed closely together. Here’s the magnetic field from a single loop of wire: Some images in this section are from hyperphysics. hyperphysics

110 Stack many loops to make a solenoid: This ought to remind you of the magnetic field of a bar magnet.

111 You can use Ampere’s “Law” to calculate the magnetic field of a solenoid.                                               B I     l N is the number of loops enclosed by our surface.

112                                               B I     l Magnetic field of a solenoid of length l, N loops, current I. n=N/ l (number of turns per unit length). The magnetic field inside a long solenoid does not depend on the position inside the solenoid (if end effects are neglected).

113 A toroid* is just a solenoid “hooked up” to itself. Magnetic field inside a toroid of N loops, current I. The magnetic field inside a toroid is not subject to end effects, but is not constant inside (because it depends on r). *Your text calls this a “toroidal solenoid.”

114 Example: a thin 10-cm long solenoid has a total of 400 turns of wire and carries a current of 2 A. Calculate the magnetic field inside near the center.

115 “Help! Too many similar starting equations!” long straight wire center of N loops of radius a solenoid, length l, N turns solenoid, n turns per unit length toroid, N loops “How am I going to know which is which on the next exam?” use Ampere’s “Law” (or note the lack of N) probably not a starting equation field inside a solenoid is constant field inside a toroid depends on position (r)

1. 2 Today’s agendum: Magnetic Fields. You must understand the similarities and differences between electric fields and field lines, and magnetic fields.

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1. 2 Today’s agendum: Magnetic Fields. You must understand the similarities and differences between electric fields and field lines, and magnetic fields.

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Presentation on theme: "1. 2 Today’s agendum: Magnetic Fields. You must understand the similarities and differences between electric fields and field lines, and magnetic fields."— Presentation transcript:

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