today’s agendum: 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

MagnetismRecall 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 SN

S N

Repel

Attract

S N

SNS N

Repel Attract

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 NS 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.

1 kg1 T = C s

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 super-conducting magnet while wearing a watch or belt buckle with iron in it!

http://liftoff.msfc.nasa.gov/academy/space/mag_field.html










10





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.

F=qv B

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):

θ F= q vB sin = q v B= q vB

+ B

v + B

vv B

Cross product as presented in Physics 103.

F= q v B F= q vB

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.html

rotate your hand until your palm points in the direction of B

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html

15

Direction of magnetic force:


+

F

Bv

F

Bv-

v

B-F?

16

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

F=qv B

ˆˆ ˆ

x y z

x y z

i j kF=q det v v v

B B B

17

Alternative* view of magnetic field units.

θF= q vB sin

θFB= q v sin

N NB =T = =C m/s A m

*“Official” definition of units coming later.

Remember, units of field are force per “something.”

18





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.

BA

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?

M B A

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…

dAB

iM i iA 0 i

lim B A

M B dA

your starting equation sheet has…

B B dA

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.

AdBM

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!

0AdBM

25

You have now learned Gauss’s “Law” for both electricity and magnetism.

These equations can also be written in differential form:

0

E

B 0

0AdBM

o

enclosedE

qAdE

26





27

Example: an electron travels at 2x107 m/s in a plane perpendicular to a 0.01 T magnetic field. Describe its path.

Motion of a charged particlein a uniform magnetic field

28

Example: an electron travels at 2x107 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 v2/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

vF

v

F

-

-

29

Motion of a proton in a uniform magnetic field

+

Bout

v

FB

+ +

v

v

FBFB

r

The force is always in the radial direction and has a magnitude qvB. For circular motion, a = v2/r so

The rotational frequency f is called the cyclotron frequency

The period T is

2mvF= q vB= r

qrB mvv= r=m qB

π π2 r 2 mT = =v qB πqB1f = =T 2 m

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

B

+

There won’t be any test problems on helical motion.

31

Lorentz Force “Law”If both electric and magnetic fields are present, .

F=q E+v B

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

Bout

+ + q

When the electric and magnetic forces balance then the charge will pass straight through. This occurs when FE = FB or

Thanks to Dr. Waddill for the use of the picture.

EF =qE

BF =qv B

EqE=qvB or v=B

33

B

S

V

+q

xr

Mass SpectrometersMass 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

+q

xr

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.

mvx=2r and r= qB

35

B

S

V

+q

xr

Conservation of energy gives speed of ion. The ions leave the source with approximately zero kinetic energy

i i ffK +U =K +U

fi i fK =K + U -U

ffi fiK =- U -U =-q V - V =-q V

21mv =-q V2-2q Vv= m

fK =-q V

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

+q

xr

-2q Vv= m

2mvx=2r= qB

2m -2q Vx= qB m

2 -2m Vx=B q

2 2B x qm=- 8 V

37

B

S

V

+q

xr

2 2B x qm=- 8 V

2 2 -190.08 T 1.7558 m 1.6022 10 Cm=- 8 -1000 V

-25m=3.9515 10 kg

m=238.04 u

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

F=qv B

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.

mvr= qBEverything 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.

EF =qE

BF =qv B

The electric force is nonzero even if v=0.

The magnetic force is zero if v=0.

BF v=0 =q 0 B=0

41

The electric force does work in displacing a charged particle.

Magnetic and Electric Forces

EF =qE

BF =qv B

The magnetic force does no work in displacing a charged particle!

EF

+D

FW =F D=FD=qED

+v B

Fds

FW =F ds=0

Amazing!

42





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.

F=qv B

Experiment, followed by theoretical understanding, gives

F=IL B.

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

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html

45

LI

F

B

F=IL B

Valid for straight wire, length L inside region of magnetic fieldconstant 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

Bds

dF

dF=I ds B

F= dF

F=I ds B

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

L LR

I

x

y

There is no magnetic force on the portions of the wire outside the magnetic field region.

48

B

L LR

I

x

y

First look at the two straight sections.

F=IL B

1 2F =F =ILBL B, so

F1 F2

49

B

L LR

I

x

y

Next look at the semicircular section.

dF=I ds B.

ds B, so

F1 F2

Calculate the incremental force dF on an incremental ds of current-carrying wire.

dF

dds

ds subtends the angle from to +d.The incremental force is

ds=R d .Arc lengthdF=I ds B.

dF=I R d B.Finally,

50

B

L LR

I

x

y

Calculate the y- component of F. F1 F2dF

ddsdFy

ydF =I R d B sin

y y0F = dF

y 0

F = I R d B sin

y 0F =I R B sin d

0

yF = -I R B cos

yF =2 I R B Interesting—just the force on a straight horizontal wire of length 2R.

51

B

L LR

I

x

y

F1 F2dF

ddsOr, you can calculate the x component of F.

Does symmetry give you Fx immediately?

xdF =I R d B cos

x 0F = I R d B cos

x 0

F =I R B cos d

0

xF = I R B sin

xF =0

dFx

52

B

L LR

I

x

y

Total force: F1 F2dF

ds

Fy

1 2 yF=F + F + F

F=ILB + ILB + 2IRB

F=2IB L + R

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).

CF =2 I R B

FC

54

B

R

I

x

y

FC

Next look at the straight section.

SF =IL B

SF =2IRBL B, and L=2R so

Fs is directed in the –y direction (right hand rule).FS

ˆ ˆ net S cF =F +F =-2IRB j+2IRB j=0

The net force on the closed loop is zero!This is true in general for closed loops

in a uniform magnetic field.

55





56

Magnetic Forces and Torques on Current LoopsWe 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.

BI

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

BI

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: FL = FR = I L

B.Because FL and FR are in opposite directions, there is no net force on the current loop, but there is a net torque.

FL

FR

58

Top view of current loop, looking “down,” at the instant the magnetic field is parallel to the plane of the loop.

FL

FRB

IL IRW2

W2

=r F .In general, torque is

R RW 1= F WILB2 2

L LW 1= F WILB2 2

net R L= =WILB=IAB

area of loop = WL

59

When the magnetic field is not parallel to the plane of the loop…

FL

FR

B

IL

IR

W2

Wsin 2

R RW 1= F sin WILB sin 2 2

L LW 1= F sin WILB sin 2 2

net R L= =WILB sin =IAB sin

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

= IA B .

60

FL

FR

B

IL

IR

W2

Wsin 2

A

= IA B

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.

= IA

= B

Your starting equation sheet has:

= N I A (N=1 for a single loop)

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:

= B p

= E p

U= - E U= - B

Electric Dipole

Magnetic Dipole

62





63

The GalvanometerNow 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.

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/galvan.html#c1

64

Electric Motors

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



http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/mothow.html

65

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

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.html#c1

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motorac.html#c1

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 welearn 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?”

F=IL B

F=qv B

F=IL B

F=qv B

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.

http://www.pa.msu.edu/courses/1997spring/PHY232/lectures/magforces/righthandrule.html

69


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

r̂

r

v

It is experimentally observed that a moving point charge q gives rise to a magnetic field

ˆμ .4π

02

qv rB= r0 is a constant, and its value is

0=4x10-7 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 v0 (v0<<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

v0

v0The electric force is

ˆ

1 2

E 2q q1F = r4 r

r̂

E

FE

ˆ

2

E 21 eF = j4 r

72

1

r

x

y

z

2

r̂

FE

At the position of proton 2 there is a magnetic field due to proton 1.

ˆ

1 1

1 2q v rB =4 r

ˆ ˆ

0

1 2ev i jB =4 r

ˆ

0

1 2evB = k4 r

B1

v0

v0

73

1

r

x

y

z

2

r̂

FE

B1

Proton 2 “feels” a magnetic force due to the magnetic field of proton 1.

B 2 2 1F =q v B

ˆˆ

0

B 0 2evF =ev i k4 r

ˆ

2 20

B 2e vF = j4 r

FB

v0

v0

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

r̂

FE

B1

Both forces are in the +y direction. The ratio of their magnitudes is

2 20

2B

2E

2

e v4 rF =F 1 e4 r

FB

v0

v0 2B0

E

F = vF

Later we will find that

21=c

75

1

r

x

y

z

2

r̂

FE

B1

Thus

FB

v0

v0

20B2

E

vF =F c

If v0=106 m/s, then

26-5B

28E

10F = 1.11 10F 3 10

Don’t you feel sorry for the poor, weak magnetic force?

76


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 dl of wire gives rise to a little bit of magnetic field.

dB

r̂

r

dl

The Biot-Savart “Law”

ˆμ4π

02

I d rdB= r

I

You may see the equation written using ̂r =r r .

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 dl because the script l does not display very well.

78

Applying the Biot-Savart “Law”

Ids

r

r̂

dB ˆμ ˆ4π

02

I ds r rdB= where r=r r

μ θ4π

02

I ds sin dB= r

B= 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.)

ˆμ4π

02

I ds rdB= rˆˆ θ

ds r=ds sin k

μ θ4π

02

I ds sindB= r

ds is an infinitesimal quantity in the direction of dx, soμ θ4π

02

I dx sindB= r

I

y

r

x

dBP

dsr̂

xz

a

L

80

2 2r= x +aθ asin = r

I

y

r

x

dBP

dsr̂

xz

μ θ4π

02

I dx sindB= r

a

μ μ4π 4π

0 03/23 2 2

I dx a I dx adB= =r x +a

μ4π

L/2 03/2-L/2 2 2

I dx aB=x +a

μ

4π L/20

3/2-L/2 2 2

I a dxB=x +a

L

81

I

y

r

x

dBP

dsr̂

xz

a

μ

4π L/20

3/2-L/2 2 2

I a dxB=x +a

look integral up in tables, use the web,or use trig substitutions

3/2 1/22 2 2 2 2dx x=

x +a a x +a

μ

4π

L/2

01/22 2 2

-L/2

I a xB=a x +a

μ

4π

01/2 1/22 22 2 2 2

I a L/2 -L/2=a L/2 +a a -L/2 +a

L

http://www.efunda.com/math/integrals/integrals.cfm

82

I

y

r

x

dBP

dsr̂

xz

a

μ4π

01/22 2 2

I a 2L/2B=a L /4+a

μ4π0

1/22 2

I L 1B= a L /4+a

μπ

02 2

I L 1B= 2 a L +4a

μπ0

2

2

I 1B=2 a 4a1+ L

83

I

y

r

x

dBP

dsr̂

xz

a μπ0

2

2

I 1B=2 a 4a1+ L

When L, μ.

π0I B=2 a

μπ0I B= 2 ror

84

I

Br

Magnetic Field of a Long Straight Wire

We’ve just derived the equation for the magnetic field around a long, straight wire* μ

π0 IB=2 r

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.

I1 I2

F12 F21

I1 I2

F12 F21

88

The magnitude of the force depends on the two currents, the length of the wires, and the distance between them. μ

π0 1 2 I I LF= 2 d

I1 I2

F12 F21

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 byμ

π0 IB=2 r

I

Br

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.

I1 I2

F12

d

Lμ ˆˆπ0 2

2IB = k2 d

12 1 1 2F =I L B

L2L1

B2μ ˆˆπ

0 212 1

IF =I Lj k2 d

x

y

μ ˆπ

0 1 2

12I I LF = i2 d

The force per unit length of wire is μ ˆπ

0 1 212 I IF = i.L 2 d

90

I1 I2

F12 F21

d

Lμ ˆπ

0 1

1IB =- k2 d

21 2 2 1F =I L B

L2L1

B1μ ˆˆπ

0 1

21 2IF =I Lj k2 d

x

y

μ ˆπ

0 1 2

21I I LF =- i2 d

The force per unit length of wire is μ ˆπ

0 1 221 I IF =- i.L 2 d

91

If the currents in the wires are in the opposite direction, the force is repulsive.

I1 I2

F12 F21

d

L

L2L1y

92

I1 I2

F12 F21

d

L2L1The 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.

μπ

0 1 212 21

I I LF =F = 2 d

ππ

-7-71 2

12 21 1 24 10 I I L LF =F = =2 10 I I2 d d

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.

dl r̂r

90-

I

y

dBy

dBx

90-

z

Complicated diagram! You are supposed to visualize the ring lying in the yz plane.dl is in the yz plane. r is in the xy plane and is perpendicular to dl. Thus

r̂

ˆ . d r =d

Also, dB must lie in the xy plane (perpendicular to dl) and is also perpendicular to r.

95

P

a

x

dB

x

dl r̂r

90-

I

y

dBy

dBx

90-

z

ˆμ4π

02

I d rdB= r

μ4π

02

I ddB= r

μ4π

02 2I ddB= x a

μ μ4π 4π

0 0x 1/22 2 2 2 2 2

I d I d adB = cos =x a x a x a

μ μ4π 4π

0 0y 1/22 2 2 2 2 2

I d I d xdB = sin =x a x a x a

By symmetry, By will be 0. Do you see why?

96

PxdBz x

dlr̂

rI

y

dBy

dBxz

When dl is not centered at z=0, there will be a z-component to the magnetic field, but by symmetry Bz will still be zero.

97

x xring

B = dB

μ4π

0x 3/22 2

I a ddB =x a

I, x, and a are constant as you integrate around the ring!

P

a

x

dB

x

dl r̂r

90-

I

y

dBy

dBx

90-

z

μ μ4π 4π

0 0x 3/2 3/22 2 2 2ring

I a I a B = d = 2 ax a x a

μ

20

x 3/22 2

I a B =2 x a

98

At the center of the ring, x=0.

P

a

x

dB

x

dl r̂r

90-

I

y

dBy

dBx

90-

z

μ 20

x,center 3/22

I a B =2 a

μ μ20 0

x,center 3 I a IB = =2a 2a

For N tightly packed concentric rings (a coil)…

μ0x,center

N IB = 2a

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

dlI

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

Br

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.

r2BdsBsdB

sd B

Ir2r2IsdB o

o

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

Br

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

Br

dspositive I negative I

IsdB o

103

Your text writes

General form of Ampere’s “Law”:

I1

ds

If your path includes more than one source of current, add all the currents (with correct sign).

I2

because the current that you use is the current “enclosed” by the closed path over which you integrate.

The reason for the 2nd term on the right will become apparent later. Ignore it for now.

encloIsdB

dtdIsdB E

oenclo

21o IIsdB

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.

IR

Cross-section of the wire:

Rr

direction of I B

2

2

o2

2

oenclo RrI

RrIIsdB

105

Rr

direction of I B

Over the closed circular path r:

Solve for B:π μ

2

0 2r2 rB= I R

μμ μπ π π

20

0 0 2 2 2Ir rB= I = I = r2 rR 2 R 2 R

B is linear in r.

r2BdsBsdB

106

R

r

direction of I

B

Outside the wire:

μπ0 IB=2 r

(as expected).

B

rR

Plot:

A lot easier than using the Biot-Savart “Law”!

Ir2BdsBsdB o

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.

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfie.html#c1

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

μ 0 B = N I N is the number of loops enclosed by our surface.

4321

sdBsdBsdBsdBsdB

encloI 0 0 0 B sdB

112

B

I

l

μ0 NB= I Magnetic field of a solenoid

of length l , N loops, current I. n=N/l (number of turns per unit length).μ0 B= n I

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.

π μ0 B 2 r = N I

μπ

0 N IB= 2 rMagnetic 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.”

NII sdB oenclo

r2BdsBsdB

114

0NB = μ I

π

-7 400T mB = 4 ×10 2 A A 0.1 m

B = 0.01 T

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!”μ

π0 IB=2 r long straight

wireμ0 N IB= 2a center of N loops of radius a

μ0 NB= I

μ0 B= n I

solenoid, length l, N turns

solenoid, n turns per unit length

μπ

0 N IB= 2 r 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 solenoid is constant

field inside a toroid depends on position (r)

today’s agendum: magnetic fields

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