Chapter 28

Sources of Magnetic Field

Physics for Scientists & Engineers, 3rd EditionDouglas C. Giancoli

© Prentice Hall

3P15-

Magnetic Fields

B q= ×F v B

Magnetic Dipoles Create and Feel B Fields:

Also saw thatmoving charges feel a force:

5P15-

Cyclotron Motion(1) r : radius of the circle

2mvqvBr

=mvrqB

⇒ =

(2) T : period of the motion

2 2r mTv qBπ π

= =

2 v qBfr m

ω π= = =(3) ω : cyclotron frequency

7P15-

B q= ×F v B

Magnetic Force on Current-Carrying Wire

( ) mcharges

= ×B

( )B I= ×F L B

charge ms

= ×B

9P15-

Magnetic Force on Current-Carrying Wire

Current is moving charges, and we know that moving charges feel a force in a magnetic field

17P15-

Electric Field Of Point ChargeAn electric charge produces an electric field:

r̂ 2

1 ˆ4 o

qrπε

=E r

r̂ : unit vector directed from q to P

18P15-

Magnetic Field Of Moving ChargeMoving charge with velocity v produces magnetic field:

2

ˆx4

o qr

µπ

=v rB

:r̂r̂

P

unit vector directed from q to P

permeability of free space70 4 10 T m/Aµ π −= × ⋅

19P15-

The Biot-Savart LawCurrent element of length ds carrying current I produces a magnetic field:

20 ˆ

4 rdI rsBd ×

=π

µ

(http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/magnetostatics/03-CurrentElement3d/03-cElement320.html)

Figure 28-1

Figure 28-3

Figure 28-5

Figure 28-15

Figure 28-16

Figure 28-19

Figure 28-17

20P15-

The Right-Hand Rule #2

ˆ ˆˆ × =z ρ φ

23P15-

Example : Coil of Radius RConsider a coil with radius R and current I

II

IP

Find the magnetic field B at the center (P)

25P15-

Example : Coil of Radius RIn the circular part of the coil…

ˆd ⊥s r ˆ| d | ds→ × =s r

r̂sd

II

I0

2

ˆ4

dIdBr

µπ

×=

s rBiot-Savart:

024

I dsr

µπ

=

024

I R dR

µ θπ

=

0

4I d

Rµ θ

π=

24P15-

Example : Coil of Radius RConsider a coil with radius R and current I

II

IP

1) Think about it:• Legs contribute nothing

I parallel to r• Ring makes field into page

2) Choose a ds3) Pick your coordinates4) Write Biot-Savart

26P15-

Example : Coil of Radius RConsider a coil with radius R and current I

0

4I ddB

Rµ θ

π=

20

0 4I dB dB

R

π µ θπ

= =∫ ∫

( )2

0 0

0

24 4

I IdR R

πµ µθ ππ π

= =∫

II

Iθ

sd

0 into page2

IR

µ=B

27P15-

Example : Coil of Radius R

Notes:•This is an EASY Biot-Savart problem:

• No vectors involved•This is what I would expect on exam

II

PI

page into2

0

RIµ

=B

29P15-

Group Problem:B Field from Coil of Radius R

Consider a coil with radius R and carrying a current I

WARNING:This is much harder than what I just did! Why??

What is B at point P?

Figure 28-39

Figure 28-37

Figure 28-38

Figure 28-43

Figure 28-42

2P18-

Review:Right Hand Rules

1. Torque: Thumb = torque, fingers show rotation2. Feel: Thumb = I, Fingers = B, Palm = F3. Create: Thumb = I, Fingers (curl) = B4. Moment: Fingers (curl) = I, Thumb = Moment

20P18-

The Biot-Savart LawCurrent element of length ds carrying current Iproduces a magnetic field:

20 ˆ

4 rdI rsBd ×

=πµ

23P18-

Ampere’s Law: The IdeaIn order to have a B field around a loop, there must be current punching through the loop

24P18-

Ampere’s Law: The Equation

The line integral is around any closed contour bounding an open surface S.

Ienc is current through S:

∫ =⋅ encId 0µsB

encS

I d= ⋅∫ J A

Figure 28-6

Figure 28-7

26P18-

Biot-Savart vs. Ampere

Biot-Savart Law

general current sourceex: finite wire

wire loop

Ampere’s law

symmetriccurrent source

ex: infinite wireinfinite current sheet

02

ˆ4I d

rµπ

×= ∫

s rB

∫ =⋅ encId 0µsB

27P18-

Applying Ampere’s Law1. Identify regions in which to calculate B field

Get B direction by right hand rule2. Choose Amperian Loops S: Symmetry

B is 0 or constant on the loop!3. Calculate 4. Calculate current enclosed by loop S5. Apply Ampere’s Law to solve for B

∫ ⋅ sB d

∫ =⋅ encId 0µsB

28P18-

Always True, Occasionally Useful

Like Gauss’s Law, Ampere’s Law is always true

However, it is only useful for calculation in certain specific situations, involving highly symmetric currents. Here are examples…

Figure 28-10

29P18-

Example: Infinite Wire

I A cylindrical conductor has radius R and a uniform current density with total current I

Find B everywhere

Two regions:(1) outside wire (r ≥ R)(2) inside wire (r < R)

30P18-

Ampere’s Law Example:Infinite Wire

I

I

B

Amperian Loop:B is Constant & ParallelI Penetrates

31P18-

Example: Wire of Radius RRegion 1: Outside wire (r ≥ R)

d⋅∫ B s

ckwisecounterclo2

0

rI

πµ

=B

B d= ∫ s ( )2B rπ=

0 encIµ= 0Iµ=

Cylindrical symmetry Amperian CircleB-field counterclockwise

32P18-

Example: Wire of Radius RRegion 2: Inside wire (r < R)

2

0 2

rIR

πµπ⎛ ⎞

= ⎜ ⎟⎝ ⎠

ckwisecounterclo2 2

0

RIr

πµ

=B

Could also say: ( )222 ; rRIJAI

RI

AIJ encenc π

ππ====

d⋅∫ B s B d= ∫ s ( )2B rπ=

0 encIµ=

33P18-

Example: Wire of Radius R

20

2 RIrBin π

µ= r

IBout πµ2

0=

Figure 28-9

34P18-

Group Problem: Non-Uniform Cylindrical Wire

I A cylindrical conductor has radius R and a non-uniform current density with total current:

Find B everywhere

0RJr

=J

35P18-

Applying Ampere’s LawIn Choosing Amperian Loop:• Study & Follow Symmetry• Determine Field Directions First• Think About Where Field is Zero• Loop Must

• Be Parallel to (Constant) Desired Field• Be Perpendicular to Unknown Fields• Or Be Located in Zero Field

39P18-

Multiple Wire Loops

Figure 28-12

Figure 28-14

Figure 28-13

Figure 28-12

40P18-

Multiple Wire Loops –Solenoid

41P18-

Magnetic Field of Solenoid

loosely wound tightly wound

For ideal solenoid, B is uniform inside & zero outside

42P18-

Magnetic Field of Ideal Solenoid

d d d d d⋅ ⋅ + ⋅ + ⋅ + ⋅∫ ∫ ∫ ∫ ∫1 2 3 4

B s = B s B s B s B s

Using Ampere’s law: Think!

along sides 2 and 4

0 along side 3

d⎧ ⊥⎪⎨

=⎪⎩

B sB

n: turn densityencI nlI=

0d Bl nlIµ⋅ = =∫ B s

00

nlIB nIl

µ µ= =/ : # turns/unit lengthn N L=

0 0 0Bl= + + +

44P18-

Group Problem: Current Sheet

A sheet of current (infinite in the y & z directions, of thickness 2d in the x direction) carries a uniform current density:

Find B everywhere

ˆs J=J k

y

45P18-

Ampere’s Law:Infinite Current Sheet

I

Amperian Loops:B is Constant & Parallel OR Perpendicular OR Zero

I Penetrates

B

B

46P18-

Solenoid is Two Current SheetsField outside current sheet should be half of solenoid, with the substitution:

2nI dJ=

This is current per unit length (equivalent of λ, but we don’t have a symbol for it)

47P18-

=2 Current Sheets

Ampere’s Law: .∫ =⋅ encId 0µsB

IB

B

X XX

X

X

XX

X

XXX

X

X

XX

X

XXXXXXXXXXXX

B

LongCircular

Symmetry(Infinite) Current Sheet

Solenoid

Torus

49P18-

Maxwell’s Equations (So Far)

0Ampere's Law:

Currents make curling Magnetic Fields

encC

d Iµ⋅ =∫ B s

Magnetic Gauss's Law: 0

No Magnetic Monopoles! (No diverging B Fields)S

d⋅ =∫∫ B A

0

Gauss's Law:

Electric charges make diverging Electric Fields

in

S

Qdε

⋅ =∫∫ E A