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Chapter 29

Magnetic Field Sources

Prof. Raymond Lee,revised 3-2-2012

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• B-field of a wire

• Use a compass to detect B-field

• If no I in wire, then no B-field dueto current & so ...

• all compass needles point towardEarth’s N pole in response to itsB-field

(SJ 2008 Fig. 30.10, p. 845)

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• B-field of a wire, 2

• But if wire carries strong I,then compass needlesdeflect tangent to circle

• Thus showing direction ofB-field produced by wire

(SJ 2008 Fig. 30.10, p. 845)

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• B-field of a wire, 3

• Tangential alignment of

iron filings shows circular

B-field produced by

current-carrying wire

(compare Fig. 29-3, p. 765)

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• Biot-Savart law

• Biot (bee-oh) & Savart (sav-arr) measured F

exerted by I on nearby magnet

• Deduced equation ! B-field at a point in space

due to I

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• Biot-Savart law, 2

• At point P, B-field is dB due to I

• Length element is ds & wire

carries steady current = I

• dB " both ds & unit vector

that points ds ! P

(compare Fig. 29-1, p. 764)

^r

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• Biot-Savart law, 3

• |dB| is proportional to: (1) 1/r

2, where r = ds!P distance

(2) current I

(3) |ds| of length element ds

(4) sin(#), for # = $ between ds &

• ! Biot-Savart law: dB = µ0 I ds x /(4%r 2)

• Resulting B-field is due to the current-carrying

conductor, not any external source

^r(Eq. 29-3,p. 765)

^r

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• Biot-Savart law, 4

• Constant µ0 is permeability of free space, where

µ0 = 4% x 10-7 Tm/A (Eq. 29-2, p. 765)

• dB is differential field from I in length segment ds

• To get total B-field, must add contributions from

all current elements Ids:

where integral is over entire I distribution

(SJ 2008 Eq. 30.3, p. 838)

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• Biot-Savart law, 5

• Law also valid for I consisting of charges flowing

in space sans conductor

• Then ds = length of small segment of space in

which charges flow (e.g., CRT’s electron beam)

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• B compared to E

• Distance

• For distance r from I source, |B| & 1/r2

• For a point charge, also have |E| & 1/r2 from

charge

• Direction

• E-field from point charge is radial in direction

• B-field created by current element Ids " both

length element ds & unit vector ^r

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• Source

• E-field can exist for 1 electric charge

• But current element Ids that yields B-field must

be part of extended I distribution, so we must

integrate over entire I distribution

• Ends

• B-field lines have no start or end, but instead form

continuous circles

• E-field lines begin on +charges & end on –charges

• B compared to E, 2

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• |B| for long, straight conductor

• Thin, straight wire carries

constant current I

• Integrating over all the

current elements gives

(see pp. 766-767)

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• |B| for long straight conductor:

Special case

• If conductor is '-long straight

wire, then #1

= 0 & #2

= %

• Then |B| is B = µ0I/(2%a)(Eq. 29-6, p. 767)

(SJ 2004 Fig. 30.3, p. 929)

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• B direction for long straight

conductor

• B-field lines: (1) are circlesconcentric with wire, &(2) lie in planes " wire

• |B| = constant on any circleof radius a

• Right-hand rule for findingB’s direction is shown

(compare Fig. 29-4, p. 766)

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• B for curved wire segment

• Find B-field at point O due

to wire segment

• ds || in straight segments, so

these add nothing

• I & R are constants, yielding

B = µ0I#/(4%R) (Eq. 29-9, p. 767)

• Note that # is in radians

• If # = 2%, then get B = µ0I 2%/(4%R) =

µ0I/(2R), the field strength at loop’s center

^r

(pp. 767-768)

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• B for a circular current loop

• Loop radius = R & loop

carries steady current = I

• Then |B| at on-axis point P

is only the Bx component:

(SJ 2008 Eq. 30.7, p. 842) (SJ 2008 Ex. 30.3, p. 841)

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• Comparison of loops

• Consider |B| far from origin along x-axis(i.e., x » R)

• Then ,

similar to |E| far from an electric dipole (Eq. 22-9,p. 584)

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• B-field lines for a loop

• (a) shows B-field lines around a current loop

• (b) shows B-field lines in iron filings

• (c) compares loop’s B-field lines to those ofbar magnet

(SJ 2008Fig. 30.7,p. 842)

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• FB between 2 || conductors

• 2 || wires each carry steady

current I

• Field B2 due to I in wire 2

exerts force on wire 1 of

F1 = I1lB2 (Eq. 28-25, p. 750)

• Substituting B2 = µ0 I2 /(2%a)

from Eq. 29-6, we get

F1= µ0 I1I2l /(2%a)(compare Eq. 29-13, p. 770)

(compare Fig. 29-9, p. 770)

F1 = I1L x B2 points down

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• FB between 2 || conductors, 2

• || conductors carrying I in same direction attract

each other

• || conductors carrying I in opposite directions

repel each other

• Write result as magnetic force between wires, FB.

• Per unit length l, this is FB/l = µ0 I1I2/(2%a)(compare Eq. 29-13, p. 770)

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• Ampere & coulomb definitions

• Use F between 2 || wires to define ampere:

If F/l between 2 long || wires 1 m apart that

carry identical Is is 2 x 10-7 N/m, then I ineach wire is defined ( 1 A

• Of course coulomb is defined in terms of

ampere: If a conductor carries steady I = 1 A,

then quantity of charge that traversesconductor’s cross section in 1 s ( 1 C

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• Ampere’s law

• Evaluate dot product B•ds for ds on circular

path defined by compass needles for a long

straight wire

• Ampere’s law sets B•ds line integral aroundclosed path = µ0I, where I is total fixed current

through any surface bounded by that path:

B•ds = µ0I (Eq. 29-14, p. 771)

units: T*m = N/A = (T*m/A)*A = T*m (Eqs. 28-4 & 29-2)

)°

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• Ampere’s law, 2

• Ampere’s law describes creation of B-fields by

any configuration of continuous I, but we

consider only highly symmetric Is

• Put right hand’s thumb in I direction through an

amperian loop that encloses I

• Then fingers curl in integration direction around

loop

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• Straight-wire B-field: Ampere’s law

• Calculate B-field at distance rfrom center of wire carryingsteady current I

• I is uniformly distributedacross wire’s cross section

(compare Fig. 29-13, p. 772)

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• Straight-wire B-field, 2

• Outside wire, r > R:

B•ds = B(2%r) = µ0I, so B = µ0I/(2%r),

meaning that B & 1/r outside wire

• Inside wire, r < R, so we need I´, current

inside amperian loop: B•ds = B(2%r) = µ0I´, where I´ = (r/R)2 I

so that B = µ0Ir/(2%R2) {Eq. 29-20, p. 773},

meaning that B & r inside wire

)°

)°

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• Straight-wire B-field, 3

• B-field & r inside wire, but

B-field & 1/r outside it

• Reassuringly, equations

are equal at r = R, where

(r/R2) = 1/r

(SJ 2008 Fig. 30.14, p. 847)

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• B-field of a toroid

• Torus is an insulating ring

around which wire is wrapped

• Find point B-field at distance r

from toroid’s center for toroid

with N turns of wire

• From Ampere’s law, B•ds

= B(2%r) = µ0NI, so that

B = µ0NI/(2%r)(Eq. 29-24, p. 778)

)°

(SJ 2008 Fig. 30.15, p. 847)

I

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• B-field of infinite sheet

• Assume thin, infinitelylarge sheet with I havinglinear current density Js

• I is in +y direction (i.e., outof screen)

• Js is I per unit lengthalong z-direction &Jsl = total I over length l

(SJ 2004 Fig. 30.15,p. 937)

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• B-field of infinite sheet, 2

• Use a rectangular amperian surface (previous slide)

• Rectangle’s w sides don’t contribute to B-field

(since B•ds = 0), but its 2 l sides (|| to surface) do

• Then from Ampere’s law:

B•ds = µ0I ! 2Bl = µ0Jsl, meaning that

B = µ0Js/2 (SJ 2004 Ex. 30.6, p. 937)

• So here B-field is independent of distance from

current sheet

)°

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• A solenoid is a long wire

wound into a helix

• Can produce ~ uniform B-field

in solenoid’s interior (space

surrounded by wire helix)

• B-field lines in interior are ~

parallel, uniformly distributed,

& close together, indicating

that field is strong & ~ uniform

• B-field of solenoid

(SJ 2008 Fig. 30.16, p. 848)

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• B-field of tightly wound solenoid

• Field distribution similar tothat of bar magnet

• As solenoid length *:• interior field ! more uniform

• exterior field ! weaker

(SJ 2008 Fig. 30.17, p. 848)

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• Ideal solenoid & Ampere’s law

• Real solenoid ~ ideal solenoid if it

has (a) tight turns & (b) length »

turn radius

• Consider rectangle with sides l ||

interior B-field & sides w " field

• Only side l inside solenoid (path

1) contributes to B-field (see

previous slide’s figure)

(SJ 2008 Fig. 30.18, p. 849)

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• Ideal solenoid, 2

• Applying Ampere’s Law gives

• Total I through rectangular path = I through

each turn x N turns, or B•ds = Bl = µ0 NI

• Solving Ampere’s law for B-field gives:

B = µ0NI/l = µ0nI (Eq. 29-23, p. 777), where

n = N/l is # turns/unit length

• Valid only near center of very long solenoid

�

B ! ds = B ! ds = B dspath1

"path1

"" = Bl

)°

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• Magnetic flux +B

• Magnetic flux associatedwith B-field is definedsimilarly to electric flux +E

• Start with area element dAon arbitrarily shaped surface

• B-field in this element is B

• dA is vector " surface, &|dA| = area of dA

(SJ 2008 Fig. 30.19, p. 850)

• Magnetic flux +B = ) B•dA (Eq. 30-1, p. 792)

• +B unit ( weber (Wb) = Tm2 (Eq. 30-3, p. 793)

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• +B through a plane

• Let uniform B-field makeangle # w.r.t. dA for a plane

• Then +B = BA cos(#)(SJ 2008 Eq. 30.19, p. 850)

• Here B-field is || plane & so+B = 0 (i.e., # = 90°)

(SJ 2008 Fig. 30.20, p. 850)

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• +B through a plane, 2

• As before, +B = BA cos(#)

• Now B-field " plane & so +B = BA,

the flux’s maximum value

(SJ 2008 Fig. 30.20, p. 850)

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• Gauss’ law in magnetism

• Unlike E-fields, B-fields don’t begin or end on a

point

• Thus # B-field lines entering a surface =

# B-field lines leaving it

• Gauss’ law in magnetism says that

B•dA = 0 (Eq. 30-1, p. 792))°

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• Earth’s B-field

• Earth’s B-field ~ that from

burying a huge bar magnet

deep in planet’s interior

• Earth’s south magnetic pole

is located near its north

geographic pole

• Earth’s north magnetic pole

is located near its south

geographic pole(compare Fig. 32-8, p. 870)

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• Dip angle of earth’s B-field

• If compass can rotate both vertically &

horizontally, it points into earth’s surface

• $ between horizontal & B-field is dip angle D

• As compass moves farther N, local B-field’s D *

• Compass needle is horizontal at equator, where D = 0°

• Compass needle points straight down at S magnetic

pole, where D = 90°

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• Earth’s B-field, 2

• D = 90° occurs just N of Hudson Bay, Canada

(i.e., location of S magnetic pole)

• Magnetic & geographic poles don’t coincide

• A site’s $ difference between true (i.e., rotational

axis) N & magnetic N is its magnetic declination

• Declination varies with latitude, longitude, & (slowly)

with time

• USNA typically has a declination of ~ 11° W,

meaning a compass here points ~ 11° W of true N

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• U. S. magnetic declinations

(SJ 2008Fig. 30.29,

p. 856)

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• Source of earth’s B-field

• Cannot be due to large masses of permanentlymagnetized materials, since core’s hightemperatures preclude permanent magnetization

• Likeliest source of earth’s B-field is convectioncurrents with ions & electrons in core’s liquidregions

• Also evidence that planet’s B-field is related to itsrotation rate

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• Reversals of earth’s B-field

• Direction of earth’s B-field reverses

every few million years

• Evidence of such reversals is found in

basalts resulting from volcanic activity

• Reversals’ origins are not well understood,

but are likely related to changes in core’s

convection currents