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ElectromagnetismElectromagnetism

University of TwenteDepartment Applied Physics

First-year course on

Part II: Magnetism: slidesPart II: Magnetism: slides

© F.F.M. de Mul

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Magnetic Force on Current WireMagnetic Force on Current WireeTdS

dl=dl.eT | j | = dI/dS j = n q v (n = part./m3)

Lorentz force on one charge: F = q v x B….. per unit of volume : f = nq v x B = j x B ….. on volume dV : dF = j x B. dV dF = j x B. dS.dl

j = j.eT dF = j.dS. eT x B . dl

dF = I . dl x B

Suppose: total current = I ;cross section S variable

B

F = I.eT x B ∫ dl = F = I. L. eT x B

Straight conductor in homogeneous field:

B

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Hall effectHall effect

G H

P

QB

I I

d

la

Suppose n charge carriers / m3

F

v

B-field causes deviation of path of charge carriers

Stationary case: Fmagn = Felec ⇒ q v B = q EHall

Build up of electric field EHall between Q and P: Q+ ; P-

With j = nqv = I/(ad) ⇒nqad

IBvBEHall ==

nqa

IBdEVVV HallPQHall ==−= . :potential Hall

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I

Magnetic field of a line currentMagnetic field of a line current

dlr

I2

0

4 :Savart &Biot rT eedB

×=π

µ

PR

Question: Determine B in P

O

dl=dzθ

dzr

IdB

20 sin

4

θπ

µ=

dzzR

R

zR

IdB

22220 1

4 : z var (1)

++==

πµ

θθθ

θπµθ

2220

sinsin

sin

1

4 : var (2)

Rd

R

IdB −== φ

0

2

:Result

eBR

IP π

µ=

er

Approach: Current line elements dl

dl=dl.eT

zr

Integration over z from -∞ to +∞

Integration over θ from 0 to π

Calculation: eT x er = eφ ;tangential component only: dB

dB

eφ

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Magnetic field of a circular circuitMagnetic field of a circular circuit

dlr

I2

0

4 :Savart &Biot rT eedB

×=π

µ

( ) yeB2/322

20

2

.

Ra

aIP +

= µ

Question: Determine B in P

PR

y

Il

a

Approach: Current line elements dlr

er

dl

dl = dl. eT

22220 2

1

4 Ra

aa

Ra

IdBy

++= π

πµ

∫=l

y dlr

IdB α

πµ

cos1

4 20

R

ra

dB

P

α αdB

y

Calculation: eT x er = 1 ;symmetry: y- component only:

dB

dB

y

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Magnetic field of a circular solenoidMagnetic field of a circular solenoidRadius: a ; Current ILength: LCoils: N , or per meter: n

Approach: Solenoid = set of circular circuits ;and for each circuit:

( ) yeB2/322

20

2

.

Ra

aIP +

= µ R is distance from circuit to P

Each circuit: strip dy; current dI = n.dy.I

Question: Determine B in P

yP

Result for L → ∞ :

B = µ0 n I ey

Result independent of a, L

P

( )∫ +=

L

aR

RdynI

02/322

20

2

...yeB

µ

O y

yP

L a

dy

y

R=yp-y

R

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1

2

I1

I2

r12

Magnetic forces between currentsMagnetic forces between currentsQuestion: determine F1→2

eT1

eT2

er1

eθ 1

Relations:

TL

rTθθ0

edlBdldF

eeeeB

.;

;2

dlIR

I

=×=

×==π

µ

F1→2

22 .. dlI 21T221 BeF →→ ×= ∫

B1→2 Calculation

121 eB θπµ

12

10

2 r

I=→

1221 FF →→ −== : If 21 LL

)(2 2

12

210r121 eF −=→ L

r

II

πµ

r11T2 eee −=× θ

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Why is the Why is the wirewire moved by Lorentz force ? moved by Lorentz force ?(since inside the wire only the conduction electrons move)

+ + + + + + ++ + + + + + ++ + + + + + +

I I

electron

Conductor:- fixed ion lattice,- conduction electrons

Hall effect: concentration of electrons (- charge) at one side of conductor - - - - - - - -

Lattice ions feel a force FE upwards +FE

Bv

FL

-q

This force (= FL ) is electrical !!

B

Magnetic field B ⊥ plane of drawing

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Ampère’s Law (1)Ampère’s Law (1)

I Long thin straight wire; current I

B

eθr

r

Idl

r

I

ll

ππ

µπ

µθθ 2

2200 =•=• ∫∫ eedlB

Ampere : 0Il

µ=•∫ dlB

r

Question: Determine the “Circulation of B-field”along circle l

∫ •dlBl

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Ampère’s Law (2)Ampère’s Law (2)

I

Beθ

ππ

µθ 2.2

.. 0 rr

IdrB

cc∫∫ ==•dlB

Ampere : 0Il

µ=•∫ dlB

rl

Question: Determine the “Circulation of B-field”along circuit c

∫ •dlB

c

Ic

0 :again and µ=•∫ dlB

Consequences:1. More currents through c add up ;2. Currents outside c do not contribute ;3. Position of current inside c is not important.

rB

dl

=r. dθ

dθ

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ar ≥

ar ≤

I

Cylinder: radius a

∫∫ •=S

I dSj :current

BB-field from a thick wire-field from a thick wire

Options for current:I: at surface II: in volume(suppose: homogeneous)

∫∫∫ •==•S

00 dSjdlB µµ I

l

IrrB 02).( µπ =

2

2

02).(:)(a

rIrrBII

ππµπ =

02).(:)( =rrBI π

r

IrB

πµ2

)( 0=

0)( =rB

20

2

.)(

a

rIrB

πµ=

Use Ampere-circuits (radius r):

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Magnetic Induction of a SolenoidMagnetic Induction of a SolenoidRadius: R ; Current: ILength: L >> RCoils: n per meterI

Result: inside: B = µ0 n I ez

outside: B = 0

ez

ereφ Components: Bz Br Bφ

GB r : Gauss-box G: Φtotal=0

Φtop =Φbottom ⇒ Φwall=0 ⇒ Br=0

Br

cc B φ : Circuit c (radius r):

Ampere: Bφ .2πr=0 ⇒ Bφ =0

Bϕ

1 2 B z : Circuit 2: B(a)=B(b)=0

Circuit 1: Ampere: Bz l=µ0 n I la

b

l

Bz

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Symmetries for Ampere’s LawSymmetries for Ampere’s Law

∫∫∫ •==•S

00 dSjdlB µµ I

l

Wire,∞ long

Plane,∞ extending

Solenoid,∞ long

Toroid,along core line

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Plane layer L with current density j

j

L

Magnetic PressureMagnetic Pressure

Magnetic pressure: P = ½µ0 j2 = ½Be2 /µ0 .

Example: this situation is met at the wall of a (long) solenoid. Then the pressure is outward, thus maximizing the cross section area.

Suppose we add an externalfield Be , with Be = BL , so that the total field behind the layer = 0

Be

j

Lorentz force on : (FL=I.L.eT × B):dFL= (j.db).dl × ½µ0 j ez dFL= ½µ0 j2.db.dl eydl

db

Be

dFL

B-field of the layer (circuit l ): 2 BL l = µ0 j l ⇒ BL = ± ½µ0 j ez

l

z

BL BL

xy

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''

×∇=⇒ ∫∫∫

v

dvr

j

4B 0

πµ

Vectorpotential Vectorpotential AA

r = f (x,y,z, x’,y’,z’ )

∫∫∫∫∫∫∫∫∫ ×

∇=

∇−×=×=

'

0

'

0

'2

0 '.1

4'

1

4'

4 vvv

dvr

μdv

r

μdv

r

μjj

ejB r

πππ

'.)(.'

∫∫∫

×∇−

×∇=⇒−=

v

dvrr

duvuvddvujj

4B 0

πµ

0)',','(;),,( =×∇⇒==∇ jj zyxfzyxf} '

'∫∫∫

×∇=

v

dvr

j

4B 0

πµ

∫

∫∫∫

=

=⇒

loop

v

r

I

dvr

circuit afor 1

and

general,in ''

dl4

A

j

4A

0

0

πµπ

µ

er

P (x,y,z)rdv’Flow

tubev’

x’,y’,z’

B = rot A

Question: determine A in P from Biot-Savart for B

j

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Magnetic Dipole (1): Far FieldMagnetic Dipole (1): Far Field

∫=r

I dlA 0

πµ4

ϕcos'.2'

1122 rrrrr

PP −+=

rP

r

r’

O

erP

dl

P

I ϕ

Far field: r’ << rP

++≈

−

+

= .....cos'

11

cos'

2'

1

112

ϕ

ϕPP

PPP

r

r

r

rr

rr

rr

∫∫ += dldlAP .cos'.44

:Thus 200 ϕ

πµ

πµ

rr

I

r

I

PP

Monopole-term =0

Dipole-term

Goal: expression in rP in stead of all r-values over circuit

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Magnetic Dipole (2): Dipole MomentMagnetic Dipole (2): Dipole Moment

∫= dlAP .cos'.4 2

0 ϕπµ

rr

I

P

Assume: Y-axis along OP’

+−

=•=

0

cos

sin

cos

sin

0

'cos'. αα

θθϕ R

R

r rer

αθ cossinR=

0

.sin4

.cossin4

)cos)(.)(cossin(4

22

0222

0

20

==

=−=

=−=

∫

∫

zy

PP

P

x

AA

Rr

IdR

r

I

dRRr

IA

πθπµααθ

πµ

αααθπµ

r0 emA ×= 24 Prπ

µ

Assume: circular circuit,with radius R << rP

P

P’

y

x

z

θ

αr’

erP

rP

O

ϕ

Define: dipole moment:m = I. Area. en = IπR2.en

en

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Magnetization and PolarizationMagnetization and Polarization

L Sen

Magnet = set of “elementarycircuits” ; n per m3

Total surface current = Itot

Total magnetic moment = Itot S en

Polarization Magnetization

Dipole moment: p [Cm] Dipole moment: m [Am2]

Polarization P = np [Cm-2]= surface charge / m2

Magnetization M = nm [Am-1]= surface current / m

Def.: Magnetization M = magnetic moment / volume = surface current / length

V=SL

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Magnetic Magnetic circuit: Tcircuit: Torqueorque and Energy and Energy

Torque: moment: τ = F . b sin θ = I B l b sin θ = I B S sin θ

Magnetic dipole moment: m = I S en

Torque: moment: τ = m x B

Potential energy: min for θ = 0 ; max for θ = π ⇒

Potential energy: Epot = -MB. cos θ = - m . B

Bm

F

F F

F

m

B

b

l

b.sin θ

θθ

I

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Electret and MagnetElectret and Magnet

E D

H B

Electret P

- - -

+++E = (D-P)/ ε0

D = ε0 E + P

∫∫∫

==•

=•

0

0

fQdSD

dlE

Magnet MH = B / µ0 -MB = µ0 H + M

0

0

=•

==•

∫∫∫

dSB

dlH fI

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B-B- and and H-H-fields at interfacefields at interface

L

A

H

B

L

A

H

B

22

2

11

1

tan

1

tan

1

αα=

2

1

2

1

tan

tan

αα

µµ =

r

r

Given: B1 ; µ1 ; µ2

Question: Calculate B2B1

B2

α1

α2

1

2

Gauss box : B1.Acos α1 - B2.Acos α2 =0 ⇒ B1⊥ = B2⊥

Circuit: : no I : H1.Lsin α1 - H2.Lsin α2 =0 ⇒ H1// = H2//

Needed: “Interface-crossing relations”:

Relation H and B: B = µ0 µr H

0 and =•=•∫ ∫∫ dABdlH freeI

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Suppose: - toroid solenoid: R, L, N, µr ;- air gap, µr =1 , width d <<R; g = gap ; m = metal

ICoreline;radius R;L = 2πR

Toroid with air gapToroid with air gap

Question: Determine Hg in gap

Relations:

0

0

=•

==•

∫∫∫

dAB

HBdlH rfI µµ

Hg d + Hm (L-d) = N IBg =Bm

Bg=µoHg ; Bm =µ0µr Hm

Result:

r

gm

r

rg

HH

dL

NIH

µµµ =

−+= and

)1(

d

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Induction: conductor moves in fieldInduction: conductor moves in fieldB

v

L1(t)

L2 (t+dt )

Suppose: circuit L moves withvelocity v through field BQuestion: Show equivalence:

∫∫∫ •−=•=SL

ind dt

dV dSBdlE

)()()(0 tddtttbottomtopwall

Φ=+Φ+Φ−=−−=⇒=• ∫∫∫∫∫∫∫∫ dSBv.dt

dlθ Area = v dt dl sinθ

∫∫∫∫∫∫ •×−=ו=•=LLwallwall

dtdtdldtv dlBvdlvBeB n )(.).(]sin...[ θ

∫∫ •+=•×−=Φ⇒LLdt

tddlEdlBv n)(

)( En = non-electrostatic field

Gauss box: top lid S1 in L1,bottom lid S2 in L2

S1

S2

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Induction: Faraday’s LawInduction: Faraday’s Law

V0

B

cSI0

Static: 0∫ =•c

dlE

Vind

Dynamic:

B

cSIind

Suppose Φ changes ⇒Vind ≠ 0 ⇒ Non-electrostatic field EN

∫∫∫ •−=Φ−=•=Sc

ind dt

d

dt

dV dSBdlEN

Consequence: Let circuit c shrink, with keeping S.

Sc c

S00 =•⇒=• ∫∫∫

Sc

dSBdlEN

For closed surface :

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Induction in rotating circuit frameInduction in rotating circuit frame

Induction potential difference :

(II) Using flux change:

tlvBtlbB

lbdt

dB

dt

d

dt

dV

extext

extind

ωωω

θ

sin2sin

]cos[

=

=−=•−=Φ−= ∫∫ dSBext

Bext

en

Bext

b

l θθv

v

v

en

ω

θ=ω t ; v=ω ½b

(I) Using Lorentz force:tlvBlvB

V

extext

extind

ωθ sin2sin2

)(

=

=•×=•= ∫ ∫ dlBvdlEN

=EN

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Electromagnetic brakesElectromagnetic brakesWhy is a conducting wire decelerated by a magnetic field?

Case I: switch open

P

Q

vB Electrons feel FLFL

P

Q

vBCase II: switch closed

Potential difference VPQ : P - , Q + (= Hall effect)

++ ++

- - - -FL

vL

FL moves electrons: vL ,

which will act on positive metal ions: Fions.

Fions

FE FL counteracted by FE

I FL2

which causes I and FL2 ,

++ ++ - - - -which causes electric field,

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Coupled circuits (1): Coupled circuits (1): M M and and LL1

2I1

Suppose: circuit 1 with current I1

Part of flux from 1 will pass through 2 : Φ21

Φ21 ~ I1

Definition: Φ21 = M21.I1

M : coefficient of mutual induction : “Mutual Inductance” : [ H ] = [ NA-1m-1.m2.A-1] = [ NmA-2 ]

Suppose: Circuit 2 has current as well: I2

I2

Flux through 2: Φ2= Φ21 + Φ22 = M21 I1 + M22 I2

L : coefficient of self-induction “(Self) Inductance”: M22 = L2

M and L are geometrical functions (shape, orientation, distance etc.)

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Coupled circuits (2): toroidCoupled circuits (2): toroid

Question: determine M21

Flux from 1 through S : ΦS = BS = µ0 µr N1 I1 S / L

Cross section S

N1 , I1 N2

Core line: L

µr

Linked flux from 1 through 2: Φ21 = N2 ΦS = µ0 µr N1 N2 I1 S / L

Coefficient of mutual induction: M21 = µ0 µr N1 N2 S / L

This expression is symmetrical in 1 and 2: M21 = M12

This result is generally valid: Mij = Mj i

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Radii: a and b (a<b)Length: lCurrent: I ; choice: inside = upward

I

Coax cable : Self inductanceCoax cable : Self inductance

Ampere: B-field tangential r

IrB r

πµµ

2)( 0=

Flux through circuit:

a

blIdr

r

Idl r

l b

a

r

S

B ln2

.

20

0

0

πµµ

πµµ ==•=Φ ∫ ∫∫∫ dSB

Self-inductance (coefficient of self-induction): per unit of length:

a

bL r ln

20

πµµ= (Compare with capacity of

coax cable, per meter:

1

0 ln2−

=

a

bC rεπε

Gauss surface

B

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Magnetic Field EnergyMagnetic Field Energy

x

z

y

I

Circuit c in XY-plane

H

H-field lines: circles around circuit.

A

A-surfaces: everywhere⊥ H (or B) -field lines

Φ through A: ∫∫ •=ΦA

dAB

Magnetic energy:Em= ½ L I2 = dv

VA ccA∫∫∫∫∫ ∫∫∫∫ •=••=••= HBdldAHBdlHdAB 2

121

21 ))((

dAdl

Compare: electric energy: Ee = dvv

∫∫∫ • ED21

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Maxwell’s Fix of Ampere’s LawMaxwell’s Fix of Ampere’s Law :Induction ∫∫∫ •−=•

Sc dt

ddSBdlEn Changing B-field

causes E-field

Question: Does a changing E-field cause a B-field ?

j

Suppose: chargeing a capacitor using jL

S1

∫ ∫∫ •=•L S1

0 dSjdlB µL encloses S1 :

S2

∫ ∫∫ •=•L S2

0 dSjdlB µL encloses S2 :

∫∫∫∫∫ •+•=•SSL dt

ddSDdSjdlB µµ :generalIn

L encloses S3 : ∫∫∫ •=•3

0

SL dt

ddSDdlB µ

S3

Volume enclosed by S2 and S3:

∫∫∫∫ •−=−=•−32 S

f

S dt

d

dt

dQdSDdSj

D