9-Magnetic Materials and Forces

8/12/2019 9-Magnetic Materials and Forces

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332:382 Electromagnetic Fields I

Instructor: Wei Jiang

Chap. 9. Magnetic Materials and Forces



2332:382 Electromagnetic Fields (W. Jiang)

9.1 Force on a moving charge

BvQF r

rr

×=

Electric force

Magnetic force

E QF rr

=

)( Bv E QF r

rrr

×+=Total force

Without other forces (no gravity, friction), the magnetic forcewill cause a charged particle to revolve or spiral

v

B

F

Centripetal forceLorentz force equation




9.2 Magnetic Force on a differential currentelement

∫ ×=×=

×==

×=

voldV B J F

dV B J F d

dV BvF d

dV dQ

BvdQF d

rrr

rrr

rr

r

rr

r

ρ

ρ

For a wire of length dL with a cross-section A carrying a current I

B L I F B

Ld B I B L Id F

B L Id F d

L Id dV J

JA I AdLdV

rrrr

rrrrr

rrr

rr

×=

×−=×=×=

===

∫∫ :recurrent wistraightaand field constantFor




9.3 Force between differential current element

2212

12102

12212

2102

202

2222

212

112

12

12

12

ˆ

4

)ˆ(

4

)(

)(

4

ˆ

Ld R

Ld a I I F

a Ld Ld

R

I I F d d

H d Bd Bd Ld I F d d

B L Id F d

R

a Ld I H d

R

R

R

r

r

r

rrr

rr

rrr

rrr

r

r

× ×

=

××=

=×=

×=

×=

∫ ∫π μ

π

μ

μ

π





ExampleTwo infinite parallel filaments with separationd and equal but opposite currents Iexperience a repulsive force F, find F.

(N/m) 2//

lengthunit perForce

ˆ2/)ˆ2/()ˆ(ˆ2/space,freeIn

ˆ2/

210

210122

100

1

d I I IB LF

ad I I Lad I a L I B L I F ad I H B

ad I H

z

π μ

π μ π π μ μ

π

ρ φ

φ

φ

==

=×−=×===

=

rrr

rr

r

H

1. In the figure, the two current are opposite in direction ( I 1=- I 2=I ), therefore the two wires repel each other

2. If the two currents are flowing in the same direction, they

attract each other.





• For two differential current elements, d(dF 2 ) ≠ -d(dF 1 )

N aaF d d

N aaaF d d

aaaa L I

a L I

a Ld Ld R I I F d d a Ld Ld

R I I F d d

z x

z y x

y z y x

y

R R

201

202

26

22

16

11

21221

1201122

12

2102

10)ˆ67.0ˆ67.4()(

10)ˆ67.2ˆ33.0ˆ33.1()(

(2,2,2);PatmAˆ)ˆ3.0ˆ4.0ˆ5.0(103

(1,0,0);PatmAˆ103

:Example

)]ˆ(4

[)( )ˆ(4

)(2112

−

−

−

−

+=

−+−=

++−×=Δ

×=Δ

××−=−≠××=

r

r

r

r

rrrrrr

π μ

π μ

Reason: the non-physical nature of current elements (only closed circuits exist,this imposes an additional constraint).

For closed circuits, total force satisfies: F 1=-F 2




9.4 Force and Torque on a closed circuit,Magnetic dipole moment

)(

)(

)2/(

)2/(

)()2/()()2/(

0

00

4

1

042

031

10

333

0

0

0

111

01

Ba IdxdyT d

a Ba B IdxdyT d T d

a IdxdyBT d T d

a IdxdyBT d T d

T d adxdyB I

F RT d

adxdyB I

Baa Idxdy Ba Idxady

F RT d

Ba IdxF d

z

x y y xi

i

y x

x y

x y

x y

x y

x y

x

rr

r

rrrr

rrr

rrr

rr

rrr

r

rrr

rrr

rrr

rr

r

×=−==

=+

−=+=−=

×=

−=××−=××−=

×=×=

∑=

sadxdyS d

BS Id T d

ˆ=×=

r

rrr

S Id md

Bmd T d r

r

rr

r

=×=

Magnetic dipole moment

(R 1: the mid-point of edge 1)




9.4 Force and Torque on a closed circuit,Magnetic dipole moment• The dipole moment formula applicable to a current

loop of any shape.• Assume the magnetic field is constant in the entire

region that the loop is present, then

BS I BmT rrr

rr

×=×=

mr

B

Non-zero torque.Will have to rotate

mr

B

zero torqueBalance

A free moment will rotate until aligned.

0=× Bmr

r




9.5 Magnetic Materials

• Microscopic origin of magnetic properties– Electron orbital motion, electron spin, nuclear spin(weak)

• Two types of electron orbital effects– Diamagnetic (always exist): If electron orbit can’t freely

“rotate” (inert gases, Ge, Si, Au, S, NaCl…), electron movesslower (if aligned), or faster (if opposite-aligned). Always

reduce the total B (<B 0 ), but this is often a weak effect.– Paramagnetic: if electron orbital can freely rotate (transition

metals, rare earth…), it always tends to produce a field alignedwith B 0 . Total B > B 0

orbit

spin





• Paramagnetic materials:– without external B 0 : electron orbital moments are randomly

aligned, therefore average B internal =0 B ≈B0

– Applied external B 0 : electron orbital moments are aligned withB0 , if this is enough to overpower diamagnetic effect,

No external field Apply external field

K, O, W, rare earth and their salts (ErCl 3, neodymium oxide, yttrium oxide…)

)dipolehavingeachV,in volumeatoms Nmaterial,shomogeneou(for1

1lim

eunit volummomentdipolemagnetic

10

mmV N

mV

mv

M vn

ii

v

rrr

rr

==

Δ==

∑

∑Δ

=→Δ





• Electron spin effect:– Spin-created field generally tends to aligned with external

field ferromagnetic– similar to paramagnetic, but much stronger (up to 10 9 )

• Ferromagnetic materials– Domains in the ferromagnetic material– Occurs below certain temperature (Curie temperature)– Above T c it’s paramagnetic, below T c, it’s ferromagnetic





• Hard drive

•Shrink bit size for more hard

drive capacity.•Too small micron domainsize is problematic• Read/write head<0.1micron is also difficult




9.6 Magnetization and Permeability

m R

R R

mm

H H B

M H B H M

M

χ μ

μ μ μ μ μ

μ χ χ

+=

==+=

=

=

1

ty) permeabilirelative: ty; permeabili:(

)(litysusceptibimagnetic:

volumematerial)/aof momentsmagnetictotal(

0

0rrr

rrr

rr

r

Hysteresis for a ferromagnetic material(Si:Fe)

Hc: coercive fieldBr : remnant field

History-dependent: increasing H anddecreasing H follow different curves.





bT

T

b

J J J

J

B

J M

J H

rrr

r

r

rr

rr

+==×

=×=×

0μ bvT

T

b

v

E

P

D

ρ ρ ρ ρ ε

ρ

ρ

+= =

−==

v

v

r

0

E & B determine the

force (force is due to thetotal charge/current)

B is usually considered a fundamental quantity (rather than H).





kA/m ˆ1.30

kA/m ˆ7.9

kA/m ˆ1.30)1(

kA/m ˆ7.9ˆ)1041.4/(05.0)/(

(c),(c) ,)( ,(a) :Find

;1.4 , ˆ05.0:Given

2r

7r 0

r

yb

y

z

z z

b

z

a M J

a H J

a x H M

a xa x B H

J J M b H

a x B

−=×=

−=×==−=

=××==

==

−

rr

rr

rr

rr

rrrr

r

μ

π μ μ

μ

Example




9.7 Magnetic boundary conditions

0

0

21 =Δ−Δ

=∫S BS B

S d B

nn

rr

21 nn B B =

LK L H L H I Ld H

t t Δ=Δ−Δ=∫

21

rr

K H H t t =− 21 K a H H N

rrr

=×−12

ˆ)( 21or

K is the surface current density (unit: A/m).If K=0 (an interface between two dielectrics), then H t1=H t2




9.9 Magnetic potential energy

∫∫∫∫

===

=

volvolvol H

vol E

dv B

dv H dv H BW

dv E DW

μ μ

22

21

21

212

1

rr

rr

Advanced problem :Think: how to apply this equation tocalculate the energy used to magnetizea ferromagnetic material?




9.10 Inductance and mutual inductance

=

=

toroid)(outside ,0

toroid)(inside ,ˆ2

Toroid

*solenoid)theinside(well ˆ

Solenoid

φ πρ a

NI H

ad

NI H z

r

r

*When the field point is more than 2 δ from the solenoidinner surface and when it is more than 2 δ from the end ofthe solenoid.

Magnetic field in Solenoids & ToroidsRefer to Sec. 8.2




9.10 Inductance and mutual inductanceR C, … now L

Φ: the total magnetic fluxFlux linkage of a N-turn toroid: N ΦInductance is defined as

0

20

0

0

0

2

2

2

toroid aFor:Example

πρ μ

πρ μ

πρ μ

φ

S N L

NIS

NI B

I N

L

=

=Φ

=

Φ=

Advanced topic:Partial flux linkage




9.10 Inductance and mutual inductanceExample:Coaxial cable

H/m) :unit length,unit pere(inductanc ln2d

H):(unit ln2

ln2

]1

[2

][

2

2

0

0

0

00

0

0

ab L

abd

I L

a

b Id

dzd I

dzd B

I B

I H

d b

a

d b

a

π μ

π μ

π

μ

ρ ρ π

μ ρ

πρ μ πρ

φ

φ

φ

=

=Φ=

=

=

=Φ

=

=

∫ ∫∫ ∫ Internal inductance:

- Due to B inside conductors- For a long, straight wire with circularcross-section and uniform J

La ,int= μ/(8π) H/m(μ ≈μ0 for non-ferromagnetic metals)*insignificant at high frequencies)

Only considered theregion between the

conductor

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9-Magnetic Materials and Forces