[PPT] - = = + = W H dB H dH H dJ H dJ i H ( t ) dB/dt PowerPoint Presentation

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MAGNETIC LOSSES: QUESTIONS & PROBLEMS 1) Electrotecnical derivation: demonstrate that the magnetic energy loss per cycle is

dJ H dJ H dH H dB H W

∫

∫ ∫ ∫

= + = = µ

Rs Rw1 RH uG iH uL u2 i2 u1 uH iH

N2 N1

1 10 100 1000 10000 0.0 0.1 0.2 0.3 0.4

Fe-Ni 30.5 DC curve

B (T) H (A/m)

iH (t) ∝ dB/dt

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U E dt dt dB t i A N dt t Ri t E

H t t H

+ = + = ∫

∫

R 1 2

) ( ) ( ) (

∫

=

) ( ) ( ) (

t H G

dt t i t u t E

t = 0 B = 0; t = t0 B = Bp

dt d t Ri t u

H G

/ ) ( ) (

Φ + =

∫ ∫

= =

p

) ( ) (

t B HdB

V dt dt dB t H V t U

) ( ) / ( ) (

1 m

t H N l t iH

=

V Al

=

m

Energy delivered to the magnetic system

dJ H dH H dB H

∫

∫ ∫

+ = µ = =

∫ ∫

dt dH H dH H

T

µ

µ

dJ H W

∫

=

Energy loss per cycle per unit volume

J H B

+ = µ

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2) Find the energy stored in the gap of a permanent magnet (the exploitable energy during magnet operation) lm lg Jm Hm B Hg

( )V

BH Eg 2 1

− =

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The whole magnetostatic energy associated with the magnet (energy of formation)

V J N dV E

d

V

ms 2

2 1 2 1

µ = ⋅ − =

∫

J H

The sum of the energy stored inside the magnet and the energy stored

utside it, possibly confined to a good

extent inside the gap.

J N

d d

µ − = = H

H

Energy stored in the material

V J N dV H E

d

V
m

2 2 2

2 1 2 1

µ µ = =

∫

( )V

N N J E E E

d d

m

ms g 2 2

2 1

− = − = µ

( )V

BH Eg 2 1

− =

Energy stored in the gap

H J B

µ

+ =

( )

V H J H V J N J J N E

d d

g

) ( 2 1 2 1

µ µ + − = − =

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lm lg Jm Hm B Hg

m m g g

l H l H

− =

S B S H

m m g g

= Φ = = Φ µ

∫

= ⋅

dl H

g 2 g

2 1 V H E

g

µ

=

Energy inside the gap

S B l H S H l H V H E

m m g g g g 2 g g

2 1 2 1 2 1

µ µ µ µ ⋅ − = ⋅ = =

( )V

BH Eg 2 1

− =

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2) Find the hysteresis loss component versus Hp and Jp in the Rayleigh region. Find the rotational contribution to the initial permeability a for isotropic distribution of easy axis (small angle rotations).

) ( 2 ) ( ) (

2 2

H H b H bH a H J

p p

− + =

 ∫

=

dH H J H W ) ( ) (

2 p p p

bH aH J

+ =

400
200

200 400

0.05

0.00 0.05

(Hp, Jp)

Ni J (T)

H (A/m)

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2a) Find the hysteresis loss component in the Rayleigh region.

) ( 2 ) ( ) (

2 2

H H b H bH a H J

p p

− + =



∫

=

dH H J H W ) ( ) (

The linear terms can be omitted.

) 2 2 2 2 ) (

p p p p p p p p

2 2 p 2 2 p

∫ ∫ ∫ ∫

− − − −

− + − =

H H H H H H H H

dH H b dH H b dH H b dH H b H W

3 p

3 4 ) ( bH H W

=

2 p p p

bH aH J

+ =

400
200

200 400

0.05

0.00 0.05

(Hp, Jp)

Ni J (T)

H (A/m)

[ ]

3 2 2 p

4 6 1 ) (

p

bJ a a b J W

+ + − = 2

p p

bH aH

> >

3 3

3 1

p

J a b W = 2 / 3

3 4

p

J b W =

2 p p

bH aH

< <

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ϕ Ms H θ K1

ϕ ϕ θ µ

2 1 s

sin ) (cos( K H M E

+ − − =

2b) Find the initial permeability associated with rotations. Minimize the energy and find the equilibrium angle ϕ

1 s s

2 cos sin sin K H M H M

+ ≅ θ µ θ µ ϕ

Small oscillations cosϕ ≅ 1

1 s

2 sin sin K H M

θ µ ϕ ≅ ϕ θ sin

sin

s

M M ≅

∆

1 s s

2 sin sin K H M H M H M

θ µ θ χ ≅ ∆ = θ µ θ χ

2 1 2 s

sin 2 ) ( K M

≅

Small oscillations

1 s

2 cos K H M

< < θ µ

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Uniform space distribution of easy directions.

π θ θ π

2 sin 2 d N dN

= θ θ θ θ

d d p sin ) (

=

1 ) (

2 /

=

∫

π

θ θ d

p

θ µ θ χ

2 1 2 s

sin 2 ) ( K M

≅ θ θ θ θ θ χ

π π

d H M d p H M

K s K s

∫ ∫

= > ≅ <

2 / 3 2 / 2

sin ) ( sin

K s

H M 3 2

rot > ≅

< χ

Fe-(3 wt%)Si, Ms = .6 ⋅106 A/m, K1= 38⋅103 J/m3, HK= 38.4⋅103 A/m

<χrot> ≅ 28.

Mn-Zn ferrite, Ms = 3.9 ⋅105 A/m , K1= 50 J/m3, HK= 205 A/m, <χrot> ≅ 1270. Examples

s K

M K H

1

2

µ > = <

θ θ θ θ

cos cos ) 3 / 1 ( sin

3 3

− =

∫

d

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3) Find: 1) The complex permeability in a linear material. 2) The relationship with the energy loss. 3) The related Q factor. 4) The equivalent R-L circuit of a ring inductor.

t H t H

ω

cos ) (

p

=

( )

t B t B t B t B

ω δ ω δ δ ω

sin sin cos cos cos ) (

p p p

+ = = − =

in-phase

ut-of-phase

       = ′ ′ = ′ δ µ δ µ

sin cos

p p p p

H B H B

The 90°-delayed component of the induction is connected with the dissipation of energy. From the definition of the energy loss per cycle W, we obtain that the power loss per unit volume is

δ π

sin ) ( ) (

p p

B H f dt dt t dB t H f Wf P

T

= ⋅ = =

∫

[W / m3]

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δ π

sin

p pJ

H W =

δ µ

sin

p p

H J

= ′ ′

µ π ′ ′ =

2 p

H W

2 2 2 p

µ µ µ π ′ ′ + ′ ′ ′ =

J W

Energy loss under given peak field value Energy loss under given peak polarization value The quality factor of the inductor is defined as the ratio between the maximum stored energy and the energy dissipated per unit volume during one cycle . [J / m3] [J / m3]

1

1

0.010
0.005

0.000 0.005 0.010

f = 1 MHz

Co67Fe4B14.5Si14.5 Jp = 10 mT

J (T) H (A/m)

R L

E E Q

π

2

=

δ

cos H B E

p p L

2 1

=

δ π

sin

p p R

H B E

=

µ µ ′ ′ ′ =

Q

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The equivalent L - R circuit

Rs Ls I RsI jωLsI δ

     = =

−

) ( p p

) ( ) (

δ ω ω

t j t j

e B t B e H t H

     ⋅ = =

−

NS e B j t V e N l H t I

t j t j ) ( p m p

) ( ) (

δ ω ω

ω      + = + =

) ( ) cos (sin ) ( ) ( ) ( ) (

p p s s

t I j I B NS t V t I L j t I R t V

δ δ ω ω

       = = δ ω ω δ ω

cos sin

p p s p p s

I B NS L I B NS R

iH (t)

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iH (t)

A ring core inductor of cross-sectional area S, mean diameter D, and number of turns N.

       = = δ ω ω δ ω

cos sin

p p s p p s

I B NS L I B NS R

       = ′ ′ = ′ δ µ δ µ

sin cos

p p p p

H B H B

D NI H

π

p p = p p

H N D I

π =

       = = δ π ω ω δ π ω

cos sin

p p 2 s p p 2 s

DH B S N L DH B S N R

       ′ = ′ ′ = µ π ω ω µ π ω

D S N L D S N R

2 s 2 s

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y z

B(t)

dy

a dy dt d dG e dP 2 ) / (

2 2

σ Φ = = Unit length a Instantaneous eddy current power loss per unit length in the infinitesimal strips Sinusoidal induction B(t) = Bp sin ωt

a dy y a t B dP 2 4 sin

2 2 2 2 p 2

σ ω ω ⋅ =

Time average

a B dy ay dP T dP

T 2 p 2 2

1

∫

⋅ = > = < ω σ

d

8 3 1

3 2 p 2 / 2

d B a dP P

d

∫

> = < = ω σ The energy loss per unit volume

f B d daf P W

2 p 2 2

6

σ π = =

The energy loss in the sheet of thickness d Thin lamination, d << width

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By the very same approach (and under the same conditions) we find the classical energy loss in cylindrical samples (sinusoidal induction B(t) = Bp sin ωt ) .

D

r dr

r dr dt d dG e dP

π σ 2

) / (

2 2

Φ = =

f B D W

2 p 2 2

16

σ π =

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Response of a lamination of unit length, thickness d, conductivity σ, permeability µ to a field step. y z

Ha(t)

dy Unit length a d

Heddy(t) M(t)

Thin lamination, d << width

1

1 2 3 4 5 6 0.0 0.2 0.4 0.6 0.8 1.0

H M

M / M0, Ha,norm t / τ

eddy a eff

H H H

− =

dt t dH dt dB t H t B t H t M ) ( ) ( ) ( ) ( ) (

eff eff eff

µ µ χ = = = Ha M We calculate the eddy current counterfield at the center of the lamination. No skin effect.

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The electromotive force on the circuit made of the two strips of thickness dy at the coordinate y and the related eddy current

ya dt dB t e 2 ) (

⋅ − =

) , ( d d 2 2 d ) , ( d

eddy

y t H y a ya dt dB G e y t i

= ⋅ ⋅ − = = σ

8 8 d d ) (

2 eff 2 2 / 2 / eddy eddy

d H d B y y dt dB H t H

d d

⋅ − = ⋅ − = − = =

∫ ∫

σ µ σ σ

  8 8

a 2 eff 2 eff

= − +

H d H d H

µ σ µ σ



= − +

b ax x  ax b z

− =

a z x /  

− =

az t z

− =

d d

t = 0; Heff = 0

) exp( ) ( at b t z

− =

)) 8 exp( 1 ( ) (

2 a eff

t d H t H

µ σ − − =

)) exp( 1 ( ) ( ) (

a eff

τ χ χ

t H t H t M

− − = =

8

2

d

µ σ τ =

µ = 10-3, σ = 107 S, d = 10-6 m, τ ∼ 10-3 s

eddy a eff

H H H

− =

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Relaxation of domain wall motion. Complex susceptibility versus frequency.

H0ejωt

Ms

Ms

t j

e H M t M t M

ω

µ α β

s

2 ) ( ) (

= +



) ( ) (

ϕ ω ϕ ω

ω

− −

= =

t j t j

e M j t M e M t M 

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t j

e H M t M t M j

ω

µ α ω β

s

2 ) ( ) (

= + ω β α µ

ω

j e H M t M

t j

+ =

s

2 ) (

ω β ω α µ ω α µ

ω ω

2 2 s 2 2 s

2 2 ) (

+ − + =

t j t j

e H M j e H M t M

2 1 2 s

/ 1 1 2

ω ω α µ χ + = ′

M

2 1 2 1 s

/ 1 / 2

ω ω ω ω α µ χ + = ′ ′

M

10

3

10

4

10

5

10

6

10

7

500 1000

χ '' χ '

ω 0 = 1 10

9 s

1

ω 1 = 1 10

6 s

1

d.w. relaxation

χ ', χ ''

Frequency (Hz)

1

/ 1 ω

τ =

10 10

1

10

2

10

3

10

4

10

5

10

6

10

7

5000 10000 15000 20000

µ '' µ '

Jp = 10 mT Amorphous ribbon Co67Fe4B14.5Si14.5 d = 5.8 µ m µ 'r, µ ''r,

Frequency (Hz)

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] / ) / ( /

s eff

t M t

∂ ∂ × ⋅ + × − = ∂ ∂

M M H M M

α γ µ

Landau-Lifshitz-Gilbert equation

10

3

10

4

10

5

10

6

10

7

10

8

10

9

500 1000 1500 2000 2500 3000

Jp = 50 mT Jp = 2 mT Jp = 2 mT Jp = 50 mT

N87 Mn-Zn ferrite

Frequency (Hz)

µ ', µ ''

Evidence for resonant precession affecting the permeability behavior versus frequency in soft ferrites.

2 2 H 2 LL 2 2 LL 2 2 H 2 2 LL 2 2 H H J ' ro

4 )] 1 ( [ ] ) 1 ( [ ) (

ω ω α α ω ω α ω ω ω ω ω χ + + − − − =

t

2 2 H 2 LL 2 2 LL 2 2 H 2 LL 2 2 H LL J ' ' ro

4 )] 1 ( [ )] 1 ( [ ) (

ω ω α α ω ω α ω ω α ω ω ω χ + + − + + =

t

10

5

10

6

10

7

20
10

10 20 30 40 50

α LL = 2 α LL = 0.1

Rotational resonance Ku = 8400 J/m

3

χ ', χ ''

Frequency (Hz)

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Bibliography

G. Bertotti, Hysteresis in Magnetism, Academic Press, 1988.
G. Bertotti, IEEE Trans. Magn. 24 (1988) 621.
G. Bertotti, J. Appl. Phys. 57 (1985) 2110.
J.B. Goodenough, IEEE Trans. Magn. 38 (2002) 3398.
F. Fiorillo, C. Appino, M. Pasquale, in The Science of Hysteresis (G. Bertotti and I.

Mayergoyz, eds., Academic Press, 2006), vol III, p.1.

R.H. Pry and C.P. Bean, J. Appl. Phys. 29 (1958) 532.
H.J. Williams, W. Shockley, C. Kittel, Phys. Rev. 80 (1950) 1090.
B.D. Cullity and C.D. Graham, Introduction to magnetic materials, Wiley, 2011.
F. Fiorillo and A. Novikov, IEEE Trans. Magn. 26 (1990) 2904.
G. Bertotti and I. Mayergoyz, eds., The Science of Hysteresis, Academic Press,

2006.