Lesson 13 Inductance, Magnetic energy /force /torquesdyang/Courses/EM/Lesson13_Slides.pdf · 2 with self-inductance L 2, mutual inductance L 21. If we maintain i 1 =I 1, while i 2

Lesson 13Inductance, Magnetic energy /force /torque

楊尚達 Shang-Da YangInstitute of Photonics TechnologiesDepartment of Electrical EngineeringNational Tsing Hua University, Taiwan

Outline

InductanceMagnetic energyMagnetic forceMagnetic torque

Sec. 13-1Inductance

1. Self & mutual inductances2. Evaluation procedures

Definition-1

Closed loop C1 carrying current I1 will create

⇒ flux: flux linkage:

1Bv

,1 111 ∫ ⋅=Φ

SsdB vv

11111 Φ=Λ N

∫×′

=' 2

0

4CR

RaldIBvv

v

πμIf , by11 rII =′

,11 BrBvv

=′⇒ ,11 1111

Φ=⋅′=Φ′ ∫ rsdBS

vv

1111 Λ=Λ′ r

⇒ “Self-inductance”of the loop C1: 1

1111 I

L Λ=

Only depend on geometry

1Bv

Definition-2

In the presence of another loop C2, will pass through C2, ⇒ mutual flux linkage:

where

1Bv

12212 Φ=Λ N

1 1122

IsdBS

∝⋅=Φ ∫vv

1

1212 I

LΛ

=

⇒ “Mutual-inductance”between the 2 loops:

Depend on geometry & material.

1Bv

1Bv

Comment

,),(4

)('

0 ∫ ′′

=C rrR

ldIrA vv

vvv

πμ

∫=⇒1

11101 4 C R

ldINAv

v

πμ

∫ ∫⋅

=1 2

21210

4 C C RldldNNvv

πμ

( ) ∫∫ ⋅=⋅×∇=Φ

=22 21

1

2 1

1

2

1

12212 CS

ldAINsdA

IN

INL

vvvv1Bv

1Av

R∫ ∫

⋅=

2 1 21210

21 4 C C RldldNNLvv

πμ

2112 LL =⇒

1. Assume current I flowing on the loop.

Evaluation of inductance (Method 1)

2. Find by Ampere’s law or Biot-Savart law:Bv

3. Find by( )I∝Λ

4. Find L by , independent of I

∫×′

=' 2

0

4CR

RaldIBvv

v

πμ,

IldH

C=⋅∫

vv

∫ ⋅=ΛS

sdBN

vv

IL Λ=

Evaluation of inductance-reference figure

R

∫×

=1 2

101 4C

R

RaldIBvv

v

πμ

1ldv

Rav

Evaluation of inductance (Method 2)


2. Find and by Method 1Hv

3. Find the stored energy

4. Find L by

Bv

( ) 2

21 IdvBHW

Vm ∝⋅= ∫ ′

vv

2

21 LIWm =

Example 13-1: Solenoid inductor (1)

Consider a hollow solenoid with cross-sectional area S, n turns per unit length. Find the inductance per unit length L.


2. By Ampere’s law: nIB 0μ=

Example 13-1: Solenoid inductor (2)

3. For unit length (l=1),

4. By definition:

( ) SnInn ⋅⋅=Φ⋅=Λ 0μ

( ) SnI

SnInI

L 020 μμ

==Λ

=

Example 13-2: Two concentric coils (1)

Consider two coils C1, C2 with N1, N2 turns and lengths l1, l2. They are wound concentrically on a thin cylindrical core of radius a with permeability μ. Find the mutual inductance L12.

1. Assume C1, C2 have currents I1, I2

Example 13-2: Two concentric coils (2)

3. Flux linkage of C2 due to C1:

4. By definition:

2. By Ampere’s law, uniform field 11

11 I

lN

B μ=

SBNN ⋅⋅=Φ⋅=Λ 1212212

2

1

2112 a

lNNL πμ=

Sec. 13-2Magnetic Energy

1. Energy of assembling current loops2. Energy of magnetic fields

Energy of assembling current loops-One loop (1)

Closed loop C1 with self-inductance L1. If the loop current i1 increases from 0 to I1 slowly (quasi-static), an emf of:

dtdi

Ldt

dv 1

111

1 =Φ

−=

will be induced on C1 to opposethe change of i1 (Faraday’s law, Lenz’s law).

Energy of assembling current loops-One loop (2)

The work done to overcome the induced v1

and enforce the change of i1 is:

211

0 111

0 11

1

0 111 21)()( 1 ILdiiLdti

dtdiLdttitvW

I==== ∫∫∫

∞∞

which is stored as magnetic energy:

2111 2

1 ILW =

one loop

Energy of assembling current loops-Two loops (1)

Insert loop C2 with self-inductance L2, mutual inductance L21. If we maintain i1=I1, while i2increases from 0 to I2 slowly, an emf of:

will be induced on C1 in an attempt to change i1

away from I1

dtdiL

dtdv 2

2121

21 =Φ

−=


The work done to maintain i1 = I1 is:

2121

0 2121

0 12

21

0 121212)( IILdiILdtI

dtdiLdtItvW

I==== ∫∫∫

∞∞


Meanwhile, an emf of:

will be induced on C2 to oppose the change of i2 (from 0 to I2).

dtdiL

dtdv 2

222

2 =Φ

−=

The work done to overcome v2 and enforce the change of i2 is:

22222 2

1 ILW =


The total magnetic energy stored in the system of two current loops is:

2222121

2112 2

121 ILIILILW ++=

two loops

Energy of assembling current loops-N loops

The total magnetic energy stored in the system of N current loops carrying currents I1, I2, ….., IN , is:

∑∑= =

=N

j

N

kkjjkm IILW

1 121

By , the flux (linkage) of loop Ck due to all the N current loops:

,1∑=

=ΦN

jjjkk IL

11212 IL Λ=

∑=

Φ=N

kkkm IW

121⇒

Energy of continuous current distributions-1

Decompose a system of continuous current distribution in a volume V' into N elementary current loops Ck , each has current ΔIk and filamentary cross-sectional area Δak

)(rJ vv

,Bv

Av

∫∫⋅=

⋅=Φ

k

k

C k

Sk

ldA

sdB

vv

vv

Energy of continuous current distributions-2

∑ ∫∑==

⋅Δ=Φ=N

kC kk

N

kkkm

k

ldAIIW1

1 2

121 vv

( ) ( ) kkkkkkk vJldaJldaJldI Δ=Δ=Δ=⋅Δvvvvvv

,21

1 ∑∫

=

Δ⋅=⇒N

kC km

k

vJAWvv

( )∫ ′⋅=

Vm dvJAW 2

1 vv

Comments

( )∫ ′⋅=

Vm dvJAW 2

1 vv

∑=

=N

kkke VQW

121 ∑

=

Φ=N

kkkm IW

121

( )∫ ′=

Ve dvVW 2

1 ρ

Electrostatics Magnetostatics

Energy of magnetic fields-1

In real applications (especially electromagnetic waves), sources are usually far away from the region of interest, only the fields are given

)(rJ vv ∞→RBHvv

,

ROIsource

( ) ( )∫∫ ′′×⋅∇−⋅=

VVm dvHAdvBHW 2

121 vvvv

( ) ( )∫∫ ′′×∇⋅=⋅=

VVm dvHAdvJAW 2

121 vvvv


)(rJ vv

V ′By vector identity:

(1)

(2)

contain all the source currents

HJvv

×∇=

( ) ( ) ( )( )HABH

HAAHHAvvvv

vvvvvv

×∇⋅−⋅=

×∇⋅−×∇⋅=×⋅∇

( ) ( )∫∫ ′′⋅×−⋅=

SVm sdHAdvBHW 2

121 vvvvv


( )∫∫ ⋅∇=⋅VS

dvAsdA

,vvv

Q

S ′

1I 2I

(3)

( ) ( )∫∫ ′′⋅×=×⋅∇⇒

SVsdHAdvHA

vvvvv

)(rJ vv

V ′


∞→RS ′

S ′

Obs. pt. 2

1R

H ∝v

RA 1∝

v


)mJ( 21 3BH

vv⋅ …energy density

( )

0111

4)()(21

21

22

2

2

→∝⋅⋅∝

⋅≈⋅×= ∫ ′

RR

RR

RRHRAsdHAIS

πvvvvv

∫ ′==⇒

V mm dvrwIW 1 )(v


∞→RV ′

∫ ′⎟⎠⎞

⎜⎝⎛ ⋅=

Vm dvBHW 2

1 vv

dvBHdWm ⎟⎠⎞

⎜⎝⎛ ⋅=

vv

21

Example 13-3: Coaxial cable inductor (1)

Cylindrical symmetry, Ampere’s law, ⇒

Find the stored magnetostatic energy and inductance per unit length of:

0μBHvv

=

⎪⎪⎩

⎪⎪⎨

⎧

<<

<=

brarIa

arraIa

B if ,

2

if ,2

0

20

πμπμ

φ

φ

v

vv


Energy density:

⎪⎪⎩

⎪⎪⎨

⎧

<<

<=⋅=

brarI

arraI

BHwm

,8

,8

21

22

20

242

20

πμπμ

vv

drrdv ⋅= π2Differential volume (L=1):


Total stored energy:

arIdrraIW

a

m <== ∫ ,164

20

0

34

20

1 πμ

πμ

braabIdr

rIW

b

am <<⎟⎠⎞

⎜⎝⎛== ∫ ,ln

41

4

20

20

2 πμ

πμ

,21 2LIWm = ⎟

⎠⎞

⎜⎝⎛+=

+=

ab

IWW

L mm ln28

)(2 002

21

πμ

πμ

internal external

Sec. 13-3Magnetic Force

1. Force on current loops2. Example: force between parallel wires

Force on current-carrying loops-1

Consider an elemental current-carrying wire of cross-sectional area S, represented by a differential displacement vector ld

v

Free charges within the wire of charge density ρmove with velocity , experiencing a force of:

( )lduvv //

( )BuldSFd m

vvvv×= ρ


lduuldvvvv =

uJ vvρ=

( )( )BldJSBlduS

BuldSFd mvvvvv

vvvv

×=×=

×=

ρ

ρ

( )BldIFd m

vvv×=⇒

∫ ×=Cm BldIF

vvv

For a current loop C :


If is created by another closed loop C2

carrying a current I2, the force exerted on the loop C1 carrying a current I1 is:

Bv

∫ ×=1 211121

CBldIFvvv

∫×

=2

21

221

22021 4 C

R

RaldI

Bvv

v

πμ

∫×′

=' 2

0

4CR

RaldIBvv

v

πμ


( )12 2

21

212121021

1 24F

RaldldII

FC C

Rvvvv

v−=

××= ∫ ∫πμ

Counterpart in electrostatics: Coulomb’s force between two charges

212

21

012 4

112 R

qqaF R πεvv

=

Example 13-4: Force between two long wires

Find the force per unit length between two infinitely long, parallel wires separated by d, carrying currents I1, I2 in the same direction.

,2

1012 d

IaB x πμvv

−= ∫ ×=1

0 122212 BldIFvvv

( )

dIIa

dIadzaI

y

xz

πμ

πμ

2

2

210

1

0 10

2

v

vv

−=

⎟⎠⎞

⎜⎝⎛−×= ∫

…attraction force

Sec. 13-4Magnetic Torque

1. Example: magnetic force & torque exerted on a current loop

Example 13-5: Force & torque on current-carrying loops (1)

Consider a circular loop on the xy-plane with radius b, current I in clockwise sense, and is placed a “uniform” magnetic filed:

||BBBvvv

+= ⊥

⊥− Bazv

||Bayv


The force exerted on a differential current element on the loop due to :⊥B

vφφbdald vv

−=

( ) ( )φ

φφ

dIbBaBabdaIFd

r

z

⊥

⊥⊥

=

−×−=v

vvv

( ) ⇒×= ,BldIFd m

vvv

bφ


The force exerted on a differential current element on the loop due toφφbdald vv

−= :||Bv

( ) ⇒×= ,BldIFd m

vvv

( ) ( )( ) ( )

φφ

φφφ

φφ

dIbBa

aaadIbB

BabdaIFd

z

yyx

y

sin

cossin

||

||

||||

v

vvv

vvv

=

×−=

×−=


The total force exerted on the loop due to :⊥Bv

⇒= ⊥⊥ ,φdIbBaFd rvv

φ

0)( 2

0

2

0

=⎥⎦⎤

⎢⎣⎡=

=

∫

∫⊥

⊥⊥

π

π

φφ daIbB

FdF

rv

vv


The total force exerted on the loop due to :||Bv

⇒= ,sin|||| φφdIbBaFd zvv

0sin 2

0 ||

2

0 ||||

=⎟⎠⎞⎜

⎝⎛ ⋅=

=

∫

∫π

π

φφ dIbBa

FdF

zv

vv


The total torque exerted on the loop due to :⊥Bv

⇒= ⊥⊥ ,φdIbBaFd rvv

φ ( )

02

0

2

2

0

=⎟⎠⎞⎜

⎝⎛ ×−=

−×=

∫

∫⊥

⊥⊥

π

π

rr

r

aaBIb

baFdT

vv

vvv


The total torque exerted on the loop due to :||Bv

⇒= ,sin|||| φφdIbBaFd zvv

( )[ ]

( )∫∫

−=

−×=π

φ

π

φφ

φφ2

0 ||2

2

0 ||2

sin

sin

daBIb

daaBIb rz

v

vv

( )∫ −×=π2

0 |||| baFdT rvvv

φφφ cossin yx aaa vvv −=−


( ) ||||2

2

0

2

0

2||

2|| cossinsin

mBaBbIa

dadaBIbT

xx

yx

vv

vvv

==

⎥⎦⎤

⎢⎣⎡ ⋅⋅−⋅= ∫∫

π

φφφφφππ

yzx BmaTTTT )(|||| −==+= ⊥vvvvv

In general, ⇒

BmTvvv

×=

22sin φ

22cos1 φ−

Documents

Lesson 13 Inductance, Magnetic energy /force /torquesdyang/Courses/EM/Lesson13_Slides.pdf · 2 with self-inductance L 2, mutual inductance L 21. If we maintain i 1 =I 1, while i 2