Chapter 7. Steady magnetic field

1

EMLAB

B (Magnetic flux density), H (Magnetic field)

• Magnetic field is generated by moving charges, i.e. current.

• If current changes with time, electric field is generated by time varying magnetic field.

• In chapter 7, we consider only steady state current. In this case, steady magnetic fields are generated and we need consider magnetic field only.

• If a charge moves in a region where magnetic flux density is non-zero, it experiences a force due to the field which is called Lorentz force.

)(

forceLorentz;q

MHB

BvF

• The force exerted on a moving charge is due to B (Magnetic flux density).

• B can be obtained from a magnet or cur-rent flowing coil.

• B due to current flowing coil only is de-fined to be H (magnetic field).

• B due to a permanent magnet is repre-sented by M (magnetization).

2

EMLAB

Biot-Savart law

This law is discovered by Biot and Savart. It enables us to predict magnetic field due to a current seg-ment.

This law is experimentally known. It is the counterpart of Coulomb’s law for electric field.

24

ˆ

R

Idd

Rs

H

'rrR 'r

r

sId

Direction of H-field

3

EMLAB

Current segment

;4

ˆ

;4

ˆ

;4

ˆ

2

2

2

S

S

C

R

d

R

da

R

Id

RJH

RKH

RsHsId

daK

dJ

Biot-Savart law : integral form4

EMLAB

Line current

surface current

Volume current

Magnetic field due to an infinitely long line current

• An infinitely long straight current flowing in the z-axis.

2

Iˆ

0sin2

sin2

Iˆ

dcos2

Iˆ

)/'z(1

/'dz

2

Iˆ

'z

'dz

4

Iˆ

'z

'dzˆ'zˆˆ

4

I

'4

'Id)(

2/

002/32

2/3222/322

'C3

zρz

rr

RsrH

dsec'dz

tan'z 2

odd function

zrρrrrR ˆ'z',ˆ,'

x

z

y

r

5

EMLAB

11

2212

z

z2/32

z

z2/322

z

z2/322

'C3

ztan,

ztansinsin

4

Iˆ

dcos4

Iˆ

)/'z(1

/'dz

4

Iˆ

'z

'dz

4

Iˆ

'z

'dzˆ'zˆˆ

4

I

'4

'Id)(

2

1

2

1

2

1

2

2

zρz

rr

RsrH

dsec'dz

tan'z 2

기함수

x

z

y

2

1

2z

1z

6

EMLAB

Magnetic field due to a finitely long current filament

x

z

y

)0for(

2ˆ

2ˆ'

4ˆ

]ˆ)'sinˆ'cosˆ(['

4

1

'

)ˆ''ˆ(''

4

1

'

)'ˆ'ˆ('ˆ''

4

1

'4

')(

2/322

22

02/322

2

2

02/322

2

02/322

2

02/322

'3

za

I

za

aId

za

aI

za

ayxzdaI

z

zdI

z

zdI

Id

C

z

zz

z

zρ

ρz

rr

RsrH

'ˆ'',ˆz,' ρrzrrrR

a

z

Magnetic field due to a loop current

• Magnetic field on the z-axis can only be found due to its simple shape. If the receiver’s position is located on the off-axis region, the integral can be evaluated.

7

EMLAB

Calculation of H of a solenoidzρrxrrrR ˆ'z'ˆ'',ˆx,'

)1(S0

S4

'

'dˆ

S0

SˆK

'

'dˆ

4

Kˆ

]'za'cosax2x[

)'cosxa('dz'da

4

Kˆ

]'za'cosax2x[

]'ˆ'zˆ)'cosxa[('dz'da

4

K

)'cosˆˆ)'cosy'sinx(ˆ'ˆ(

]'za'cosax2x[

)'ˆ'zˆaˆ'ˆx('dz'da

4

K

'

)ˆ'z'ˆaˆx('ˆ'dz'da

4

K

'

)ˆ'z'ˆ'ˆx('ˆ'dz'd'

4

K

'4

'ds)(

S2

S2

2

02/3222

2

02/3222

2

02/3222

2

03

2

03

'C3

적분

r

r

rr

aR

r

rz

rr

aRz

z

ρz

zxx

ρzx

rr

zρx

rr

zρx

rr

RKrHzH ˆKin

0extH

K

Surface current density

If a copper wire is wound around a cylinder N times in the length d and current I is flowing through it, it can be approximated by a surface current along direction with a magnitude K = NI/d.

d

a

8

EMLAB

14

sinˆˆ

4

')ˆ(ˆ

'4

'ˆ

'near is)2(

)'(0

''4

'ˆ

)'(0

')1(

','4

ˆ

'4

1)(

2

2

22

2

2

2

r

rr

RR

RR

rr

aR

rr

rr

arr

aR

rr

rr

rrRrr

R

rrr

R

ddR

R

dad

ford

dd

for

V

SSVr

(1) If r is outside V, the integral becomes zero in that Laplacian ϕ becomes zero.

S

r

(2) If r is inside V, the integral can be changed into a surface integral over a enclos-ing sphere, which has non-zero value.

integral volume:'r

면적분:'r

r

R'da

r 근방에서의 적분을 계산하기 위해서 r 을 중심으로 하고 반지름이 인 구면에서의 면 적분을 하면 결과는 1 이 나와서 앞 장의 결과가 나온다 .

Calculation of integral 19

EMLAB

Ampere’s law

• Ampere law facilitates calculation of mangetic field like the Gauss law for electric field..

• Unlike Gauss’ law, Ampere’s law is related to line integrals.

• Ampere’s law is discovered experimentally and states that a line integral over a closed path is equal to a current flowing through the closed loop.

• In the left figure, line integrals of H along path a and b is equal to I because the paths enclose cur-rent I completely. But the integral along path c is not equal to I because it does not encloses com-pletely the current I.

path closed

IdsH

10

EMLAB

Example- Coaxial cable

I

I

a b cI

I

002

)4(

2ˆˆ

1

ˆˆ

ˆ2

)3(2

ˆˆ2

)2(2

ˆˆ

ˆ

2ˆ

)1(

22

22

22

22

0

2

0

2

0

0

2

0

2

2

2

0

2

0

2

0

H

H

zJzJ

zJ

H

H

zJaJ

Hs

rH

cr

bc

rc

r

IH

Ibc

br

dddd

ddrH

crbr

IHIrH

braa

rIH

Ia

rddd

rHrddH

ar

a r

b

outin

r

in

r

in

S

C

zJ ˆa

I2in

zJ ˆ)bc(

I22out

H

• The direction of magnetic fields can be found from right hand rule.

• The currents flowing through the inner conductor and outer sheath should have the same magnitude with different polarity to minimize the magnetic flux leakage

11

EMLAB

Example : Surface current

nK

x

xH

s

ˆ20z

2

Kˆ

0z2

Kˆ

KL)L)(H(LHdH x

C

x

• The direction of magnetic field con be conjectured from the right hand rule.

12

EMLAB

Example : Solenoid

)0H(

ˆK

KL)L(HLHdH

out

out

C

in

zH

s

• The direction of magnetic field con be conjectured from the right hand rule.

• If the length of the solenoid be-comes infinite, H field outside be-comes 0.

d

I

13

EMLAB

Example : Torus14

EMLAB

A wire of 3-mm radius is made up of an inner material (0 < ρ < 2 mm) for which σ = 107

S/m, and an outer material (2mm < ρ < 3mm) for which σ = 4×107S/m. If the wire carriesa total current of 100 mA dc, determine H everywhere as a function of ρ.

100mA

2

2mm1

Example problem 8.18

E

1J

2J

EJ 11

]/[10133

)(

]mA[100)(

4

22

12

212

1

22

12

212

1

mVE

ErrEr

JrrJrI

]/[6652

2

1

12

12

mAE

H

EJH

mm2

mm3mm2

EEr

H

ErErH

221

21

22

12

12

1

2

1

2

)(

)(2

mm3

Err

H

ErrErH

2

)(

)(2

22

2212

1

22

12

212

1

15

EMLAB

Example problem 8.9

22

P

x

y

16

EMLAB

Example problem 8.6

S

]rad/s[z

17

EMLAB