1

MAGNETOSTATIC FIELD(MAGNETIC FORCE, MAGNETIC MATERIAL

AND INDUCTANCE)

CHAPTER 8

8.1 FORCE ON A MOVING POINT CHARGE

8.2 FORCE ON A FILAMENTARY CURRENT

8.3 FORCE BETWEEN TWO FILAMENTARY CURRENT

8.4 MAGNETIC MATERIAL

8.5 MAGNETIC BOUNDARY CONDITIONS

8.6 SELF INDUCTANCE AND MUTUAL INDUCTANCE

8.7 MAGNETIC ENERGY DENSITY

8.1 FORCE ON A MOVING POINT CHARGE

EQFe Force in electric field:

BUQFm Force in magnetic field:

Total force:

me FFF BUEQF or

Also known as Lorentz force equation.

EQ

BUQ

Charge Condition

Field

Stationary -

Moving

E

EQ

B Field

BUEQ

Combination

E Band

EQ

Force on charge in the influence of fields:

8.2 FORCE ON A FILAMENTARY CURRENT

The force on a differential current element , due to the uniform magnetic field, :

__

dlIB

BlIdFd

____

dlBIBdlIF

0ldBIF

It is shown that the net force for any close current loop in the uniform magnetic field is zero.

Ex. 8.1: A semi-circle conductor carrying current I, is located in plane xy as shown in Fig. 8.1. The conductor is under the influence of uniform magnetic field, . Find:

(a) Force on a straight part of the conductor.

(b) Force on a curve part of the conductor.

0ˆByB

r

I

B

x

ySolution:

(a) The straight part length = 2r. Current flows in the x direction.

(N) 2ˆˆ)2(ˆ 001 IrBzByIrxF

BdlIF __

(N) 2ˆsinˆ 0

0

0

0

2

IrBzdrBIz

BldIF

(b) For curve part, will be in the –ve z direction and the magnitude is proportional to sin

Bdl__

12 FF Hence, it is observed that and it is shown that

the net force on a close loop is zero.

r

I

B

x

y

8.3 FORCE BETWEEN TWO FILAMENTARY CURRENT

P1(x1,y1,z1)P2(x2,y2,z2)

Loop l1

Loop l2

__

11 dlI

2

__

2 dlI

I1

I2

12ˆRa R12

x

y

z

The magnetic field at point P2 due to the

filamentary current I1dl1 :

P1(x1,y1,z1)P2(x2,y2,z2)

Loop l1

Loop l2

__

11 dlI

2

__

2 dlI

I1

I2

12ˆRa

R12

x

y

z

212

112

4

ˆ x 12

R

aldIHd R

(A/m)

BldIFd x (N)

2

12

11222

4

ˆ x x 12

R

aldIldIFdd Ro

222212

11222 x

4

ˆ x x

1

12 BldIR

aldIldIFd

l

Ro

We have :

where is the force due to I2dl2 and due to the magnetic field of loop l12Fd

Integrate:

2 1

12

212

11222

4

ˆ x x

l l

Ro

R

aldIdlIF

2 1

12

2212

1212 x

xˆ

4 l l

Ro ldR

ldaIIF

2

s

22 dsBJF s

dvBJFv

222

For surface current :

For volume current :

Ex. 8.2: Find force per meter between two parallel infinite conductor carrying current, I Ampere in opposite direction and separated at a distance d meter.

d

z

2B

21FI1

I2

I1 = I2 = I

x

y

Solution:

2B at position conductor 2

d

Ix

r

IHB

c

2

ˆ

2

ˆ101

0202

(N/m) 2

ˆ2

ˆ

2

ˆ)ˆ(

2

ˆ

2021

0

1

0

102

102

1

0

22

d

Iy

d

IIy

d

IxdzzI

d

IxdIF

Hence:

Ex. 8.3: A square conductor current loop is located in z = 0 plane with the edge given by coordinate (1,0,0), (1,2,0), (3,0,0) and (3,2,0) carrying a current of 2 mA in anti clockwise direction. A filamentary current carrying conductor of infinite length along the y axis carrying a current of 15 A in the –y direction. Find the force on the square loop.

Solution:

(1,0,0)

(1,2,0)

(3,0,0)

(3,2,0)

x

y

z

2 mA15 A

T ˆ103

104 ∴

A/m ˆ2

15ˆ

26

7-0 z

xHHB

zx

zx

IH

-

Field created in the square loop due to filamentary current :

zxy

aa Rl

ˆˆˆ

ˆˆˆ

Hence:

____

dlBIBdlIF

1

3

0

2

3

1

2

0

63

ˆ1

ˆˆ

ˆ

ˆ3

ˆˆ

ˆ103102

x y

x y

ydyz

xdxx

z

ydyz

xdxx

zF

nN ˆ8

ˆ2ˆ3

1lnˆ

3

2ˆ3ln106

ˆˆlnˆ3

1ˆln106

9

0

2

1

3

2

0

3

1

9

x

xyxy

xyyxxyyxF

(1,0,0)

(1,2,0)

(3,0,0)

(3,2,0)

x

y

z

2 mA15 A

8.4 MAGNETIC MATERIAL The prominent characteristic of magnetic material is magnetic polarization – the alignment of its magnetic dipoles when a magnetic field is applied.

Through the alignment, the magnetic fields of the dipoles will combine with the applied magnetic field.

The resultant magnetic field will be increased.

8.4.1 MAGNETIC POLARIZATION (MAGNETIZATION)

Magnetic dipoles were the results of three sources of magnetic moments that produced magnetic dipole moments : (i) the orbiting electron about the nucleus (ii) the electron spin and (iii) the nucleus spin.

The effect of magnetic dipole moment will produce bound current or magnetization current.

Magnetic dipole moment in microscopic view is given by :

______

dsIdm Am2

___

dmwhere is magnetic dipole moment in discrete and I is the bound current.

In macroscopic view, magnetic dipole moment per unit volume can be written as:

where is a magnetization and n is the volume dipole density when v -> 0.

M

vn

iidm

vM

1

___

0v

1lim A/m

Am-1______

dsnIdmnM

If the dipole moments become totally aligned :

___

idm

ds

___

idm

0

0___

M

dmi

Macroscopic v

Microscopic base

___

idm

0aB0aB

0M

0

0___

M

dmi

Magnetic dipole moments in a magnetic material

tend to align themselves

sdm'___

msm JJ and

8.4.2 BOUND MAGNETIZATION CURRENT DENSITIES

x

y

z

I

aB

M

___

dm

smJ

___

dm

M

M

aB

aB

M

Alignment of within a magnetic material under uniform conditions to form a non zero on the slab surfaces, and a within the material.

sdm'___

aB

smJ0mJ

8.4.3 TO FIND msm JJ and

y

Bound magnetization current :

vInddIm ________

)()()( ldsdnIldsdnIdIm

Am-1______

dsnIdmnM We have:

ldMdIm Hence:

__

ldM = I__

m

s

mm dsJ = I

through the loop l’

on the surface bound by the loop l’

s l s

mm dsMldMdsJ = I

Using Stoke’s Theorem:

Hence:

dIm = Mtan dl'

MJ m (Am-2) is the bound magnetization current density within the magnetic material.

And to find Jsm :

smJ

M

n

loop l’

On the slab surface

From the diagram :

smm Jld

dIM

tan

nMJ sm is the surface bound magnetization current density

(Am-1)

8.4.4 EFFECT OF MAGNETIZATION ON MAGNETIC FIELDS

HBJB

JH

oo

;

charge) (free

Due to magnetization in a material, we have seen the formation of bound magnetization and surface bound magnetization currents density.

due to free charges and bound magnetization currents

Maxwell’s equation:

M

BH

o

JH

Define:

m

o

JJB

mJM

MJB

o

JMB

o

)( MHB o

HM m litysusceptibi magnetic m

)1( mo HB

)1( mr

HB

typermeabiliro

Hence:

Magnetization in isotropic material:

Hence:

Ex. 8.4: A slab of magnetic material is found in the region given by 0 ≤ z ≤ 2 m with r = 2.5. If in the slab, determine: 2mWb/m ˆ5ˆ10 yxxyB

Ja )( mJb )( 0zon )( smJdMc )(

Solution:

236

70

kA/m ˆ775.4ˆ1010510

ˆ5.2104

1 )(

zz

zdy

dB

dBx

dBBHJa xy

r

23 kA/m ˆ163.710ˆ775.45.11 )( zzJJJb rmm

kA/m ˆ387.2ˆ775.4

5.2104

10ˆ5ˆ105.1

)(

7

3

0

yxxy

yxxy

BHMc

rmm

nM ˆsmJ

kA/m ˆ 775 . 4 ˆ 2.387

ˆ )ˆ 387 . 2 ˆ 775 . 4(

yy xx

z yx xy

smJ

Because of z = 0 is under the slab region of 0 ≤ z ≤ 2 , therefore zn ˆˆ

(d)

kA/m ˆ387.2ˆ775.4 )( yxxyMc

Ex. 8.5: A closely wound long solenoid has a concentric magnetic rod inserted as shown in the diagram.In the center region, find: in both air and magnetic rod, (b) the ratio of the in the rod to the in the air, (c) on the surface of the rod and within the rod. Assume the permeability of the rod equals

MBHa and , )(B

smJmJ

B

5o .

r

0

0

ab

z

magnetic rod

Solution:

(a) Using Ampere’s circuital law to the closed path P1 - P2 - P3- P4 . If using

path P1 - P2’ - P3’ -P4 - P1, Hz in the rod will be the same as in the air since

Ampere’s circuital law does not include any Im in its Ien term.

P3’P2’

P4

P3P2

P1

dNI

IdHdzzHzdH enz

P

P

z

3

2

)ˆ()ˆ(

sz JNI

H

Hence:

P3’P2’

P4

P3P2

P1

0

0

ab

zmagnetic rod

airin 0

rod in the )ˆ(ˆ

M

HzMzM zmz

)ˆ(0 zHzB

)ˆ(0 zr HzB

in air

in the rod

(since m of air is zero)

5ˆ

ˆ5 )(

zo

zo

air

rod

Hz

Hz

B

Bb

zczsm MrMznMJc ˆ)ˆ(ˆ )(

0)ˆ( zm MzMJ

flux Js

flux Jsm

z

smJ

M

rn ˆˆ

aa

0

b

sJ smJ

Js=Hz=NI/l

50Hz

Bz=0Hz

Jsm=4Hz

Mz

JsJsm

Bz

Hz

Mz=4Hz

Hz=NI/l

Plots of H, B and M, Js and Jsm

along the cross section of the solenoid and the magnetic rod

8.4.5 MAGNETIC MATERIAL CLASSIFICATION

Magnetic material can be classified into two main groups:

Group A – has a zero dipole moment

0dm diamagnetic material eg. Bismuth

9999834.0,1066.1 5 rm

Group B – has a non zero dipole moment

(a) Paramagnetic material - 0dm 0M;

aBWhen is applied, there will be a slight alignment of the atomic dipole moment to produce 0M

00002.1,102 5 rm μχEg. Aluminum -

dmaB

(b) Ferromagnetic material : has strong magnetic moment in the absence of an applied field.

Eg: metals such as nickel, cobalt and iron.

8.5 MAGNETIC BOUNDARY CONDITIONSMHB and ,To find the relationship between

s

HB /21n

l

2/h

sJ

2/h

Boundary2/h2/h

ab

cd

11 , m

22 , mRegion 2:

Region 1:

HB and To find normal component of at the boundary

Consider a small cylinder as 0h 0__

dsBand use

0

21

21

__

nn

nn

BB

sBsBdsB

2211 nn HH

s

HB /21n

l

2/h

sJ

2/h

Boundary2/h2/h

ab

cd

11 , m

22 , mRegion 2:

Region 1:HB and To find tangential component of

Consider a closed abcd as 0h

and use enc

l

IdlH __

at the boundary

stt

enctt

JHH

IlHlH

21

21

sJ tH1 tH 2where is perpendicular to the directions of and

sn JHHa 2121ˆ

21ˆna=

In vector form :

21ˆna is a normal unit vector from region 2 to region 1

x xy

z

yxz ˆˆˆ

stt J

BB

2

2

1

1

mJM

stt JHH 21

We have:

mv v m IdVJdVM

mlIdlM

__

and

smtt JMM 21

HM m

sm

t

m

t JMM

2

2

1

1

2

22

1

11

m

n

m

n MM

Hence:

We have:

Hence:

2211 nn HH

We have:We have:

Hence:

Hence:

2

2

1

121 or

tt

tt

BBHH

If Js = 0 :

If the fields were defined by an angle normal to the interface

222111 coscos BBBB nn

22 , mRegion 2:

11 , mRegion 1:1

2

x

y

Boundary at y = 0 plane

11 HorB

22 HorB

1B

1tB

1nB

22

2211

1

1 sinsin

BHH

Btt

2

1

2

1

2

1

tan

tan

r

r

(1)

(2)

Divide (2) to (1):

Ex. 8.6: Region 1 defined by z > 0 has 1 = 4 H/m and 2 = 7 H/m in region 2 defined by z < 0. A/m on the surface at z = 0. Given

, find

xJ sˆ80

mT ˆˆ3ˆ21 zyxB 2B

Solution:

Z = 0

#1

#2

1B

1tB

1nB

2B 2tB 2nB

21n z

z12n

xJ sˆ80

z

yx

tH 2

tH11BNormal component

ˆ

ˆ

ˆˆˆˆ3ˆ2

ˆ)ˆ(

12

121211

zBB

z

zzzyx

nnBB

nn

n

A/m ˆ750ˆ500

104

10ˆ3ˆ2

ˆ3ˆ2

6

3

1

11

111

yx

yxBH

yxBBB

tt

nt

A/m ˆ670ˆ500

ˆ80ˆ750ˆ500

ˆ80ˆˆ750ˆ500

ˆ1212

yx

yyx

xzyx

JnHH stt

mT ˆˆ69.4ˆ5.3

ˆ69.4ˆ5.3ˆ670ˆ500107

222

6222

zyxBBB

yxyxHB

nt

tt

Ex. 8.7: Region 1, where r1 = 4 is the side of the plane y + z < 1 . In region 2 , y + z > 1 has r2 = 6 . If find yxB ˆˆ21 22 and HB

Solution:

2/)ˆˆ(ˆ zyn The unit normal:

zyxHB

HzyxH

zyxBBB

BzynB

zyyxB

trt

tt

nt

nn

n

ˆ75.0ˆ75.0ˆ3

ˆ125.0ˆ125.0ˆ5.01

ˆ5.0ˆ5.0ˆ2

ˆ5.0ˆ5.0ˆ2

12

1

2

ˆˆˆˆ2

2202

20

1

111

2121

1

121211 ˆ)ˆ( nnBB n na

y

z

r1 = 4

r2 = 6#1

#2

O

12n

(A/m) ˆ04.0ˆ21.0ˆ5.01

(T) ˆ25.0ˆ25.1ˆ3

02

2

zyxH

zyxB

nt BBB 222

y + z = 1

8.6 SELF INDUCTANCE AND MUTUAL INDUCTANCE

+

_

VL

Coil

I

Magnetic flux

From circuit theory the induced potential across a wire wound coil such as solenoid or a toroid :

Simple electric circuit that shows the effect of energy stored in a magnetic field of an inductor :

dt

dILVL

where L is the inductance of the coil, I is the time varying current flowing through the coil – inductor.

2

2

1CVWE

In a capacitor, the energy is stored in the electric field :

In an inductor, the energy is stored in the electric field, as suggested in the diagram :

+

_

VL

Coil

I

Magnetic flux

switch

Idtdt

dILIdtVW

tt

t

tt

t

Lm 00

00

)( 2

1 2

0

0

JouleLIdILItt

t

HenryI

L

Define the inductance of an inductor :

where (lambda) is the total flux linkage of the inductor

Nm

)( HenryHI

L

I

NL m

Weber turns

Hence :

H

Two circuits coupled by a common magnetic flux that leads to mutual inductance.

I1 Circuit 1N1 turns

Circuit 2N2 turns

I2

1

1212 I

M

Mutual inductance :

12 is the linkage of circuit 2 produced by I1 in circuit 1

For linear magnetic medium M12 = M21

Ex. 8.8: Obtain the self inductance of the long solenoid shown in the diagram.

Solution:

mψAssume all the flux links all N turns and that does not vary over the cross section area of the solenoid.

B flux

N turns

22

22

2

have We

al

IN

Nal

NINaH

HB

NaBNm

l

aN

IL

22

I

SNb

NII

Nac

B

I

N

IL m

2

4

2

Ex. 8.9: Obtain the self inductance of the toroid shown in the diagram.

a

cb

m

0

avel Bads ˆ __

Mean path

Cross sectional area S

Feromagnetic core

N turns

I

Solution:

area sectional cross toroidal-

radiusmean - where2

2

S

bb

SNL

Ex. 8.10: Obtain the expression for self inductance per meter of the coaxial cable when the current flow is restricted to the surface of the inner conductor and the inner surface of the outer conductor as shown in the diagram.

Solution:

a

b

mψ

mψThe will exist only between a and b and will link all the current I

a

b

I

dzdr

r

I

I

dzdrH

IIL

b

a

c

c

b

a

cm

ln2

2

1

0

1

0

Ex. 8.11: Find the expression for the mutual inductance between circuit 1 and circuit 2 as shown in the diagram.

Let us assume the mean path :

2b >> (c-a)

Solution:

Two circuits coupled by a common magnetic flux that leads to mutual inductance.

I1 Circuit 1N1 turns

Circuit 2N2 turns

I2

a

b

c

b

SNN

I

SNbIN

I

Nac

B

I

N

IM m

22

4

21

1

211

1

2

2

12

1

2)12(

1

1212

8.7 MAGNETIC ENERGY DENSITY

Henry I

L

We have :

Joule 2

1

2

1

2

1 22 III

LIWm

BSNINIW mm 2

1

2

1

Consider a toroidal ring : The energy in the magnetic field :

Multiplying the numerator and denominator by 2b :

bSb

NIBWm

222

1

a

cb

m

IS

N turns

toroid theof volume theis )2( and 2

where VbSHb

NI

BHVWm 2

1

Hence :

32 2

1

2

1 JmHBHV

Ww m

m

HBwm 2

1In vector form :

dvHBI

WI

Lv

m 2

12222

Hence the inductance :

Ex. 8.12: Derive the expression for stored magnetic energy density in a coaxial cable with the length l and the radius of the inner conductor a and the inner radius of the outer conductor is b. The permeability of the dielectric is .

(J) ln4

21

8

1

82

1

2

2

22

2

22

22

a

blI

drrlr

IW

dvr

IdvHW

r

IH

b

a

m

vvm

Solution:

a

b

mψ