EMT Electromagnetic Theory MODULE III

MODULE III

Magnetostatics

Compiled by MKP for CEC S5 batch August 2008

SyllabusMagnostatics and Maxwell’s equations:

Magnostatic fields - Biot Savart law - Ampere’s circuital law -Applications of Ampere’s circuital law - Magnetic flux density -Magnetic scalar and vector potentials. Magnetic forces, materials and devices - Forces due to magnetic fields - Magnetic torque and moment - Magnetic dipole - Magnetization in materials -Classification of magnetic materials - Magnetic boundary conditions - Inductors and inductances - Magnetic energy - Magnetic circuits -Faraday’s law - displacement current. Time harmonic fields -Maxwell’s equations for static fields and time varying fields - word statement.

Compiled by: MKP for CEC S5 - July 2008

ReferencesText Books:

1. Mathew N.O. Sadiku, Elements of Electromagnetics, Oxford University Press

2. Jordan and Balmain, Electromagnetic waves and radiating systems,Pearson Education PHI Ltd.References:

1. Kraus Fleisch, Electromagnetics with applications, McGraw Hill2. William.H.Hayt, Engineering Electromagnetics, Tata McGraw Hill3. N.Narayana Rao, Elements of Engineering Electromagnetics, Pearson

Education PHI Ltd. 4. D.Ganesh Rao, Engineering Electromagnetics, Sanguine Technical

Publishers.5. Joseph.A.Edminister, Electromagnetics, Schaum series-McGraw Hill6. K.D. Prasad, Electromagnetic fields and waves, Sathya Prakashan

Compiled by: MKP for CEC S5 - July 2008

Concept of Current elementDirect currents flows only in closed loops. We can find out the contributions to the magnetic field due to differential lengths of such current carrying conductors.A current element is the current I flowing through a differential vector length of a filamentary conductor.A filamentary conductor is the limiting case of a cylindrical conductor of circular cross section as the radius approaches zero.

dL

IdL

I


Biot-Savart’s LawBiot-Savart’s law states that the magnetic field intensity dHproduced at a point P by a differential current element Idl is proportional to the product Idl and the sine of the angle αbetween the element and the line joining P to the element and is inversely proportional to the square of the distance Rbetween P and the element and its direction can be obtained by right handed screw rule.

2

sinIdldHR

α∝ 2

sinIdldH kR

α=

1 / 4k is a proportionality constant whose value is π

2

sin4

IdldHRα

π=


Biot-Savart’s Law

2

sin4

IdldHRα

π=

This equation can be modified by incorporating the direction of the magnetic field intensity

2

ˆ4

RI dl adHRπ×

=34

I dl RRπ×

= ˆ and RRR R aR

= =

Idl Current element⇒ˆ Ra Unit vector directed from current element to P⇒

R Distance from current element to point P⇒

dH Magnetic field intensity at P⇒


Biot-Savart’s Law

•

α R

P

I

dl ˆRa

⊗dH

2

ˆ4

RI dl adHRπ×

=

2

sin4

IdldHRα

π=

2

sin ˆ4 n

IdldH aRα

π=


Current elements

I

I dl KdSJdV

J

K

Line current Surface current Volume current


Concept of surface current densityWhen current flows through a sheet of vanishingly small thickness we cannot measure current density in amperes per square meter as it becomes infinite.In this case surface current density is measured in amperes per meter width and is designated as

If the surface current density is uniform, the total current in any width b is I=Kb where the width b is measured perpendicular to the direction of current flow.

J

K

b

I

K


Concept of surface current densityFor a non-uniform surface current density, we have to integrate K over the path of interest.

Where dN is a differential element of the path across which the current is flowing.Let ΔS be the cross sectional area of the wire. Then

Let dv=dSdh, where dh is the thickness of the surface current. Then

I KdN= ∫

IdL J SdL Jdv= Δ =

( )Jdv JdSdh Jdh dS KdS= = =

IdL Jdv KdS⇒ ⇒


Biot-Savart’s LawIn terms of distributed current sources, the Biot-Savart’s law may be expressed in the following ways:

2

ˆ

4R

Lfor line currentI dl aH

Rπ×

= ∫

2 ˆ

4

RS

for surKdS a face cu e tR

rr nHπ×

= ∫

2

ˆ

4 R

VforJdv a voH

Rlume current

π×

= ∫


Magnetic field of a linear conductor

A

R

P

dl

I

α

ρ

z

yx

2α

B

0

1α

z

0,0,z

cotz ρ α=

aρρ

ˆzza

ˆ ˆzR a zaρρ= −

2 2 R zρ= +



dl Small current element in the conductor AB⇒P Point where the magnetic field is required.⇒

Perpendicular distance between the conductor and the point Pρ ⇒

1 2, Angles subtended by the lower and upper ends of AB.α α ⇒

Let

By Biot - Savart's law the contribution dH at P due to an element (0,0,z)dl at is

34I dl RdH

Rπ×

=

ˆ ˆ ˆz zBut andl = dza R zd a aρρ= −



( )ˆ ˆ ˆz zdl R = dza z a aρρ× −× ˆ= dzaφρ

3 ˆ4I dzdH a

R φρπ

=

3 ˆ4I dzH a

R φρπ

= ∫

cotBut z = ρ α 2cosecdz = dρ α α−

2 2 3/2 ˆ4 ( )

I dz az φ

ρπ ρ

=+∫

( )2

1

2 2

3/22 2 2

ˆ4I cosec dH a

+ cot

α

φα

ρ α απ ρ ρ α

= − ∫ ( )2

1

2 2

3/22 2

ˆ4I cosec d a

1+cot

α

φα

ρ α απ ρ α

= −⎡ ⎤⎣ ⎦

∫



2

1

2 2

3 3

ˆ4I cosec d a

cosecα

φα

ρ α απ ρ α

= − ∫2

1

2 2

3/22 2

ˆ4 cos

I cosec dH aec

α

φα

ρ α απ ρ α

= −⎡ ⎤⎣ ⎦

∫

2

1

ˆ sin 4

I a dα

φ αα α

πρ= − ∫

( )2 1 ˆcos cos4

I aφα απρ

= −

( )2 1 ˆcos cos4

IH aφα απρ

= − ˆ ˆla a aφ ρ= ×

[ ] 2

1ˆ cos

4I a α

φ αα

πρ=


Magnetic field of a linear conductorWhen the conductor is semi-infinite, so that point a is now at O(0,0,0) while B is at

When the conductor is of infinite length, point A is at

1 2(0,0, ) 90 , 0α α∞ = =and

( ) ˆcos0 cos904

IH aφπρ= − ˆ

4I aφπρ

=

ˆ4

IH aφπρ=

1 2 (0,0, ) 180 , 0B is at α α∞ = =while and

(0,0, ) −∞

( ) ˆcos0 cos1804

IH aφπρ= − ˆ

2I aφπρ

=ˆ

2IH aφπρ

= ˆ ˆla a aφ ρ= ×

ˆ ˆla a aφ ρ= ×


Magnetic field at the centre of a circular current loop

90α = •I dl

ˆRaX

Y

Z

O

P

2

sin4

ˆnIdldH

Raα

π=

Ia



The magnetic field intensity at O is given by where is the field intensity at O due to any current element The direction of at any point P on the circular wire is given by the tangent at P in the direction of current flow.The unit vector at P directed towards O is along the radius PO so that α = 90o.

Total field intensity at the centre of the circular wire is obtained by integrating dH around the circular path.

H dH= ∫dH Idl

2

ˆ4

RI dl adHRπ×

=

2

sin4

ˆnIdldH

Raα

π=

2

sin904

ˆnIdl

aa

π= 24

ˆzdl aIaπ

=

2 ˆ4 z

IH d alaπ

= ∫

Idl



2 ˆ4 z

IH d alaπ

= ∫

2 ˆ4 z

IH d alaπ

= ∫ 2 2 ˆ4 z

Ia

aaππ

= ˆ2 zIa

a=

ˆ2 zIHa

a=


Magnetic field at a line through the centre of a circular current loop

P

X

Y

Z

O

•

I

Rh

a

φφ

'dl dl

'dH

'ydH ydH

dH

'zdH zdH

dH

• •

R

2 2sin a

a hφ =

+

2 2R a h= +

φ

ˆRa



Let P be a point at a distance h from the centre of a circular current loop.Consider two diametrically opposite elements of the loop dl and dl’. The field intensity at P distant R from the current element is given by

Since and are perpendicular

2

ˆ4

RI dl adHRπ×

= 2

sin4

ˆndl aI

Rα

π=

dl ˆRa

24ˆn

IdldHR

aπ

=

ˆ ˆ n Ra A unit vector perpendicular to the plane containing dl and a⇒

24IdldH

Rπ=



The field is oriented at an angle Ф to the plane of the loop. The diametrically opposite element dl’ will also produce a field of magnitude equal to dH.Its component parallel to the plane of the loop gets cancelled.The components along the z axis gets added up.

24IdldH

Rπ=

2 sin4z

IdldHR

φπ

=

2 2 2

2 2 sin aBut and R a h

a hφ = = +

+

( )2 2 2 24zIdl adHa h a hπ

=+ + ( )3/22 24

Ia dla hπ

=+



The resultant field intensity at the point P is given by integrating the Z components of the field contributions of all the current elements.

( )3/22 24z

IadH dla hπ

=+

( )3/22 24z

IaH dH dla hπ

= =+

∫ ∫

( )3/22 24

Ia dla hπ

=+

∫ ( )3/22 22

4

Ia aa h

ππ

=+ ( )

2

3/22 22

Ia

a h=

+

( )2

3/22 2ˆ

2z

IaH aa h

=+ ( )

2

3/22 2ˆ

2z

IaH aa h

=+


Example 1Find the magnetic field at (0,0,5) due to side OA and BO of the triangular loop carrying a current of 10A and lying in the xy plane.

Y

Z

2

1

10AO

AB

1

X


Example 1

X X

Z

OA2

( )2 1 ˆcos cos4

IH aφα απρ

= −

5

ρ

1α 2αaρ

ˆ ˆ ˆla a aφ ρ= ×

ˆ ˆl xa a=

5ρ = 1cos cos90 0α = =

22cos29

α =

10 2 ˆ4 5 29

H aφπ⎛ ⎞= ⎜ ⎟⋅ ⎝ ⎠

10I =

ˆ0.059.1aφ=

ˆ ˆ ˆx ya a a aφ ρ= × = −

ˆ0.059.1 yH a= −

To find H due to OA


Example 1

Y

Z

2

5

10AO

A

B

1

2

5ρ =27

( )2 1 ˆcos cos4

IH aφα απρ

= −


5ρ =

1cos cos105.7 0.272α = = −

2cos cos90 0α = =

10I =

( )1

1 22

5180 sin5 2

α −

⎡ ⎤⎢ ⎥

= − ⎢ ⎥⎢ ⎥+⎣ ⎦

2 90α =

1 2105.7 , 90α α= =

To find H due to BO

1α


Example 1( )2 1 ˆcos cos

4IH aφα απρ

= −

( )10 ˆ0 0.2724 5

H aφπ= +

⋅0.272 ˆ ˆ0.0433

2a aφ φπ

= =


ˆ ˆ ˆcos45 cos45l x ya a a= − − ˆ ˆ0.707 0.707x ya a= − −

( )ˆ ˆ ˆ ˆ0.707 0.707x y za a a aφ = − − ×

ˆ ˆl za a= ×

ˆ ˆ0.707 0.707y xa a= −

( )ˆ ˆ0.0433 0.707 0.707y xH a a= −

ˆ ˆ0.0306 0.0306x yH a a= − +


Example 2 A circular loop located on x2+y2= 9 carries a direct current of 10 A along . Determine magnetic field at (0,0,4) and (0,0,-4)aφ

( )2

3/22 2ˆ

2z

IaH aa h

=+

3a = 4h = 10I =

( )2

3/22 2

10 3 ˆ2 3 4

zH a×=

+90 ˆ250 za= ˆ0.36 za=

(0,0,4)at

3a = 4h = − 10I = (0,0, 4)at −

( )( )2

3/222

10 3 ˆ2 3 4

zH a×=

+ −90 ˆ250 za= ˆ0.36 za=

•

•

(0,0,4)

(0,0, 4)−

3a =

H ⇑

H ⇑

10I = 4

Z


Ampere’s Current lawAmpere’s circuital law states that the line integral of the tangential component of around a closed path is the same as the net current enclosed by the path.

Line integral of the magnetic field around a closed path is called magneto motive force.So ampere’s law can be stated as : The magneto motive force around a closed path is equal to the current enclosed by that path.

enclH dl I⋅ =∫

H

H

enc SBut I J dS= ⋅∫

SlJ dSH dl⋅ = ⋅∫ ∫

SlJ dSH dl⋅ = ⋅∫ ∫


Ampere’s Current law

enclH dl I⋅ =∫

I


Ampere’s Current lawApplying Stokes theorem to the LHS of the above equation

This is the Ampere’s law in differential form or point form and is the third of the Maxwell’s equations.

( )S S

J dSH dS∇× ⋅ = ⋅∫ ∫H J∇× = H J∇× =


Magnetic field of an infinite line current

Z

Y

X

ρ ⋅PI

O

Amperian Path

dl


Magnetic field of an infinite line currentLet an infinitely long filamentary current element be placed along the z axis.To determine H at a point P we select a closed amperian path that passes through P.Since this path encloses the whole current,

enc lI H dl= ⋅∫

ˆ ˆl

I H a dlaφ φ φ= ∫l

H dlφ= ∫ 2Hφ πρ=

2IHφ πρ

= ˆ2

IH aφπρ=

ˆ2

IH aφπρ=


Magnetic field of a coaxial cable

a

b

t

1L

2L

3L

4L

⊗I− I+

Amperian pathsZ

ρ


Magnetic field of a coaxial cableConsider an infinitely long transmission line consisting of twoconcentric cylinders having their axes along the Z axis.The inner conductor has radius a and carries current I while the outer conductor has inner radius b and thickness t and carries return current –I.The magnetic field is to be evaluated for the four regions separately.

0 aa bb b t

b t

ρρρ

ρ

≤ ≤≤ ≤≤ ≤ +≥ +



1 0 ' For the region a apply Ampere s law to path Lρ≤ ≤

2encII J dS d da

ρ φ ρπ

= ⋅ =∫ ∫∫

Current is assumed to be uniformly distributed over the sec .cross tion

2 ˆzIJ aaπ

= ˆ zand dS d d aρ φ ρ=

1encL

H dl I J dS⋅ = = ⋅∫ ∫

2

I d da

φ ρ ρπ

= ∫ ∫


Magnetic field of a coaxial cable2

2 22enc

IIa

ρππ

= × ×2

2

Iaρ

=

1

2

2LH dl I

aρ

=⋅∫

1

2

2ˆ ˆL

H a dla Iaφ φ φρ

=⋅∫

1

2

2LH d

al I

φρ

=∫2

22H Iaφ πρ ρ

=

2

2 2I

aHφ

ρπρ

=

22I

aHφ

ρπ

=

22IH

aφρπ

= 0 aρ≤ ≤



2 ' For the region a b apply Ampere s law to path Lρ≤ ≤

2encL

H dl I J dS⋅ = = ⋅∫ ∫

2

ˆ ˆL

H a dl Iaφ φ φ⋅ =∫2 Since whole current is enclosed by the path L

1LH dl Iφ =∫

2H Iφ πρ =

2H I

φ πρ=

2IHφ πρ

= a bρ≤ ≤



3 ' For the region b b t apply Ampere s law to path Lρ≤ ≤ +

3LH dl I J dS⋅ = + ⋅∫ ∫

3

ˆ ˆL

H a dla I J dSφ φ φ⋅ = + ⋅∫ ∫

3LH dl I J dSφ = + ⋅∫ ∫

2H I J dSφ πρ = + ⋅∫ The J in this case is the current density of the outer

conductor



( )2 2ˆz

IJ ab t bπ

= −⎡ ⎤+ −⎣ ⎦

2H I J dSφ πρ = + ⋅∫

( )2 2ˆ2 ˆz zS

II a d d ab t b

Hφ ρ φ ρπρπ

− ⋅+ − ⎦

=⎡ ⎤⎣

∫

( )2 2 S

II d db t b

ρ φ ρπ

= −⎡ ⎤+ −⎣ ⎦

∫

( )

2

2 20 b

II d db t b

ρπ

ρ

φ ρ ρπ =

= −⎡ ⎤+ −⎣ ⎦

∫ ∫



( )

2

2 22 2

2bH II

b t b

ρ

φ πρρ

ππ

⎡ ⎤⎣ ⎦− × ×⎡ ⎤+ −⎣ ⎦

=

( )2 2

2 2 22I b

Ib bt t b

ρ −= −

⎡ ⎤+ + −⎣ ⎦

( )2 2

22I b

Ibt tρ −

= −⎡ ⎤+⎣ ⎦

( )2 2

212b

It btρ⎧ ⎫−⎪ ⎪= −⎨ ⎬⎡ ⎤+⎪ ⎪⎣ ⎦⎩ ⎭ ( )2 2

212 2

bIHt btφ

ρπρ

⎧ ⎫−⎪ ⎪= −⎨ ⎬⎡ ⎤+⎪ ⎪⎣ ⎦⎩ ⎭

b b tρ≤ ≤ +


Magnetic field of a coaxial cable4 ' For the region b t apply Ampere s law to path Lρ ≥ +

4

0L

H dl I I⋅ = − =∫2 0Hφ πρ =

0Hφ =

0H φ =

b tρ ≥ +

2 0 2

I aHaφ ρρ

π= ≤ ≤

2

IH a bφ πρρ= ≤ ≤

( )2 2

212 2

bIH

t btb b tφ

ρπρ

ρ⎧ ⎫−⎪ ⎪= −⎨ ⎬⎡ ⎤

≤ ≤⎪ ⎪⎦⎩ ⎭

++⎣

0 tH bφ ρ ≥ +=



Hφ

0

12 aπ

12 bπ

a b b t+ρ

H Vsφ ρIMPORTANT:

Outside the co-axial cable magnetic field is zero. This is the basic principle behind electro magnetic shielding


Magnetic field by a toroidA toroid can be considered as a solenoid “bent” into a circle as shown. We can apply Ampere’s law along the circular path inside the toroid.

enclLH dl I⋅ =∫

enclI NI=

N is the number of loops in the toroid, and I is the current in each loop

ˆ2

NIB aφμπρ

=

enclLHdl I=∫

enclLH dl I=∫

2 enclH Iπρ =

2 H NIπρ =

ρ

2NIHπρ

=

ˆ2NIH aφπρ

=


Magnetic field by a solenoidWhen the coils of the solenoid are closely spaced, each turn can be regarded as a circular loop, and the net magnetic field is the vector sum of the magnetic field for each loop. This produces a magnetic field that is approximately constant inside the solenoid, and nearly zero outside the solenoid.


Magnetic field by a solenoidThe ideal solenoid is approached when the coils are very close together and the length of the solenoid is much greater than itsradius. Then we can approximate the magnetic field as constant inside and zero outside the solenoid.

H

Apply Ampere’s law for the closed path 1234

N turns in a length of L

ˆxa


Magnetic field by a solenoid

12 23 34 41L

H dl H dl H dl H dl H dl⋅ = ⋅ + ⋅ + ⋅ + ⋅∫ ∫ ∫ ∫ ∫

enclLH dl I⋅ =∫

0 0H L H L= + + =

= C urren t enc losed by the pa th is N I

H L N I=

N IHL

=

N IBL

μ=

ˆ xN IB aL

μ=

times number of turns per unit lengthNB I IL

μ μ= =

ˆ xN IH aL

=


Magnetic field by a long cylindrical conductor

Find the magnetic field inside and outside an infinitely long cylindrical conductor having radius R and carrying a current I

ρ

ρ

ˆ za



A long straight wire of radius R carries a steady current I that is uniformly distributed through the cross-section of the wire. In region where ρ> R choose a circle of radius ρ centered on the wire as a path of integration. Along this path, H is again constant in magnitude and is always parallel to the path.

enclLH dl I=∫

enclLH dl I⋅ =∫

( )2 enclH Iπρ =

2 H Iπρ =

2IHπρ

=

ˆ2

IH aφπρ=

ˆ2

IB aφμπρ

=

ˆ2

IH aφπρ=

ˆ2

IB aφμπρ

=



In region where ρ < R choose a circle of radius ρ centered on the wire as a path of integration. Along this path, H is again constant in magnitude and is always parallel to the path.

enclLH dl I=∫

enclLH dl I⋅ =∫

( )2 enclH Iπρ =

2

enclIHπρ

=

encl SI J dS= ⋅∫ 2

2IRρ

=

2 ˆzIJ aRπ

=

2 ˆ êncl z zS

II a dSaRπ

= ⋅∫

ˆzdS dSa=

2 S

I dSRπ

= ∫2

2IR

πρπ

=



2

2 2

IHR

ρπρ

=

2 2

IHRρ

π=

2 2

IBR

μ ρπ

=

2 ˆ 2

IB aR φ

μ ρπ

=

2 ˆ 2

IB aR φ

μ ρπ

=

2 ˆ 2

IH aR φρ

π=

2 ˆ 2

IH aR φρ

π=



H

ρR

2 ˆ for 2

IB a RR φ

μ ρ ρπ

= <

ˆ for 2

IB a Rφμ ρπρ

= ≥

ρ

ρ

2 ˆ for 2IH a R

R φρ ρ

π= <

ˆ for 2

IH a Rφ ρπρ

= ≥


Magnetic flux densityElectric flux density and field intensity is relates asSimilarly we can define magnetic flux density by the relation

Magnetic flux through a surface is obtained by integrating flux density throughout the surface

0B Hμ=

0D Eε=

70 =4 10 /Permiability of free space H mμ π −⇒ ×

B Magneic flux density⇒

H Magnetic field intensity⇒

SB dSψ = ⋅∫


Magnetic flux density

B

BB

B

I




Magnetic flux densityMagnetic flux lines is a path to which is tangential at every point. The magnetic flux lines due to a long straight conductor is shown in figure 1.Each flux line is closed and has no beginning or end.In an electrostatic field the flux crossing a closed surface is the same as the charge enclosed.

So it is possible to have an isolated electric charge and the flux lines produced by it need not be closed.It is not possible to have an isolated magnetic pole.As a result, magnetic flux lines always close upon themselves incontrast to electric flux lines.

B

D dS Qψ = ⋅ =∫



•Q+

Q over the closed surfaceψ =

Electric flux lines need not be closed. So it is possible to have an isolated electric charge.

D dS Qψ = ⋅ =∫



N S

Magnetic flux lines are always closed. It is not possible to have an isolated magnetic pole.


Magnetic flux densityN

S

N

S

N

S

N

S S

N

S

N

S

N

NS

NS

NS

NS

NS

NS

NS

NS

It is not possible to isolate the north and south poles of a magnet. An isolated magnetic pole cannot exist


Gauss’s Law for magnetostatic fieldsThe total flux through a closed surface in a magnetic field is zero.

By applying divergence theorem,

This is the Maxwell’s fourth equation. It states that magnetostaticfields have no sources or sinks.

0B dS⋅ =∫

= 0V

B dS B dv⋅ = ∇⋅∫ ∫= 0B∇⋅


Maxwell’s Equations for static EM fields

VD ρ∇⋅ =

0B∇⋅ =

0E∇× =

H J∇× =

VS VD dS dVρ⋅ =∫ ∫

0SB dS⋅ =∫

0LE dl⋅ =∫

L SH dl J dS⋅ = ⋅∫ ∫

Differential form Integral form Derived from

' Gauss s Law

Nonexistance of magneticMonopole

Conservativeness of Electrostatic field

' Ampere s Law


Magnetic scalar and vector potentialsIn electrostatics electric field intensity and potential are related as

Similar to this we can relate magnetic field intensity with two magnetic potentials:

1. Magnetic scalar potential2. Magnetic vector potential

Magnetic scalar potential Vm is related to by the relation

According to Maxwell’s equation For any scalar

E V= −∇

H

0 (1) m ifH V J= −∇ = − − − −

(2)H J∇× = − − − −( ) 0 (3)φ∇× ∇ = − − − −

( )mH V∇× = ∇× −∇ (0 4)− − − −= 0 mH V if J=−∇ = (4) 0Eq is true only if J =


Magnetic scalar and vector potentialsThe definition of Magnetic scalar potential according to equation (1) must satisfy equations (2) and (4).For this, the condition must be satisfied.Vm satisfies Laplace’s equation also.

0 J =

2 0 0 mV if J∇ = =2 0 0 mV if J∇ = =


Magnetic scalar and vector potentialsFor a magnetostatic field

For any vector

We can define another potential called magnetic vector potential that satisfies equations (5) and (6) simultaneously

In many EM problems it is more convenient to first find and then find from it.

0 (5)B∇⋅ = − − − −

( ) 0 (6)A∇⋅ ∇× = − − − −

Let B A= ∇×

A is called magnetic vector potentialA

B


Derivation of Magnetic vector potentials from Biot-Savart’s law

From Biot-Savart’s law, 034 L

I dl RBR

μπ

×= ∫

O

( , , )x y z

( ', ', ')x y z 'dl

r

'r

'R r r= −



Let the source dl’ be located at (x’,y’z’) and let (x,y,z) be the point where we want to find the magnetic potential.

0 1 (14

' )L

dlR

B Iμπ

⎛ ⎞×∇⎜ ⎟⎝ ⎠

= − − − −∫

1/22 2 2' ( ') ( ') ( ')R r r x x y y z z⎡ ⎤= − = − + − + −⎣ ⎦

3/22 2 2

ˆ ˆ ˆ( ') ( ') ( ')1

( ') ( ') ( ')x y zx x a y y a z z a

R x x y y z z

− + − + −⎛ ⎞∇ = −⎜ ⎟⎝ ⎠ ⎡ ⎤− + − + −⎣ ⎦

ˆ ˆ ˆx y za a ax y zφ φ φφ ∂ ∂ ∂

∇ = + +∂ ∂ ∂

3

1RR R

⎛ ⎞= −∇⎜ ⎟⎝ ⎠

3

RR

= −

3

1 RR R

⎛ ⎞∇ = −⎜ ⎟⎝ ⎠



Using the vector identity ( ) ( )F F Fφ φ φ∇× = ∇× + ∇ ×

( ) ( )F F Fφ φ φ∇ × = ∇× − ∇×1 1 1' '' dd d

Rl

Rl l

R⎛ ⎞= ∇× −∇×⎜

⎛ ⎞× ∇⎜ ⎟⎝ ⎠

⎟⎝ ⎠

( ) ( )F F Fφ φ φ× ∇ = ∇× −∇×

. . ( , , ) ' ( ', ', ')operates w r t x y z while dl is a function of x y z∇

' 0So dl∇× =

1 1' ' (2)dl dlR R

⎛ ⎞ ⎛ ⎞× ∇ = −∇× − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(1)Substituting in

( , , ) ( ', ', ')R is a function of x y z and x y z



0 ' 4 L

I dlBR

μπ

⎛ ⎞= ∇×⎜ ⎟

⎝ ⎠∫ 0 '

4L

I dlR

μπ

= ∇× ∫

Comparing with B A= ∇×

'

0 4

S

for surKdSAR

face currentμπ

= ∫

0 4L

I dlAR

μπ

= ∫

Similarly

0 ' 4L

I dlAR

μπ

= ∫

'0 4

V

for volumJdvA e currenR

tμπ

= ∫

0 4S

KdSAR

μπ

= ∫

0 4V

JdvAR

μπ

= ∫


Magnetic flux from vector potential

.SB dSψ = ∫ ( ).

SA dS= ∇×∫

( ) . ' S L

By Stoke s Theorem A dS A dl∇× = ⋅∫ ∫

LA dlψ = ⋅∫

.S LB dS A dlψ = = ⋅∫ ∫

LA dlψ = ⋅∫


Magnetic force on a charged particleA magnetic field can exert a force only on a moving charge.The magnetic force Fm experienced by a charge Q moving with a velocity in a magnetic field is given by

An electric field can exert a force on a stationary charge and the force is given by

For a moving charge Q in the presence of both electric and magnetic fields the total force on the charge is given by

This is called Lorentz force equation

u B

mF Qu B= ×

eF QE=

( )e mF F F Q E u B= + = + ×


Magnetic force on a charged particleWhen the mass of the charged particle moving in an electromagnetic field with velocity is m,

The above equation relates mechanical force to electrical force.

u

( )duF m Q E u Bdt

= = + ×

Independent of velocity It can perform work on the charge It can change the kinetic energy of the charge

eElectric force F is


Magnetic force on a charged particle

Dependent of velocity

Force is normal to velocity and magnetic field

It cannot perform work on the charge since the motion of the particle is at right angles to the force

mMagnetic force F is

It cannot change the kinetic energy of the charge


Magnetic force on a current elementThe differential magnetic force that a differential element charge, dQmoving with a velocity experiences in a magnetic field of flux density is

The current in a conductor, or in a beam of electrons or ions can be expressed as charge per unit time.

Where t is the time taken for the charge to move a distance l.

Rewriting the equation for incremental charges and distances andusing vector notation

uB

-----(1)dF dQu B= ×

QIt

=

. . QI l lt

= = . lQt

=Qu

----(2)I dl dQu=


Magnetic force on a current elementEquation(2) shows that an elemental charge dQ moving with velocity . thereby producing convection current element dQ is equivalent to a conduction current element Substituting equation(2) in (1)

If the current I is through a closed path L, the force on the whole loop is

We have the following relations between current elements.

dF I dl B= ×

uuI dl

L

F I dl B= ×∫

Idl KdS Jdv= =

L

F I dl B= ×∫


Magnetic force on a current elementUsing these relations

The total force in these cases is given by

=dF KdS B K BdS= × ×

=dF Jdv B J Bdv= × ×

SF K BdS= ×∫

VF J Bdv= ×∫


Magnetic force between two current elements

1dl

2dl

1I2I

21R

1C2C


Magnetic force between two current elementsConsider two current loops C1 and C2 with currents I1 and I2. The loops are divided in to very small vector line segmentsThe current elements in the loops are denoted byConsider two current elements in the two loops andAccording to Biot-Savart’s law, current elements produce magnetic fields.The magnetic field produced by at is

Hence the force on current element due to the field produced by the current element is

dl

1Idl 2IdlIdl

1Idl2Idl

2120 22 2

21

ˆ4

RI dl adBR

μπ

×=

1Idl 2dB2Idl

1 1 21dF I dl dB= ×211 20 1 2

221

ˆ

4RI dl I dl a

R

μ

π

⎡ ⎤× ×⎣ ⎦=


Magnetic force between two current elements

The total force on current loop 1 due to current loop 2 is

Similarly the total force on loop 2 due to magnetic field produced by loop 1 is obtained by interchanging the subscripts

211 20 1 21 2

21

ˆ

4RI dl I dl a

dFR

μ

π

⎡ ⎤× ×⎣ ⎦=

21

1 2

1 20 1 2

1 221

ˆ

4R

l l

dl dl aI IFR

μπ

⎡ ⎤× ×⎣ ⎦= ∫ ∫

1F

2F 1B

12

1 2

2 10 1 2

2 212

ˆ

4R

l l

dl dl aI IFR

μπ

⎡ ⎤× ×⎣ ⎦= ∫ ∫


Example 1A charged particle of mass 2Kg and charge 3C starts at point(1,-2,0) with velocity in an electric field At time t=1s determine its acceleration, velocity, kinetic energy and position.Solution

ˆ ˆ4 3 /x za a m s+ ˆ ˆ12 10 /x ya a V m+

F ma QE= = QEam

=3 ˆ ˆ12 102 x ya a a⎡ ⎤= +⎣ ⎦

2ˆ ˆ18 15 m/sx ya a= +

( ) ˆ ˆ, , 18 15 x y z x ydu da u u u a adt dt

= = == +

18xdudt

= 118xu t C= +


Example 1

15ydudt

= 215yu t C= +

0zdudt

= 3zu C=

ˆ ˆ 0 4 3x zAt t u a a= = +

118xu t C= +

1 1( 0) 4 4 0 4xu t C C= = +⇒ ⇒= =

2 2( 0) 0 0 0 0yu t C C= = +⇒ ⇒= =

215yu t C= + 3zu C=

3( 0) 3 3 3zu t C⇒ ⇒= = =


Example 1

, ,( )x y zu u u u= (18 4,15 ,3)t t= +ˆ ˆ ˆ( 1 ) 22 15 3 m/sx y zu t s a a a= = + +

21 2

Kinetic energy m u=

( )( )2 2 21 2 22 15 32

= + + 718J=

( , , )dl du x y zdt dt

= = (18 4,15 ,3)t t= +

2118 4 9 4x

dx u t x t t Cdt

= = + ⇒ = + +


Example 1

2215 7.5y

dy u t y t Cdt

= = ⇒ = +

33 3zdz u z t Cdt

= = ⇒ = +

t=0, (x,y,z)=(1,-2,0)At

1 1( 0) 1 1 0 1x t C C= = = ⇒+ =⇒

2 2( 0) 2 2 0 2y t C C⇒ ⇒= = − − = + = −

3 3( 0) 0 0 0 0z t C C= = = ⇒+ =⇒

( )2 2( , , ) 9 4 1, 7.5 2, 3x y z t t t t= + + −


Magnetic torque and momentWhen a current loop is placed in a magnetic field, it experiences a force tending to rotate it.The rotating tendency of the loop is expressed in terms of torque.The torque on the loop is the vector product of the moment arm and the force.

T r F= ×

F

Fr α

α•


Magnetic torque and moment

w

l0F

0 'F

BI

z

12

3 4

⊗ w0F

0F


Magnetic torque and moment

l0F 0 'F

BI

z

12

3

4

⊗

w

0F 0Fα

T

αα


Magnetic torque and momentLet a rectangular loop of length l and width w is placed in a uniform magnetic field

is parallel to along sides 1-2 and 3-4 of the loop.So and so the force acting on these sides is zero.The force acting on lengths 2-3 and 3-4 are

No force is exerted on the loop as a whole.

BIdl B

0Idl B× =

3

0 2F I dl B= ×∫

( )1

4ˆzI dz a B= − ×∫

0 0 'F F F= + = 0IBl IBl= −

0 0 'F F F= +

1

0 4'F I dl B= ×∫

3

2ˆzI dza B= ×∫ IBl=

IBl= −


Magnetic torque and momentWhen the axis of the loop is pivoted at the centre axis, the forces are acting at different points with reference to the axis, and this creates a couple. If the normal to the plane of the loop makes an angle α with (force makes an angle α with moment arm) the magnitude of the torque on the loop is

B

0T F sin wα= BIl w sinα=

BIS sinα= , Area of the plw S loo=

ˆ ,nDefining as the dm I ipole moment oSa f the loop=

ˆ n Unit vector normal to the plane ofa the loop⇒

T m B= × ˆnm ISa=ˆ nT IS a B= ×

IS Bsinα=


Magnetic torque and momentThe magnetic dipole moment is the product of current and area ofthe loop and its direction is normal to the loop.Torque direction is the direction of the axis of rotation. In this case it is along the z direction.

T m B= ×


Magnetic dipoleA bar magnet or a small filamentary current loop is a magnetic dipole.While isolated electric charges exists, it is not possible to have an isolated magnetic pole.But we can obtain an equivalent effect by considering a long barmagnet such that the poles have little interaction.Each pole can, then be treated as isolated poles.Using this concept the Coulomb’s law of force between magnetic poles can be written as

is the pole strength expressed as force per unit flux density.

0 1 2122

12

ˆ4

m mQ QF aR

μπ

=

mQ

mFQB

=


Magnetic dipolePole strength is unity if the pole experiences a force of 1N when situated in a field of magnetic flux density 1 Tesla.An iron bar may be regarded as composed of a large number of very tiny magnets produced by atomic spin and orbital motion.A small bar magnet with pole strength Qm and length l may be treated as a magnetic dipole with magnetic moment Qml . Such a dipole may be compared to a current loop with area S and carrying a current I.This loop has a magnetic moment m=IS.

.mQ l IS=


Magnetic dipole

N

S

Magnet Current Loop

.mQ l IS⇔

s


Magnetic dipoleA bar magnet is composed of a large number of magnetic dipoles.If A is the area of cross section of the bar magnet and l its length, the volume of the magnet is Al.If Qm is the pole strength, the dipole moment is Qm l.Net dipole moment per unit volume is

Dipole moment per unit volume is called magnetisation and it is represented as a vector

Magnetization at any point in the magnetized bar is obtained by

mmQ lAl

QA

=

ˆmn

QM aA

=

ˆ .n Unit vector directed from south to north pa ole⇒

0lim ampere/meterv

mMvΔ →

=Δ


Magnetic dipoleThe magnetization is the net magnetic dipole moment per unit volume If there are N atoms in a given volume Δv and the kth atom has a magnetic moment

A medium for which is not zero everywhere is said to be magnetized.

M

km

M

10

lim

N

kk

v

mM

v=

Δ →=

Δ

∑


Magnetization in materialsAny material is composed of atoms.Each atom is composed of electrons orbiting about a central positive nucleus.The electrons spin about their own axes.So an internal magnetic field is produced by electrons orbiting around the nucleus and electrons spinning about their axes.Both of the electron motion produces internal magnetic field similar to the one produced by a current loop.

+ --


Magnetization in materials

The equivalent loop has a magnetic moment of Without an external field applied to the material the sum of the magnetic moments is zero due to the random orientation of the moments.

iB

bI

ˆb nm I Sa=


Magnetization in materials

When an external magnetic field is applied the magnetic moments of the electrons align themselves with so that the net magnetic moment is not zero.

B

B

vΔvΔ


Magnetic vector potential of a volume distribution of dipoles

O

( , , )x y z

( ', ', ')x y z

'm M dv=

r

'r

'R r r= −

02

ˆ'

4RM adA dv

Rμ

π×

=

V

'dv

03 '

4M R dv

Rμπ×

=



For a differential volume dv’, the magnetic moment isThe vector magnetic potential due to is

'dm Mdv=dm

02

ˆ'

4RM adA dv

Rμ

π×

= 03 '

4M R dv

Rμπ×

=



Let the source dv’ be located at (x’,y’z’) and let (x,y,z) be the point where we want to find the magnetic potential.

1/22 2 2' ( ') ( ') ( ')R r r x x y y z z⎡ ⎤= − = − + − + −⎣ ⎦

3/22 2 2

ˆ ˆ ˆ( ') ( ') ( ')1

( ') ( ') ( ')x y zx x a y y a z z a

R x x y y z z

− + − + −⎛ ⎞∇ = −⎜ ⎟⎝ ⎠ ⎡ ⎤− + − + −⎣ ⎦

ˆ ˆ ˆx y za a ax y zφ φ φφ ∂ ∂ ∂

∇ = + +∂ ∂ ∂

3

1RR R

⎛ ⎞= −∇⎜ ⎟⎝ ⎠

3

RR

= −

3

1 RR R

⎛ ⎞∇ = −⎜ ⎟⎝ ⎠



1/22 2 2' ( ') ( ') ( ')R r r x x y y z z⎡ ⎤= − = − + − + −⎣ ⎦

3/22 2 2

ˆ ˆ ˆ( ') ( ') ( ')1'( ') ( ') ( ')

x y zx x a y y a z z aR x x y y z z

− + − + −⎛ ⎞∇ =⎜ ⎟⎝ ⎠ ⎡ ⎤− + − + −⎣ ⎦

ˆ ˆ ˆ'' ' 'x y za a a

x y zφ φ φφ ∂ ∂ ∂

∇ = + +∂ ∂ ∂

3

1' RR R

⎛ ⎞∇ =⎜ ⎟⎝ ⎠

3

RR

=

3

1' RR R

⎛ ⎞∇ =⎜ ⎟⎝ ⎠



We know that 3

1'RR R

= ∇

0 1' '4

dA M dvR

μπ⎡ ⎤= ×∇⎢ ⎥⎣ ⎦

0 1' '4

A M dvR

μπ

= ×∇∫

1' MR

But ×∇

( ) ( ) F F Fφ φ φ∇ × = ∇ × + ∇ ×

( ) ( ) F F Fφ φ φ∇ × = ∇ × − ∇ ×

( ) ( ) F F Fφ φ φ×∇ = ∇ × − ∇ ×

1 1' 'M MR R

= ∇ × −∇ ×

0 0' ' 4

' '4

,S M MA dubstituting v dvR R

μ μπ π

∇ ×= − ∇ ×∫ ∫



Applying the following vector identity to the second term we get

( )' 'V S

F dv F dS∇ × = − ×∫ ∫( )0 0

' '

ˆ' ' '4 4

n

V SA dv ds

R RM aMμ μ

π π= +

×∇ ×∫ ∫

0 ' A=4 S

KdSComparing withR

μπ ∫0 ' A=

4 S

JdVR

μπ ∫

⎫⎪⎪⎬⎪⎪⎭

bJ M= ∇×

ˆb nK M a= ×

0 0' '

' '4 4V S

b bJ KA dv dsR R

μ μπ π

= +∫ ∫


Bound surface and volume current densities

ˆb nK M a= ×

bJ M= ∇ × b Bound volume currentJ density⇒

b Bound surface current denK sity⇒

Magnetic potential of a magnetic body is due to a volume currentdensity throughout the body and a surface current density on the surface of the body

bKbJ


Bound surface and volume current densities

We have seen that the magnetic properties may be derived by regarding the material as being made up of many small current loops.In the interior of the material, incomplete cancellation of currents in adjacent loops results in a net volume current density.At the surface of the material, the currents are not cancelled, giving a net equivalent surface current density.

bJ

bK


Magnetization in materialsFrom Maxwell’s equation

In a material medium

H J∇× =

f Free volume currentJ density⇒

0f b

B J Jμ

⎛ ⎞∇× = +⎜ ⎟

⎝ ⎠

H M= ∇× +∇× ( )H M= ∇× +

( )0B H Mμ∇× = ∇× +

( )0 (1)B H Mμ= + − − −( )0B H Mμ= +

f bJ J J= +

0

B Jμ

⎛ ⎞∇× =⎜ ⎟

⎝ ⎠

b Bound volume currentJ density⇒


Magnetization in materialsdepends linearly on for linear materials. So we can write

Substituting in equation (1)

M H

mM Hχ=

m Magnetic Susceptibility of the mediumχ ⇒

( )0 mB H Hμ χ= + ( )0 1 m Hμ χ= + Hμ=

( )0 0 1r mμ μ μ χ= +

0 rHμ μ=

1r mμ χ= +

1r mμ χ= +

m Magnetic Susceptibility of the mediumχ ⇒ r Relative Permeability of the mediumμ ⇒


Classification of magnetic materialsMagnetic materials may be classified in to three:

1.Diamagnetic2. Paramagnetic3.Ferromagnetic

Magnetic Materials

Linear Non Linear

Diamagnetics Paragnetics Feromagnetics0, 1m rχ μ< < 0, 1m rχ μ> > 0, 1m rχ μ

1r mμ χ= + m Magnetic Susceptibility of the mediumχ ⇒

r Relative Permeability of the mediumμ ⇒


Properties of diamagnetic materialsDiamagnetism occurs in materials where the magnetic fields due to orbital and spin motion of electrons completely cancel.When placed inside a magnetic field, it gets feebly magnetized in direction opposite to that of the magnetic field and reduces the net field.The magnetic susceptibility is small.A diamagnetic material is feebly repelled by a magnet.A rod of the material suspended in a magnetic field gets slowly set at right angles to the direction of the field.

has a small negative value.μr is just less than 1Behaviour is not affected by temperature.Does not obey Curies law,

M

1/m Tχ ∝


Properties of paramagnetic materialsParamagnetism occurs in materials where the magnetic fields produced by orbital and spin motion of electrons do not completely cancel. On application of external magnetic field it gets magnetised along the direction of the applied field and increases the net field.When subjected to non uniform magnetic field it moves from weaker part of the magnetic field to stronger part.A rod of paramagnetic material suspended in a magnetic field slowly gets along the direction of external magnetic field.A paramagnetic material is feebly attracted by a magnet.

has a small positive value.μr is just less than 1Susceptibility has a small positive value.It tends to lose its magnetism if subjected to temperature.Obeys Curies law,

M

1/m Tχ ∝


Properties of ferromagnetic materialsFeromagnetism occurs in materials whose atoms have relatively large permanent magnetic moment.On application of external magnetic field it gets strongly magnetisedalong the direction of the applied field and retains its magnetism when removed from the fieldThey are strongly attracted by a magnet.On being subjected to non uniform magnetic field it moves from weaker part of the magnetic field to stronger part.

has a large positive value.μr is very large.Value of flux density inside the material is very much greater than in vacuum.Susceptibility has a large positive value.Ferromagnetism decreases with temperature. At a critical temperature called Curie temperature, ferromagnetic properties disappears and the material becomes paramagnetic.

M


Magnetisation curveFor all materials the relationship between can be expressed as

For a ferromagnetic material, relationship between depends on previous magnetization.So it can only be represented by a magnetization curve.

( )0B H Mμ= +

B Hand

B Hand


Magnetisation curve

maxB

rB

maxB−

rB−

B

HmaxHcHcH−maxH−

O

P

Q

Initial magnetizationcurve

Hysteresis loop

- Typical B H curve


Magnetization curveAssume that initially the ferromagnetic material is unmagnetized.If H is increased by increasing current, from 0 to Hmax, curve OP is produced.This curve is the initial magnetization curve.After reaching saturation at P, if H is decreased, B does not follow the initial curve but lags behind H.This phenomenon is called hysteresis.If H is reduced to zero, B is not reduced to zero but to Br, which is called remnant flux density.The existence of Br is the reason for the formation of permanent magnets.If H increases negatively, B becomes zero when H becomes Hc, which is known as coercive field intensity.


Magnetization curveFurther increase of H in the negative direction to reach Q and then to P gives a closed curve called hysteresis loop.The shape of hysteresis loops varies from one material to another.Some ferrites have an almost rectangular hysteresis loop and areused in digital computers as magnetic memories.The area of the hysteresis loop gives the energy loss called hysteresis loss per unit volume per cycle of magnetization.This energy loss is in the form of heat.So materials used in magnetic devices must have tall but narrow hysteresis loops, so that hysteresis losses are minimum.


Magnetic boundary conditions

1Medium 1 ( )μ

2Medium 2 ( )μ

1nB

2nB

1tB

2tB

1B

2B

1nB

2nB

ˆNaSΔ

2h⎫ Δ

⎬⎭

2h⎫ Δ

⎬⎭


Magnetic boundary conditionsConsider the boundary between two isotropic homogeneous linear materials with permeabilities μ1 and μ2. At the boundary, the magnetic flux density may be decomposed in to a normal component Bn and a tangential component Bt.Applying Gauss’s law for magnetic field on a small pill box shaped volume,

0SB dS⋅ =∫

B

1 2 0n nB S B SΔ − Δ =

1 2n nB B=

Normal component of B is continuous accross the boundary .between the two adjacent media

1 2n nB B=

( )2 1 ˆ. ., 0Ni e B B a− ⋅ = ( )2 1 ˆ 0NB B a− ⋅ =



2 1B B−ˆNa

( ) ( )2 1 2 1 .1. sˆ coN BB a BB α− ⋅ = −

1 2 0n nB B= =−

α



The normal components of are not continuous across the boundary The normal components becomes continuous only when μ1 = μ2. In that case there is no boundary between the two materials.

1 2n nB B=

1 1 2 2n nH Hμ μ=1 2

2 1

n

n

HH

μμ

=

12 1

2n nH Hμ

μ=

12 1

2

. ., n ni e H Hμμ

=

H



1Medium 1 ( )μ

2Medium 2 ( )μ

ˆNa1tH

2tH

C

2h⎫ Δ

⎬⎭

2h⎫ Δ

⎬⎭

K

lΔ

ûa

1nH

2nH

1tH

2tH

1H

2H


Magnetic boundary conditionsNow construct a closed contour C across the boundary and apply Ampere’s circuital law for this closed path.

enclCH dl I⋅ =∫

2 2 1 1 1 22 2 2 2t n n t n n uh h h hH l H H H l H H K lΔ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ + + − Δ − − = Δ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 1t t uH l H l K lΔ − Δ = Δ

2 1t t uH H K− =

Tangential component of H is discontinuous across the boundary .between the two adjacent media

uK Component of K normal to the⇒ .plane of the closed path

2 1t t uH H K− =



The above equation may be written in vector notation as

If both the media across the boundary are not conductors, then

2 1t t uH H K− =

( )2 1 ˆNH H a K− × = ( )2 1 ˆ NH H a K− × =

0K =

2 1 0t tH H− =

2 1t tH H=

Tangential component of H is continuous accross the boundary 0 between the two adja icent med fia K =

2 1t tH H=



2 1H H−ˆNa

α

( ) ( )2 12 1 ˆ.1.ˆ .N uH H a H H sin aα= −− ×

( )2 1 ˆ ˆt t u u uH H a K a K= − = =

ûa


Magnetic boundary conditions2 1

2 1

t tB Bμ μ

=

22 1

1t tB Bμ

μ=

22 1

1t tB Bμ

μ=

Tangential component of B is discontinuous accross the boundary 0 between the two adja icent med fia K =


Magnetic boundary conditions1Medium 1 ( )μ

2Medium 2 ( )μ

1tB

2tB

1B

2B

1nB

2nB

1θ

2θ

1 2 ----(1)n nB B= 2 1t tH H=

1 2

1 2

----(2)t tB Bμ μ

=

1 2

1 1 2 2

(2) / (1) ----(3)t t

n n

B BB Bμ μ

⇒ =

1 1 2 2

1 1 1 2 2 2

sin sin( cos ) ( cos )B BB B

θ θμ θ μ θ

=

1 2

1 2

tan tanθ θμ μ

=

1 1

2 2

tantan

θ μθ μ

=

1 1

2 2

tantan

θ μθ μ

=

1tB


Inductance

N

I


InductanceA closed loop carrying a current I produces a magnetic field which causes a flux Ψ according to

If the loop has N turns the total flux linkage λ is defined as

According to Biot-Savart’s law B is proportional to the current I and hence flux Ψ is proportional to I.Flux linkage λ is thus proportional to current I.

Here L is a proportionality constant called the inductance of the circuit.

SB dSψ = ⋅∫

Nλ ψ=

Iλ ∝ LIλ =


Inductance

Inductance L of an inductor is defined as the ratio of magnetic flux linkage λ to the current I through the inductor. Its unit is Henry.

The inductance defined thus is the self inductance of the coil since the flux linkages are produced by the inductor itself.The magnetic energy stored in an inductor is given by

So inductance may be considered as a measure of the magnetic energy that can be stored in an inductor.

LIλ = LIλ

=

NLI Iλ ψ

= =

212mW LI= 2

2 mWLI

=


Mutual Inductance

N

I

N

I

1I

2I

1N2N11ψ 22ψ

12ψ

21ψ

Circuit 1 Circuit 2

12 1 12Nλ ψ=21 2 21Nλ ψ=

1212

2

1 12

2

MI

NI

λ ψ= = 21

211

2 21

1

MI

NI

λ ψ= =

( )111 2

11

1

1

11LI

NI

ψ ψλ += =

( )222 2

22

2

2

21LI

NI

ψ ψλ += =


Mutual InductanceIf there are two circuits carrying currents I1 and I2 a magnetic interaction exists between these circuits.Four component fluxes are produced

If is the field due to I2 and S1 is the area of the circuit 1, then

11 1 1 Flux passing through circuit due to current Iψ ⇒




112 2S

B dSψ = ⋅∫2B


Mutual Inductance

Mutual inductance M12 is defined as the ratio of the flux linkage

on circuit 1 due to current I2.

Similarly

If the medium is linear M12=M21

221 1S

B dSψ = ⋅∫

12 1 12Nλ ψ=

1212

2

1 12

2

MI

NI

λ ψ= =

2121

1

2 21

1

MI

NI

λ ψ= =

1 1212

2

NMIψ

=

2 2121

1

NMIψ

=


Self Inductance of coupled coilsThe self inductances of circuits 1 and 2 are given by

( )111 2

11

1

1

11LI

NI

ψ ψλ += =

1 11

1

NLIψ

=

( )222 2

22

2

2

21LI

NI

ψ ψλ += = 2 2

22

NLIψ

=

1 11 12= ψ ψ ψ+

2 21 22 =ψ ψ ψ+


Procedure for calculating Inductance1. Choose an appropriate coordinate system2. Assume that the inductor carries a current I3. Find the magnetic flux density using Ampere’s circuital

law if symmetry exists; otherwise use Biot-Savart’s law.4. Find the flux linking each turn using5. Find the inductance using

SB dSψ = ⋅∫NL

Iψ

=


Magnetic field by a solenoidWhen the coils of the solenoid are closely spaced, each turn can be regarded as a circular loop, and the net magnetic field is the vector sum of the magnetic field for each loop. This produces a magnetic field that is approximately constant inside the solenoid, and nearly zero outside the solenoid.


Magnetic field by a solenoidThe ideal solenoid is approached when the coils are very close together and the length of the solenoid is much greater than itsradius. Then we can approximate the magnetic field as constant inside and zero outside the solenoid.

H

Apply Ampere’s law for the closed path 1234

N turns in a length of l

ˆxa

l


Magnetic field by a solenoid

12 23 34 41L

H dl H dl H dl H dl H dl⋅ = ⋅ + ⋅ + ⋅ + ⋅∫ ∫ ∫ ∫ ∫

enclLH dl I⋅ =∫

0 0H l H l= + + =

= C urren t enc losed by the pa th is N I

H l N I=

N IHl

=

N IBl

μ=

ˆ xN IB al

μ=

times number of turns per unit lengthNB I Il

μ μ= =

ˆ xN IH al

=


Self inductance of a solenoidFind the self inductance per unit length of an infinitely long solenoid.

S

N IB per unit leng thl

μ=

l


Self inductance of a solenoidThe magnetic field inside an infinitely long solenoid is

If S is the cross sectional area of the solenoid, then the total flux through this solenoid is

Hence the inductance per unit length is

N IBl

μ=

SB dSψ = ⋅∫ S

B dS= ∫ B S=

N I Sl

ψ μ=

NLIψ

=2N I S

lIμ=

2N Sl

μ=

2L N Sμ=

2L N Sμ=


Magnetic energyConsider a differential volume Δv in a magnetic field of flux density as shown below.Let the top and bottom faces be covered with thin metal foils that carry a current ΔI

B

xy

z

xΔyΔ

zΔB

IΔ

IΔ


Magnetic energyThe inductance of this differential volume is

The energy stored in this elemental inductance is

Substituting (1) in (2)

But

LIψΔ

Δ =Δ

B x zI

Δ Δ=

Δ-----(1)H x z

Iμ Δ Δ

=Δ

21 -----(2)2mW L IΔ = Δ Δ

2

21

mx zW H I

Iμ Δ Δ

Δ = ΔΔ

1 2

H x z Iμ= Δ Δ Δ

I H l H yΔ = Δ = Δ

21 2mW H x y zμΔ = Δ Δ Δ

21 2mW H vμΔ = Δ


Magnetic energyMagnetic energy density wm is defined as

Or

lim mm v

WwvΔ →∞

Δ=

Δ21 lim

2m v

Hw vv

μΔ →∞

= ΔΔ

212

Hμ=

12mw HHμ= 1

2B H= ⋅

( )2 31 /2mw H J mμ=

( )31 /2mw B H J m= ⋅

1 .2m

Bw Bμ

=2

2Bμ

=2

3( / )2mBw J mμ

=


Magnetic energyThe total energy in a magnostatic field is

m mVW w dv= ∫

21 2m V

W H dvμ= ∫1 2m V

W B Hdv= ⋅∫

21 2m V

W H dvμ= ∫

1 2m V

W B H dv= ⋅∫21 2 2m V

BW dvμ

= ∫21

2 2m V

BW d vμ

= ∫


Example 1Find the self inductance of a coaxial cable of inner radius a and outer radius b.

a

b

1l m=

I

I

a

b


Example 1

212mW LI=

2

2, mo WLrI

=

1 2m V

W B Hdv= ⋅∫2

2V

B dvμ

= ∫

1 22IBa

μ ρπ

=

2 2IB μπρ

=

The inductance produced by the flux internal to the conductor2

12

1in V

L B dvI μ

= ∫2 2 2

2 2 4

14V

I d d dzI a

μ ρ ρ ρ φμ π

= ∫3

2 44 Vd d dz

aμ ρ ρ φπ

= ∫2 3

2 4 0 0 04l a

d dz da

πμ φ ρ ρπ

= ∫ ∫ ∫ 8lμπ

= 8intlL μπ

=

2

2

22V

BL dvI μ

= ∫


Example 1

222

1ext V

L B dvI μ

= ∫2 2

2 2 2

14V

I d d dzI

μ ρ ρ φμ π ρ

= ∫

2

14 V

d d dzμ ρ φπ ρ

= ∫2

2 0 0

14

l b

adz d d

πμ φ ρπ ρ

= ∫ ∫ ∫ ln2

l ba

μπ

⎛ ⎞= ⎜ ⎟⎝ ⎠

ln2ext

l bLa

μπ

⎛ ⎞= ⎜ ⎟⎝ ⎠

int extL L L= + 1 ln2 4

l ba

μπ⎡ ⎤⎛ ⎞= + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ 1 ln

2 4l bL

aμπ⎡ ⎤⎛ ⎞= + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦


Magnetic circuitsMagnetic devices such as toroids, transformers, motors, generators and relays may be considered as magnetic circuits.The analysis of such circuits is made simple if an analogy between magnetic and electric circuits is established.

turnsNI

B

SΔmmf NI=F


Magnetic circuitsThe electrostatic potential is related to electric field intensity by

Magnetic potential is related to magnetic field intensity by

Ohm’s law in point form for electric circuit has the form

Magnetic flux is analogous to electric current and magnetic fluxdensity is analogous to current density.Corresponding to ohm’s law in electric circuits, we have the relation

To find the total current we useTo find the total flux we use

E V= −∇

mH V= −∇

J Eσ=

B Hμ=S

I J dS= ⋅∫

SB dSψ = ⋅∫


Magnetic circuitsIn electric circuits we defined resistance as the ratio of voltage and current.

Similarly, in magnetic circuits reluctance is the ratio of magneto motive force and the total flux.

The resistance R is given byCorrespondingly reluctance

The magneto motive force is

VRI

=

mVψ

ℜ =lRAσ

=

lAμ

ℜ =

NIF =


Magnetic circuitsIn electric circuits we have the voltage source is part of the closed path here.But in magnetic circuitsHere magneto motive force is linked to the magnetic circuit without becoming part of the closed circuit. Based on this we can write Kirchhoff’s voltage and current laws for electric and magnetic circuits:

For n magnetic circuits in seriesFor n magnetic circuits in parallel

0LE dl⋅ =∫

enclLH dl I⋅ =∫

LH dl Nor I⋅ =∫

0V IR− =∑ ∑ 0NI ψ− =ℜ∑ ∑0I =∑ 0ψ =∑

1 2 3 ...... nψ ψ ψ ψ= = = =1 2 3 ...... nψ ψ ψ ψ ψ= + + + +


Example 1The torroidal core of a coil has a=10cm and a circular cross section with b=1cm . If the core is made of steel (μr=1000) and has a coil with 200 turns,calculate the amount of current that will produce a flux of 0.5mWb in the core.

a

2b

ψI

200 turns

NI

ψ

ℜ


Example 1In the analogous magnetic circuit

NI ψ= ℜlNIA

ψμ

= 20

2

r

ab

πψμ μ π

=

20

2

r

aIN bπ ψ

μ μ π=

3

7 2

2 0.2 0.5 104 10 1000 200 (0.01)

Iπ

−

−

× × ×=

× × × ×7.96A=


Faraday’s lawWhenever the magnetic flux linking with a closed circuit varies with time an emf is induced in that circuit and the induced emf is equal to the time rate of change of the magnetic flux linkage by the circuit.

The negative sign shows that the induced emf acts in such a way as to oppose the flux producing it. This is called Lenz’s law.It shows that the direction of current in the circuit is such that the induced magnetic field produced by the induced current opposes the original magnetic field.

emfdV Ndtψ

= −

Number of turns in the c tN ircui⇒ Flux through each turnψ ⇒


Faraday’s law

For a single turn

The variation of flux with time may be produced in the followingways:

A stationary loop area in time varying magnetic field.A time varying loop area in a in a static magnetic field.A time varying loop area in a in a time varying magnetic field

emfdV Ndtψ

= −

emfdVdtψ

= −S

d B dSdt

= − ⋅∫

emf L S

dV E dl B dSdt

= ⋅ = − ⋅∫ ∫

=S

Since B dSψ ⋅∫


Stationary loop in time varying magnetic field.

( )Increasing B t ( )Decreasing B t

I I

'B 'B


Stationary loop in time varying magnetic field.

In the case of stationary loop and time varying magnetic field, we can modify the above equation.

This equation is called transformer equation since the induced emf is due to transformer action.Applying Stoke’s theorem

This is one of Maxwell’s equations for time varying fields.

emf L S

dV E dl B dSdt

= ⋅ = − ⋅∫ ∫

emf L S

BV E dl dSt

∂= ⋅ = − ⋅

∂∫ ∫

( )S S

BE dS dSt

∂∇× ⋅ = − ⋅

∂∫ ∫BEt

∂∇× = −

∂

BEt

∂∇× = −

∂


Moving loop in a static magnetic field.When a conducting loop is moving in a static magnetic field, an emfis induced in the loop.The force on a charge moving with uniform velocity in a magnetic field is

Motional electric field is

The emf induced in the moving loop is

This type of emf is called motional emf.

mF Qu B= ×m

mFE u BQ

= = ×mE

uB

( )emf mL LV E dl u B dl= ⋅ = × ⋅∫ ∫


Moving loop in a static magnetic field.

( ) ( )( )mS SE dS u B dS∇× ⋅ = ∇× × ⋅∫ ∫

' Applying Stoke s theorem

( )mE u B∇× = ∇× ×( )mE u B∇× = ∇× ×

( )emf mL LV E dl u B dl= ⋅ = × ⋅∫ ∫


Moving loop in a time varying magnetic field.In this case both the transformer and motional emf are present .

Applying Stoke’s theorem,

( )emf L S L

BV E dl dS u B dlt

∂= ⋅ = − ⋅ + × ⋅

∂∫ ∫ ∫

( ) ( )( )S S S

BE dS dS u B dSt

∂∇× ⋅ = − ⋅ + ∇× × ⋅

∂∫ ∫ ∫

( )BE u Bt

∂∇× = − +∇× ×

∂

( )BE u Bt

∂∇× = − +∇× ×

∂


Displacement currentMaxwell’s equation for magnetic fields derived from Ampere’s circuital law is given by

For any vector, divergence of the curl is always zero.

But the continuity equation states that

Equations 1 and 2 are incompatible for time varying conditions. So we have to modify equation 1 to agree with equation 2.For this we will add an additional term to equation 1 so that itbecomes

H J∇× =

( ) 0 (1)H J∇⋅ ∇× = ∇⋅ = − − −

(2)vJtρ∂

∇⋅ = − − − −∂

(3)dH J J∇× = + − − −


Displacement currentApplying the condition that divergence of the curl is always zero to the equation 3,

( ) ( ) =0dH J J∇⋅ ∇× = ∇⋅ +

( ) = =0d dJ JJ J ∇⋅ +∇⋅∇ ⋅ +

vdJ J

tρ∂

∇⋅ = −∇⋅ =∂

( )Dt∂

=∂

∇⋅Dt

∂⋅=∂

∇

dDJt

∂=∂ 3 Equation is now modified as

(4)DH Jt

∂∇× = + − − −

∂

DH Jt

∂∇× = +

∂


Displacement currentEquation 4 is Maxwell’s equation derived from ampere’s law for a time varying field.The term is known as displacement current density. Based on displacement current density, we can define the displacement current as

At low frequencies is very small when compared with At high frequencies the two terms are comparable.Without the term electromagnetic wave propagation would be impossible.

/D t∂ ∂

d dI J dS= ⋅∫D dSt

∂= ⋅

∂∫dJ J

dJ


Displacement current

I

1S

I

2SL L


Displacement currentDisplacement current is a result of time varying electric field.An example is the current through a capacitor, when an alternating voltage source is applied across its plates.Consider that we are applying the unmodified form of Maxwell’s equation to calculate the current through a capacitor considering a closed path as shown in figure. In the first case

In the second case

since no conduction current flows through S2

1enclL S

H dl J dS I I⋅ = ⋅ = =∫ ∫

2

0enclL SH dl J dS I⋅ = ⋅ = =∫ ∫


Displacement currentThis anomaly is corrected if we include the displacement current in the above equation.

( )2

dL SH dl J J dS⋅ = + ⋅∫ ∫

2dS

J dS= ⋅∫2S

D dSt

∂= ⋅

∂∫

2S

d D dSdt

= ⋅∫dQdt

= I=


Maxwell’s Equations in final forms

VD ρ∇⋅ =

0B∇⋅ =

BEt

∂∇× = −

∂

VS VD dS dVρ⋅ =∫ ∫

0SB dS⋅ =∫

L S

dE dl B dSdt

⋅ = − ⋅∫ ∫

L S

DH dl J dSt

⎛ ⎞∂⋅ = + ⋅⎜ ⎟∂⎝ ⎠

∫ ∫

Differential form Integral form Derived from

' Gauss s Law

Nonexistance of magneticMonopole

' Faraday s Law

' Ampere s Law DH J

t∂

∇× = +∂ modified by continuity eqn


Time-harmonic fieldsA time harmonic field is one that varies periodically or sinusoidallywith time.In most practical applications EM fields are time harmonic.Also, any wave form can be expressed in terms of sinusoids usingFourier transform techniques.Sinusoids are easily expressed in phasors, which are more convenient to work with.A phasor z is a complex number that can be written as

Rectangularz o mx rjy f= + ⇒

2 2 , r z xTh yen = = +

( )cos r j sinφ φ= +jz re φ=-1, a ytan

xnd φ =

Polar oz r f rmφ= ∠ ⇒


Time-harmonic fieldsAddition and subtraction of phasors are better performed in rectangular form while multiplication and division are better done in polar form.Let 1 1 1z x j y= + 2 2 2z x j y= +

( ) ( ) ( )1 2 1 2 1 2 z z x x j y y+ = + + +

( ) ( ) ( )1 2 1 2 1 2 z z x x j y y− = − + −

( )1 2 1 2 1 2z z r r φ φ= ∠ +

( )1 11 2

2 2

z rz r

φ φ= ∠ −

( )1 1 1 / 2z r φ= ∠


Time-harmonic fieldsTo introduce the time element let

A sinusoidal current equals the real part of

The sinusoidal current equals the imaginary part of

tφ ω θ= +( ) j jt tjjNow re re re eω ωθ θφ + ==

{ } ( )Re cosjre r tφ ω θ= +

{ } ( )Im jre rsin tφ ω θ= +

( )0( ) cosI t I tω θ= +

0j j tI e eθ ω

( )0( )I t I sin tω θ= +

0j j tI e eθ ω


Time-harmonic fieldsThe complex term which results from dropping the time factor in I(t) is called the phasor current and is denoted by Is.

0jI e θ

j te ω

0 0j

sI I e Iθ θ= = ∠

0( ) cos( ) I t I tω θ= + can be expressed as { }( ) j tSI t Re I e ω=


Time-harmonic fieldsA phasor can be a scalar or a vector.If a vector is a time harmonic field, the phasor form of , is

As an example let

Comparing (1) and (2), the phasor form of is

( , , , )A x y z tA ( , , )SA x y z

{ } (1)j tSA Re A e ω= − − −

0 ˆcos( ) yA A t x aω β= −

{ }0 ˆ (2)j x j tyA Re A e a eβ ω−= − − −

A

0 ˆj xS yA A e aβ−=


Time-harmonic fields

This shows that taking the time derivative of the instantaneousquantity is equivalent to multiplying its phasor form by jω

Similarly,

{ }j tSA Re A e ω=

{ }j tS

A Re A et t

ω∂ ∂=

∂ ∂{ } j t

SRe j A e ωω=

SA j At

ω∂=

∂

SAA tjω

∂ =∫


Time-harmonic fieldsWe can apply the concepts of phasors to time varying EM fields. The following field quantities and their derivatives can be expressed in phasor form.

Based on these quantities we can rewrite the Maxwell’s equations for time-harmonic EM fields.

( , , , )E x y z t( , , , )D x y z t

( , , , )H x y z t( , , , )B x y z t

( , , , )J x y z t( , , , )V x y z tρ


Maxwell’s Equations for time harmonic fields

S VSD ρ∇⋅ =

0SB∇⋅ =

S SE j Bω∇× = −

S VSS VD dS dVρ⋅ =∫ ∫

0SSB dS⋅ =∫

S SL SE dl j B dSω⋅ = − ⋅∫ ∫

( )S S SL SH dl J j D dSω⋅ = + ⋅∫ ∫

Differential form Integral form

S S SH J j Dω∇× = +

Date post:	10-Apr-2015
Category:	Documents
Upload:	aravind
View:	751 times
Download:	12 times

EMT Electromagnetic Theory MODULE III

Documents