Field Theory Magnetostatics

ENEL2FT

FIELD THEORY

Magnetostatic Fields

Dr. Thomas Afullo Ukzn, Durban

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ENEL2FT

FIELD THEORY

Magnetostatic Fields

REFERENCES 1. M.N. Sadiku: Elements of Electromagnetics, Oxford University Press, 1995, ISBN 0-19-510368-8. 2. P. Lorrain, D. Corson: Electromagnetic Fields and Waves, W.H. Freeman & Co, 1970, ISBN: 0-7167-0330-0. 3. David T. Thomas: Engineering Electromagnetics, Pergamon Press, ISBN: 08016778-0.

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ENEL2FT

FIELD THEORY

Magnetostatic Fields

INTRODUCTION As we have noticed, an electrostatic field is produced by static or stationary charges. If the charges are moving with a constant velocity, a static magnetic field (or magnetostatic field) is produced. There are two major laws governing magnetostatic fields: - The Biot-Savart Law - Ampere’s Circuital law.

Like Coulomb’s law, the Biot-Savart law is the general law of magnetostatics. Just as in Gauss’s law, Ampere’s circuital law is a special case of Biot-Savart law and is easily applied in problems involving symmetrical current

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FIELD THEORY

Magnetostatic Fields

MAGNETIC FORCES It is common experience that circuits carrying electric currents exert forces on each other. For example, the force between two straight parallel wires carrying currents Ia and Ib is proportional to IaIb/ρ, where ρ is the distance between the wires. Ia

dla

r

Ib

dlb

4

MAGNETIC FORCES The force is attractive if the currents flow in the same direction, and it is repulsive if they flow in opposite directions. For the more general case shown in the above figure, the force between the current-carrying conductors is given by:

 µ I I dl x( dl xrˆ ) F = ∫∫ 4π r o

ab

a

b

b

a b

a

1

2

This is the force exerted by current Ia on current Ib, and the line integrals are evaluated over the two circuits. This is the magnetic force law. The vectors dla and dlb point in the direction of current flow, r is the distance between the two elements dla and dlb, and r1 is the unit vector pointing from dla to dlb. Magnetostatic Fields ENEL2FT FIELD THEORY -7 The constant, µo=4πx10 H/m, is the permeability of

5

MAGNETIC FORCES The force Fab can be expressed in symmetrical form by expanding the triple vector product under the integral sign:

dl x( dl xrˆ ) dl ( dl .rˆ ) rˆ ( dl .dl = − r r r b

a

1

a

2

b

1

1

a

2

b

)

2

To show that the double integral of the first term on the right is zero, we note that:

( dl ( dl .rˆ ) dl .rˆ ) = ∫ dl ∫ ∫∫ r r a

a b

b

2

1

b

a

a

b

1

2

This is the integral of dr/r2 around a closed curve, circuit b; which implies that the upper and lower limits of integration are identical. It is therefore zero. Magnetostatic Fields

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FIELD THEORY

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MAGNETIC FORCES We are thus left with the double integral of only the second term for the triple vector product. Thus:

 µ I I rˆ ( dl .dl F =− ∫∫ 4π r o

a

b

1

ab

a

b

)

2

a b

Despite the fact that the above integral for Fab is simpler and more symmetrical than that involving the triple vector product, it is not as useful. This is because in the above integral, the force cannot be expressed as the interaction of the current b with the field in a.  µo I a I b dlb x dla xrˆ1  Thus we use Fab =the earlier I b ∫obtain: dlb xBa ∫ ∫ relationship= to

(

4π r  µ I ( dl xrˆ ) B = ∫ 4π r a b

o

a

Magnetostatic Fields

a

a

a

)

2

b

1

2

ENEL2FT

FIELD THEORY

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MAGNETIC FORCES The vector Ba is called the magnetic induction due to the circuit a at the position of the element dlb of circuit b. Therefore the element of force, dF, on an element of wire of length dl carrying a current I in a region   is B is given by:  induction where the magnetic

dF = Idl xB

If the current I is distributed in space with a (free)  dv’ the elemental current density Jf, then, with  µo J f xrˆ1 volume, we have:

B=

Magnetostatic Fields

4π

∫∫∫ v'

(

r2

ENEL2FT

)dv'

FIELD THEORY

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MAGNETIC FLUX As in electrostatics, where we use lines of force to describe an electric field, we can describe a magnetic field by drawing lines of B that are everywhere tangent to the direction of B. It is convenient to use the concept of flux, the flux of the magnetic induction B through a surface S being defined as the normal component of B integrated over S:

  Φ = ∫ B.dS S

The flux Φ is expressed in webers. Magnetostatic Fields

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FIELD THEORY

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FIELD THEORY

Magnetostatic Fields

BIOT-SAVART’S LAW The Biot-Savart’s law states that the magnetic field intensity, dH, produced at a point P shown in the figure below by the differential current element Idl is proportional to the product Idl and the sine of the angle α between the element and the line joining P to the element, and is inversely proportional to the square of  the distance R between P and the element. dl

α Current I

 R P

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FIELD THEORY

Magnetostatic Fields

BIOT-SAVART’S LAW That is: dH ∝

Idl sin α kIdl sin α ⇒ dH = R2 R2

Here, k is the constant of proportionality. In SI units, the above equation becomes: dH =

Idl sin α 4πR 2

From the definition of cross product, it is seen that, in    vector form,

( )

( )

 Idl xaˆ R Idl xR dH = = 2 4πR 4πR 3

Here, aR is the unit vector in the direction of vector R. 11

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FIELD THEORY

Magnetostatic Fields

BIOT-SAVART’S LAW Just as we have different charge configurations, we can have different current distributions: the line current, the surface current, and the volume current, as shown below.   K

I dl  I

Line current density

 J  KdS

Surface current density

 Jdv

Volume current density

If we define K as the surface current density (A/m2), and    3 J as the volume current density I dl = KdS = J(A/m dv ), then we have: 12

ENEL2FT

FIELD THEORY

Magnetostatic Fields

BIOT-SAVART’S LAW Thus in terms of the distributed source currents , the Biot-Savart law becomes:

  Idl xaˆ R H=∫ (line current ) 2 L 4πR   ( KdS ) xaˆ R H=∫ ( surface current ) 2 4πR S   ( Jdv ) xaˆ R H =∫ (volume current ) 2 v 4πR

( )

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FIELD THEORY

Magnetostatic Fields

FIELD DUE TO A STRAIGHT CURRENTCARRYING CONDUCTOR z

B dl

z

α2 α

I

Aα

 R

1

O

ρ

P

Consider the field due to a straight current-carrying filamentary conductor of finite length AB, as shown above. 14

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FIELD THEORY

Magnetostatic Fields

FIELD DUE TO A STRAIGHT CURRENTCARRYING CONDUCTOR Let us assume the conductor is along the z-axis, with its upper and lower ends respectively subtending angles α1 and α2 at P, the point at which the magnetic field strength, H, is to be determined. If we consider the contribution dH at P due to an   element dl at (0,0,z), from Biot-Savart’s law:  we Idl have, xR dH = 4πR 3   dl = dzzˆ; R = ρρˆ − zzˆ   ⇒ dl xR = ρdzϕˆ  Iρdz ∴H = ∫ ϕˆ 3 / 2 4π ρ 2 + z 2

( )

(

)

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FIELD THEORY

Magnetostatic Fields

FIELD DUE TO A STRAIGHT CURRENTCARRYING CONDUCTOR Let us make the following substitutions: Let z = ρ cot α ⇒ dz = − ρ cos ec 2αdα  I α 2 ρ 2 cos ec 2αdα 1 α2 ∴H = − ϕˆ = ∫ ∫ sin αdαϕˆ 3 3 4π α1 ρ cos ec α 4πρ α1  I ( cosα 2 − cosα1 )ϕˆ ∴H = 4πρ Note that H is always along the unit vector ϕ, irrespective of the length of the wire or the point of interest P.

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FIELD THEORY

Magnetostatic Fields

FIELD DUE TO A STRAIGHT CURRENTCARRYING CONDUCTOR – EXAMPLE The conducting triangular loop below carries a current of 10 A. Find the magnetic filed intensity H at (0,0,5) due to side 1 of the loop. y

1 3

2 10A

1

1

2

x

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FIELD DUE TO A STRAIGHT CURRENTCARRYING CONDUCTOR – EXAMPLE Solution: The problem can be solved using the following figure:

 R

x

1

 Idl

z

γ

α

P(0,0,5 )

y

Magnetostatic Fields

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FIELD THEORY

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FIELD DUE TO A STRAIGHT CURRENT-CARRYING CONDUCTOR – EXAMPLE From the Biot-Savart law, we obtain:   Idl xaˆ R dH = 4πR 2  dl = xˆdx; aˆ R = − xˆ cos α + zˆ sin α

sin α =

z

(z2 + x2 )

; cos α = 3/ 2

x

( z 2 + x 2 )3 / 2

z ⇒ z = x tan α x Idxxˆx( − xˆ cos α + zˆ sin α )

∴ tan α =  ∴ dH =

(

4π z 2 + x 2

)

=−

I sin αdxyˆ

(

4π z 2 + x 2

)

=−

(

Izdxyˆ

4π z 2 + x 2

)3 / 2

x ⇒ x = z tan γ z dx d z 2 = z ( tan γ ) = = z sec γ dγ dγ cos 2 γ

tan γ =

∴ dx = z sec 2 γdγ Magnetostatic Fields

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FIELD THEORY

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FIELD DUE TO A STRAIGHT CURRENT-CARRYING CONDUCTOR – EXAMPLE Therefore we obtain:

( z 2 + x 2 )3 / 2 = ( z 2 + z 2 tan 2 γ )3 / 2 = z 3 (1 + tan 2 γ )3 / 2 = z 3 sec3 γ  ∴ dH = −

(

Izdxyˆ 2

4π z + x

)

2 3/ 2

I z 2 sec 2 γdγ I = − yˆ = − yˆ cos γdγ 3 3 4π z sec γ 4πz

−1  tan ( 2 / z ) I ˆ H= −y cos γdγ = − ∫ 4πz 0  I H = − yˆ sin tan −1 ( 2 / z ) 4πz  I H = − yˆ sin tan −1 ( 2 / 5) ( 0, 0 ,5) 4πz

(

)

(

Magnetostatic Fields

yˆ

I sin γ 4πz

tan −1 ( 2 / z ) 0

) = −0.0591yˆ A / m

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FIELD THEORY

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FIELD DUE TO A STRAIGHT CURRENT-CARRYING CONDUCTOR – EXAMPLE 1. Find H due to side 3 of the rectangular loop

Ans :  H = −0.03063xˆ + 0.03063 yˆ A / m

Magnetostatic Fields

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FIELD THEORY

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FIELD THEORY

Magnetostatic Fields

FIELD DUE TO A STRAIGHT CURRENT-CARRYING CONDUCTOR As a special case, when the conductor is semi-infinite (with respect to P), point A is now at O(0,0,0), while point B is at (0,0,∞); then α1=90o, α2=0o. Then we have:

   µI I H= ϕˆ ; B = µH = ϕˆ 4πρ 4πρ Another special case is when the conductor is infinite in length. For this case, point A is at (0,0,-∞),while B is at (0,0,∞); then α1=180o,α2=0o. Then  we have:

I µI H= ϕˆ ; B = ϕˆ 2πρ 2πρ

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FIELD THEORY

Magnetostatic Fields

FORCE BETWEEN TWO LONG PARALLEL CURRENT-CARRYING CONDUCTORS Consider two long parallel conductors, separated by distance ρ, carrying current in the same direction as shown in the figure below. Ia

Ib

dlb

Ba

dF

ρ

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FORCE BETWEEN TWO LONG PARALLEL CURRENT-CARRYING CONDUCTORS The current Ia produces a magnetic induction Ba, as shown, at the position Ib. The force acting on an element    of current F = I b dlbis: xBa Idlb of thisdcurrent  µI Ba = a ϕˆ 2πρ  µI    µI a   ∴ dF = I b  dlb x ϕˆ  = I b  zˆdlb x a ϕˆ  2πρ  2πρ      µI I dl  ⇒ dF = − ρˆ  a b b   2πρ 

(

)

 µI I dl  dF  µI a I b  dF =  a b b  ⇒ =  2 πρ dl 2 πρ     The last expression is theENEL2FT force per unit length of the wire. FIELD THEORY

Magnetostatic Fields

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ENEL2FT

FIELD THEORY

Magnetostatic Fields

FORCE BETWEEN TWO LONG PARALLEL CURRENT-CARRYING CONDUCTORSEXAMPLE Consider a current-carrying conductor of finite length L placed a distance d from another current-carrying conductor of infinite length. Determine the magnetic force per unit length acting on the z finite conductor. I dlb

Infinitely long wire dF

d

I Ba ρ Finite wire of length L

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FORCE BETWEEN TWO LONG PARALLEL CURRENTCARRYING CONDUCTORS-EXAMPLE THE Solution is as follows:  µI B= ϕˆ 2πd   µI  ∴ dF = I  zˆdzx ϕˆ  2 π d     µI 2   µI 2  dF dz ⇒  dF = ρˆ  =  2πd   dz  2πd     L  µI 2   µI 2  dz = ρˆ  L F = ∫ ρˆ   2πd   2πd  0       µI 2  F  N /m ∴ = ρˆ   2πd  L   Magnetostatic Fields

ENEL2FT

FIELD THEORY

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FIELD THEORY

Magnetostatic Fields

FIELD DUE TO A CIRCULAR CURRENT-CARRYING LOOP Consider the circular loop shown below, with the loop z having radius ρ.

P

  R1 dl1

 R

h ρ

ρ

 dl

y

x The magnetic field intensity dH at point P(0,0,h) contributed by current element Idl is given by BiotSavart’s law:

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FIELD THEORY

Magnetostatic Fields

FIELD DUE TO A CIRCULAR CURRENT-CARRYING   LOOP  Idl xR dH = 4πR 3 Further, from the diagram, we note that:

dl = ρdϕϕˆ ;  R = − ρρˆ + hzˆ   ∴ dl xR = ( ρdϕϕˆ ) x( − ρρˆ + hzˆ ) = ρhdϕρˆ + ρ 2 dϕzˆ

Therefore we determine the elemental magnetic field intensity due to current element Idl to be:   Idl xR I 2 ˆ dH = = ρ hd ϕ ρ + ρ dϕzˆ 3 3/ 2 2 2 4πR 4π ρ + h

[

]

(

)

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ENEL2FT

FIELD THEORY

Magnetostatic Fields

FIELD DUE TO A CIRCULAR CURRENT-CARRYING LOOP Similarly we determine the elemental magnetic field dH1 intensity due  to current element Idl1 to be: R1 = ρρˆ + hzˆ; dl = − ρdϕϕˆ   ∴ dl1 xR1 = ( − ρdϕϕˆ ) x( ρρˆ + hzˆ ) = ρhdϕρˆ + ρ 2 dϕzˆ    Idl1 xR1 I 2 ˆ ( dH 1 = = − ρ hd ϕ ρ + ρ dϕzˆ ) = dH ρ ρˆ + dH z zˆ 3 2 2 3/ 2 4πR1 4π [ ρ + h ]

Therefore, by symmetry, the H contributions along ρ add up to zero because the radial components produced bypairs of current elements 180o apart cancel each 2π I I 2πρ 2 Iρ 2 2 ( ρ dϕzˆ ) = zˆ H = zˆ ∫ Thus: dH z = ∫ = zˆ other. 2 2 3/ 2 2 2 3/ 2 2 2 3/ 2 0

4π [ ρ + h

]

4π [ ρ + h

]

2[ ρ + h

]

29

FIELD DUE TO A CIRCULAR CURRENT-CARRYING LOOP The magnetic induction is thus given by:

  B = µH = zˆ

[

µIρ 2 2

2ρ +h

]

2 3/ 2

Thus the magnetic induction is maximum in the plane of the current-carrying loop (h=0), and it drops off as h →∞ or h»ρ.

Magnetostatic Fields

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FIELD THEORY

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Magnetostatic Fields

FIELD DUE TO A CIRCULAR CURRENT-CARRYING LOOP – EXAMPLE A circular loop located on x2+y2=9, z=0, carries a direct current of 10 A. Determine the magnetic field intensity at: A) (0,0,4) B) (0,0,-4)

31

FIELD DUE TO A CIRCULAR CURRENT-CARRYING LOOP – EXAMPLE SOLUTION: For the circular loop, the radius ρ=3. therefore, at a height h above the x-y plane, the field strength is given by:

 H = zˆ  ∴H  ∴H

Magnetostatic Fields

[

Iρ 2 2

2ρ +h

( 0, 0, 4 )

= zˆ

( 0,0, −4)

]

2 3/ 2

(10)32

[

2 32 + 4

= zˆ

[

]

2 3/ 2

(10)32

2 32 + ( − 4 ) ENEL2FT

= 0.36 zˆ A / m

]

2 3/ 2

= 0.36 zˆ A / m

FIELD THEORY

32

FIELD DUE TO A CIRCULAR CURRENTCARRYING LOOP – EXAMPLE A thin loop of radius 5 cm is placed on the plane z=1 cm so that its centre is at (0,0,1 cm). The loop carries a 0.05A current. Determine H at: A) (0,0, -1 cm) B) (0,0, 10 cm) Answer:

 A) H = 0.4 zˆ A / m  B ) H = 0.0573 zˆ A / m

Magnetostatic Fields

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FIELD THEORY

33

FIELD DUE TO A SOLENOID Consider a solenoid of length L, consisting of N turns of wire carrying a current I, whose cross-section is shown in the figure below. The number of turns per unit length, n, is given by n=N/L L z

ρ

θ1

dz

θ

θ2 z P

Cross-section of a solenoid

Magnetostatic Fields

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34

FIELD DUE TO A SOLENOID For a single loop centered about the z-axis, the value of H a distance z is: 2 

H = zˆ

[

Iρ

2

2ρ +z

]

2 3/ 2

For an arbitrary point z along the length of the solenoid, the incremental electric field intensity for small length dz is: Iρ 2 ndz dH = zˆ 2 2 3/ 2

[

2ρ +z

Magnetostatic Fields

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]

FIELD THEORY

35

FIELD DUE TO A SOLENOID From the sketch, we obtain: z = ρ / tan θ = ρ cot θ

(

)

dz = − ρ cos ec 2θ ⇒ dz = − ρ cos ec 2θ dθ dθ

( ρ 2 + z 2 ) = ρ 2 (1 + cot 2 θ ) = ρ 2 cos ec 2θ  Iρ 2 ndz Iρ 2 n  − ( ρ cos ec 2θ )dθ  ∴ dH = zˆ = zˆ 2  ρ 3 cos ec 3θ  2 2 3/ 2 2[ ρ + z ]  nI dH = − sin θdθzˆ 2  θ 2  nI nI  H = ∫  − sin θdθ  zˆ = zˆ ( cosθ 2 − cosθ1 ) 2  θ1  2  NI ( cosθ 2 − cosθ1 ) H = zˆ 2L

In the middle of a very long solenoid, θ1=0o, θ2=180o; therefore:

 NI H = zˆ = zˆnI L Magnetostatic Fields

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FIELD THEORY

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Magnetostatic Fields

FIELD DUE TO A TORROID For a toroidal coil of radius R, with N coils of wire carrying a current I, the total values of B and H are obtained from the figure below.

37

FIELD DUE TO A TORROID Note that µ is the magnetic permeability of the material around which the current-carrying coil is wound. It is assumed that the coil lies on the x-y plane. Note that with the length, L, of the torroid, the B and H fields are given by: 

NI NI ϕˆ = ϕˆ L 2πR  µNI B= ϕˆ 2πR H=

It is assumed that the coil lies on the x-y plane. If the coil has a circular cross-section with radius ρ, and if ρ<
Φ = BA =

Magnetostatic Fields

µNI µNIρ (πρ 2 ) = 2πR 2R ENEL2FT

FIELD THEORY

38

FORCE ON A POINT CHARGE MOVING IN A MAGNETIC FIELD – THE LORENTZ FORCE Let us determine the force F on a single charge Q moving at a velocity v in a magnetic field B. The force on a current element Idl is:

   dF = Idl xB

I = n( dav ) Q

If the cross-sectional area of the wire is da, the current I is given by:

Here n is the number of carriers per unit volume, v is their average drift velocity, and Q is the charge per carrier. Magnetostatic Fields

ENEL2FT

FIELD THEORY

39

FORCE ON A POINT CHARGE MOVING IN A MAGNETIC FIELD – THE LORENTZ FORCE Therefore the total charge flowing per second is the charge on the carriers that are contained in a length v of the wire.      dFon = Id l xBelement = ( ndadlQ v xB Then the force the dl ) becomes:

   F charge = Qv xBQ moving at a velocity v The force on a single in a field B is:

[

(

)]

    = QisEalso + van xBelectric field E, the More generally, if F there force is: Magnetostatic Fields

ENEL2FT

FIELD THEORY

40

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FIELD THEORY

Magnetostatic Fields

AMPERE’S CIRCUITAL LAW – MAXWELL’S EQUATION Ampere’s circuital law states that the line integral of the tangential component of H around a closed path is the same as the net current enclosed by the path.  In other words:

∫ H .dl

= I enclosed

Ampere’s circuital law is similar to Gauss’s law, and is easily applied to determine H when the current distribution is symmetrical. Ampere’s circuital law is a special case of the Biot-Savart law. 41

AMPERE’S CIRCUITAL LAW – MAXWELL’S EQUATION By applying Stoke’s theorem, we obtain:

I encl

     = ∫L H .dl = ∫S ( ∇xH ).dS

We can further simplify this, with J the current density, to obtain:

I encl

     = ∫S J .dS ⇒ ∇xH = J

This is the third Maxwell’s equation, which is Ampere’s law in differential form.

Magnetostatic Fields

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42

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Magnetostatic Fields

APPLICATION OF AMPERE’S CIRCUITAL LAW – INFINITE LINE CURRENT Consider an infinitely long filamentary current I along the z-axis, as shown below. z Amperian path

ρ

y

 dl

x This path, on which ampere’s law is to be applied, is known as the Amperian path (analogous to the Gaussian surface). 43

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FIELD THEORY

Magnetostatic Fields

APPLICATION OF AMPERE’S CIRCUITAL LAW – INFINITE LINE CURRENT We choose a concentric circle as the Amperian path in view of the previous considerations. Since this path encloses the entire current I, according to Ampere’s law,

I = ∫ H φ aˆφ .ρdφaˆφ = H φ ∫ ρdφ = H φ .2πρ I ∴H = aˆφ 2πρ This is the result expected, from the application of the Biot-Savart law.

44

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Magnetostatic Fields

AMPERE’S CIRCUITAL LAW – INFINITELY LONG COAXIAL TRANSMISSION LINE Consider an infinitely long transmission line consisting of two concentric cylinders having their axes along the z-axis. The cross-section of the line is shown below.

Amperian path (one of 4)

a

b

t

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FIELD THEORY

Magnetostatic Fields

AMPERE’S CIRCUITAL LAW – INFINITELY LONG COAXIAL TRANSMISSION LINE The inner conductor has radius a and carries a current I, while the outer conductor has inner radius b and thickness t, and carries return current –I. We want to determine H everywhere, assuming the current is uniformly distributed in both conductors. Since the current distribution is symmetrical, we apply Ampere’s law along the Amperian path for each of the four possible regions: Region Region Region Region

1: 2: 3: 4:

0<ρb+t 46

AMPERE’S CIRCUITAL LAW – INFINITELY LONG COAXIAL TRANSMISSION LINE For Region 1, we apply Ampere’s law, giving:

    ∫ H .dl = I = ∫ J .dS enc

Re g 1

Since the current is uniformly distributed over the cross section, we have:  I

J=

πa

2

zˆ; dS = ρdφdρzˆ

  I Iρ I = ∫ J .dS = ∫∫ ρdφdρ = πa a   Iρ H . d l = H 2 πρ = ∫ φ a Iρ ∴ Hφ = 2πa enc

2

2

2

2

2

Re g 1

2

Magnetostatic Fields

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FIELD THEORY

47

AMPERE’S CIRCUITAL LAW – INFINITELY LONG COAXIAL TRANSMISSION LINE For Region 2: a<ρ
Re g 2

H φ 2πρ = I ∴ Hφ =

I 2πρ

For Region 3, b<ρ
    ∫ H .dl = I = I + ∫ J .dS

Re g 3

 J= Magnetostatic Fields

enc

(− I) π [( b + t ) − t 2

Re g 3

2

]

zˆ

ENEL2FT

FIELD THEORY

48

AMPERE’S CIRCUITAL LAW – INFINITELY LONG COAXIAL TRANSMISSION LINE Thus we obtain:   I = I + ∫ J .dS = I − enc

Re g 3

(− I) π [( b + t ) − t 2

2π

2

]∫

ρ

∫ ρdρ

φ = 0 ρ =b

 ρ −b  I = I 1 −  t + 2bt  I  ρ −b  ∴ Hφ = 1−  2πρ  t + 2bt  2

enc

2

2

2

2

2

For Region 4, ρ>b+t, we obtain: I  ρ −b  H = 1− =0   2πρ  t + 2bt  2

φ

2

2

ρ =b

Magnetostatic Fields

ENEL2FT

FIELD THEORY

49

AMPERE’S CIRCUITAL LAW – INFINITELY LONG COAXIAL TRANSMISSION LINE Therefore from the equations we have, for a coaxial cable:

Iρ ˆ  φ 0≤ ρ ≤a  2πa  I ˆ   φ a≤ ρ ≤b H = 2πρ  I  ρ −b  ˆ  2πρ 1 − t + 2bt φ b ≤ ρ ≤ b + t    0 ρ >b+t  2

2

2

2

Thus Ampere’s law can only be used to find H due to symmetric current distributions for which it is possible to find a closed path over which H is constant in magnitude. Magnetostatic Fields

ENEL2FT

FIELD THEORY

50

MAGNETIC VECTOR POTENTIAL The magnetic flux through a surface S is given by:

  Ψ = ∫ B.dS s

However, unlike electric flux lines, magnetic flux lines always close upon themselves. This is due to the fact that it is not possible to have isolated magnetic poles or magnetic charges. Thus the total flux through a closed surface in a magnetic field must be zero; that is:

  Ψ = ∫ B.dS = 0

Magnetostatic Fields

ENEL2FT

FIELD THEORY

51

MAGNETIC VECTOR POTENTIAL This is Gauss’s law for magnetostatic fields. By applying the divergence theorem, we obtain:

    Ψ = ∫ B.dS = ∫ ∇.Bdv = 0   ∴ ∇.B = 0 V

This is the fourth Maxwell’s equation. In order to satisfy the above equation, we define the vector magnetic potential, A, such that:

     ∇.B = 0 ⇒ B = ∇xA

Magnetostatic Fields

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FIELD THEORY

52

MAGNETIC VECTOR POTENTIAL Just as we defined the electric scalar potential, V as:

dQ V =∫ 4πε r o

We can define:

  µ Idl ( line current ) A=∫ 4πε r   µ KdS ( surface current ) A=∫ 4πε r   µ Jdv ( volume current ) A=∫ 4πε r o

o

o

o

o

o

Magnetostatic Fields

ENEL2FT

FIELD THEORY

53

MAGNETIC VECTOR POTENTIAL Let us apply Stoke’s theorem to the flux equation:

  Ψ = ∫ B.dS       B = ∇xA ⇒ Ψ = ∫ ∇xA.dS      ∫ ∇xA.dS = ∫ A.dl   ∴ Ψ = ∫ A.dl s

s

s

L

L

Magnetostatic Fields

ENEL2FT

FIELD THEORY

54

MAGNETIC VECTOR POTENTIAL - EXAMPLE: Given the magnetic vector potential,

 A = − ρ / 4 zˆ 2

Calculate: The total magnetic flux density The total magnetic flux crossing the surface: φ=π/2, 1<ρ <2 m, 0
Magnetostatic Fields

ENEL2FT

FIELD THEORY

55

MAGNETIC VECTOR POTENTIAL - EXAMPLE: SOLUTION

   ∂A ˆ ρ ˆ B = ∇xA = − φ= φ ∂ρ 2   Ψ = ∫ B.dS  dS = dρdzφˆ z

  1 5 ∴ Ψ = ∫ B.dS = ∫ ∫ ρdρdz = ρ 2 ρ 4 Ψ = 3.75Wb 5

z =0

Magnetostatic Fields

ENEL2FT

2

=1

FIELD THEORY

2 2 1

56

MAGNETIC VECTOR POTENTIAL EXAMPLE: A current distribution gives rise to a vector magnetic potential:

 A = x yxˆ + y xyˆ − 4 xyzzˆ 2

2

Calculate: A) B at (-1, 2, 5) B) The flux through the surface defined by z=1, 0<x<1, -1
 B = 20 xˆ + 40 yˆ − 3 zˆ Ψ = 20Wb ENEL2FT

FIELD THEORY

57

MAGNETIC VECTOR POTENTIAL Let us consider the curl of the magnetic flux density:

   ∇xB = µ J          ∇xB = ∇x∇xA = ∇ ( ∇. A) − ∇ A      ∴ ∇ ( ∇. A) − ∇ A = µ J o

2

2

o

However, for a static magnetic field, we have:

    ∇. A = 0 ⇒ ∇ A = − µ J 2

o

∴∇ A = −µ J 2

x

o

∇ A = −µ J

y

∇ A = −µ J

z

x

2

y

o

2

z

Magnetostatic Fields

o

ENEL2FT

FIELD THEORY

58

MAGNETIC BOUNDARY CONDITIONS The magnetic boundary conditions are defined as the conditions that B and H must satisfy at the boundary between different media. To derive these conditions, we make use of Gauss’s law for magnetic fields, namely:

  ∫ B.dS = 0   ∫ H .dl = I

We also use Ampere’s circuital law, namely:

Magnetostatic Fields

ENEL2FT

FIELD THEORY

59

ENEL2FT

FIELD THEORY

Magnetostatic Fields

MAGNETIC BOUNDARY CONDITIONS Consider the boundary between two magnetic media 1 and 2, characterized respectively by µ1 and µ2, as shown below. Medium 1, µ 1

B1

B1n

∆S

B1t

∆h

B2t B2n

B2

Medium 2, µ 2

60

MAGNETIC BOUNDARY CONDITIONS Applying Gauss’s law to the pillbox and allowing ∆h→0, we obtain:

B ∆S − B ∆S = 0 1n

2n

∴B = B 1n

2n

µH =µH 1

1n

2

2n

The above equation shows that the normal component of B is continuous at the boundary. It also shows that the normal component of H is discontinuous at the interface; that is, H undergoes Magnetostatic Fields ENEL2FT FIELD THEORY some change at the interface.

61

ENEL2FT

FIELD THEORY

Magnetostatic Fields

MAGNETIC BOUNDARY CONDITIONS Similarly, we apply Ampere’s circuital law to the closed path abcda shown in the figure below, where surface current K on the boundary is assumed normal to the path. Medium 1, µ 1

H1

H1n H1t

a

b ∆h

K

H2t d H2n

∆w

c

H2

Medium 2, µ 2

62

MAGNETIC BOUNDARY CONDITIONS We obtain:

K∆w = H ∆w + H 1t

1n

∆h ∆h ∆h ∆h +H − H ∆w − H −H 2 2 2 2 2n

2t

2n

1n

As ∆h-0, the above equation becomes:

H −H =K 1t

2t

This shows that the tangential component of H is discontinuous. The above equation may be written in terms of B as:

B B − =K µ µ 1t

Magnetostatic Fields

1

2t

2 ENEL2FT

FIELD THEORY

63

MAGNETIC BOUNDARY CONDITIONS If the boundary is free of current or the media are not conductors, K=0, and we have the tangential components of H being equal:

B B H =H ; = µ µ 1t

1t

2t

1

2

2t

If the fields make an angle θ with the normal to the interface, then, from the normal and tangential B1 cos θ1 = we B1n =have: B2 n = B2 cosθ 2 components of B (with K=0)

B B sin θ = H = H = sin θ µ µ 1

2

1

1t

2t

1

2

tan θ µ ∴ = tan θ µ Magnetostatic Fields

2

1

1

2

2

ENEL2FT

FIELD THEORY

64

ENEL2FT

FIELD THEORY

Magnetostatic Fields

MAGNETIC CIRCUITS Consider a current-carrying conductor formed into a coil of N turns around a doughnut-shaped iron core (magnetic material), the electrically-induced magnetic flux lines will be largely concentrated inside the iron core. This is the example of a simple magnetic circuit. Flux φ

I

Iron core

Crosssectional area A, mean length L

N-turn coil, with current I I

65

MAGNETIC CIRCUITS The flux φ in this magnetic circuit is analogous to the electric current in an electric circuit. In the figure, there are N turns, each having a current I; thus the cause of the induced flux φ is the current flow NI. The analogy in an electric circuit is the fact that a voltage potential difference is the cause of the flow of the current carriers (that is, V causes I). The quantity NI is called the magnetomotive force (MMF). It is the driving force behind the existence of the magnetic flux φ. Fm = MMF = NI Ampere − turns Thus:

(

Magnetostatic Fields

ENEL2FT

FIELD THEORY

)

66

ENEL2FT

FIELD THEORY

Magnetostatic Fields

MAGNETIC CIRCUITS If the same magnetomotive force (MMF) is applied to similar iron cores, each with a different mean length Li, the resulting magnetic flux density for the i-th coil is:

NI B=µ ; φ = BA Li

One therefore expects that stronger magnetic fields will result in iron cores with shorter length L. Therefore the other B quantity NI Fm of interest in H = is =the magnetizing = magnetic circuits force, or µ Li Li magnetic field intensity: 67

MAGNETIC CIRCUITS: RELUCTANCE When the current I or the number of turns N is increased in the simple magnetic circuit, the magnetomotive force, Fm, is increased, resulting in a higher flux φ in the magnetic core. Thus, we have:

Fm ∝ φ

∴ Fm = kφ Based on the analogies previously established, the electric voltage, E or V is analogous to the magnetomotive force Fm(=NI), and the electric current I is analogous to the magnetic flux φ. The equation relating Fm to φ is thus similar to Ohm’s Magnetostatic Fields electric circuits: ENEL2FT 68 law for V=RI.FIELD THEORY

MAGNETIC CIRCUITS: RELUCTANCE Thus the constant of proportionality, k, is actually a measure of the opposition to the establishment of magnetic flux. This quantity is called the reluctance, R of the magnetic circuit, and is analogous to the resistance R of an electric circuit. Hence Ohm’s law for the magnetic circuit can be expressedFas:= Rφ ⇒ R = Fm A / Wb m

Magnetostatic Fields

φ

ENEL2FT

FIELD THEORY

69

MAGNETIC CIRCUITS: PERMEABILITY It is easier to establish or set up the magnetic flux lines in some materials (e.g. iron) than it is in other materials (e.g. air). The magnetic lines of force, like electric current, always try to follow the path of least resistance. Permeability is the property of materials that measures its ability to permit the establishment of magnetic lines of force. It is analogous to conductivity in electric circuits. Air is taken as the reference material, with its permeability called µo. The permeability µ of any r other material is given by:o

µ=µ µ

Magnetostatic Fields

ENEL2FT

FIELD THEORY

70

MAGNETIC CIRCUITS: PERMEABILITY Where µr is called the relative permeability. Nonmagnetic materials (e.g. air, glass, copper and aluminum) are characterized by their µr which is approximately unity. Magnetic materials such as iron, steel, cobalt, nickel, and their alloys are called ferromagnetic materials, as characterized by high values of µr(100 to 100,000 or more) From the definitions of reluctance, R of the magnetic circuit, and permeability µ of the material, it is clear that one is the opposite of the other. Thus, we have the relationship between reluctance Fm NI NI L and permeability:

R=

Magnetostatic Fields

φ

=

BA

=

=

( µNI L ) A µA

ENEL2FT

FIELD THEORY

71

ENEL2FT

FIELD THEORY

Magnetostatic Fields

MAGNETIC CIRCUITS: EXAMPLES 1. The simple magnetic circuit with one coil would around a doughnut-shaped iron core has a cross-sectional area of 50cm2 and a mean length of 2 m. The relative permeability of the magnetic material of the core is 80. If the current coil has 150 turns and the resulting flux is 80 µWb, determine: The reluctance of the magnetic circuit (Ans: 3.98x106 A/Wb) The value of the current flowing in the coil (Ans: 2.123 A) 2. The same magnetic circuit in example 1 has a magnetomotive force of 200 A. If the length of the coil is 40cm, and the permeability of the core is 6x10-4 Wb/m2, determine: The magnetizing force, H (Ans: 500 A/m) The flux density B in the iron core (Ans: 0.3 Wb/m2)

72

MAGNETIC CIRCUITS – AMPERE’S CIRCUITAL LAW The figure below shows an example of a simple series magnetic circuit, made up of three different types of materials, including the air gap. Since there is only one path for the magnetic flux lines φ, it must be the same in all parts of this series magnetic circuit. This is similar to a series electric circuit where the current is the same in all series components. Air Gap

I

I

1

2

N

N

1

2

Mate rial 1 Magnetostatic Fields

ENEL2FT

FIELD THEORY

Materi al 2

73

MAGNETIC CIRCUITS – AMPERE’S CIRCUITAL LAW

As E and V are analogous to Fm (=NI) and φR or HL, a law similar to KVL (algebraic sum of voltage rise and drop in a closed loop =0) also applies to the closed-loop series magnetic circuit. This is Ampere’s circuital law. Ampere’s circuital law states that the sum of the magnetomotive force (MMF or Fm) rises equals the sum of the MMF drops around any closed path of a magnetic circuit. In general, Ampere’s circuital law can be stated as follows:

A lg ebraic sum of applied MMF ' s = φR1 + φR2 + φR3 + .. = φRT

= H1L1 + H 2 L2 + .H 3 L3 + .. Here, φ is the same amount of magnetic flux in the series magnetic circuit, R1 is the reluctance of part 1 of the circuit (similarly for R2 and R3), H1 is the magnetic field intensity of part 1 of the circuit (similarly for H2 and H3), and L1 is the length of this part of the circuit (similarly for L2 and L3).

Magnetostatic Fields

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FIELD THEORY

74

MAGNETIC CIRCUITS – AMPERE’S CIRCUITAL LAW From the above equation, one can easily see that in a series magnetic circuit,

R T = T o ta l s e r ie s r e lu c ta n c e = R 1 + R 2 + R 3 That is, the net algebraic sum of the applied MMFs in the assumed positive direction of φ is the sum of φ times the sum of the reluctances of each part of the series circuit, consisting of materials 1 and 2 and the air gap (material 3). The above form of Ampere’s circuital law is used if the dimensions and the permeabilities of each portion of the circuit are known, so that the reluctances can be calculated. Magnetostatic Fields

ENEL2FT

FIELD THEORY

75

ENEL2FT

FIELD THEORY

Magnetostatic Fields

MAGNETIC CIRCUITS – AMPERE’S CIRCUITAL LAW In the magnetic circuit shown in the figure below, the flux φ induced by the MMF (NI) can be split into two parts, since there are two different paths of the magnetic flux lines, in either branch a (φa) or branch b (φ b). φb

φ

I

N φa

φ

φb

76

ENEL2FT

FIELD THEORY

Magnetostatic Fields

MAGNETIC CIRCUITS – AMPERE’S CIRCUITAL LAW The sum of the two magnetic fluxes must be the same as the total flux, φ:

φ = φa + φb This is the law of conservation of flux. It is similar to KCL at a node of an electric circuit. If the reluctance of branch a is Ra, and the reluctance of branch b is Rb, then the equivalent reluctance of branches a and b is: R R

R

eq

=

a

b

Ra + R

b 77

ENEL2FT

FIELD THEORY

Magnetostatic Fields

MAGNETIC CIRCUITS – AMPERE’S CIRCUITAL LAW – EXAMPLE 1 Consider the magnetic circuit for a relay shown below. The average length of the iron core is 40 cm, and the length of the air gap is 0.2 cm, while the average cross-sectional area is 2.5 m2. The number of turns of the coil is 50, and µr for iron is 200. Determine the current required to produce a flux density of 0.1 Wb/m2 in the air gap. I

φ φ

Iron N Air Gap

φ

78

MAGNETIC CIRCUITS – AMPERE’S CIRCUITAL LAW – EXAMPLE 1 Solution: µo = 4πX 1 0

−7

= 1 .2 5 7 X 1 0 − 6 W b / A m

(

µ ir o n = µ o µ r = 1 .2 5 7 X 1 0 ∴ R R

ir o n

a ir − g a p

=

1

µ ir o n

−6

)X

2 0 0 = 2 .5 1 3 X 1 0 − 4 W b / A m

L ir o n 0 .4 = A 2 .5 1 3 X 1 0 − 4 X 2 .5 X 1 0

(

) (

−4

1 L a ir − g a p 0 .0 0 2 = = µo A 1 .2 5 7 X 1 0 − 6 X 2 .5 X 1 0

∴ RT = R

(

ir o n

+ R

a ir − g a p

) (

= 6 .3 6 6 X 1 0 6 A / W b

)

−4

)

= 6 .3 6 4 X 1 0 6 A / W b

= 1 .2 7 3 X 1 0 7 A / W b

M M F = F m = φ R T = B A R T = ( 0 .1 ) ( 2 . 5 X 1 0

−4

(

) 1 .2 7 3 X 1 0

7

) = 3 1 8 .2 4 A

∴ M M F = N I = 3 1 8 .2 4 A = 5 0 I ∴ I = 6 .3 6 5 A Magnetostatic Fields

ENEL2FT

FIELD THEORY

79

ENEL2FT

FIELD THEORY

Magnetostatic Fields

ELECTROMAGNETIC INDUCTION A current-carrying conductor produces a magnetic field. The next question is: can a magnetic field result in a current flow, or induced voltage? In other words, is the electromagnetic induction process reversible? Faraday observed that when the magnetic lines of force (flux, φ) linking a conductor are changed, a voltage will be induced across the terminals of the conductor. The magnetic flux linkage can be changed by either moving the conductor or the magnetic filed itself in such a way that the conductor cuts across the magnetic lines of force. The induced voltage and the resulting induced current are produced only if the cutting action is exhibited.

80

ENEL2FT

FIELD THEORY

Magnetostatic Fields

ELECTROMAGNETIC INDUCTION

In the figure below, φorig is the original magnetic flux of the stationary horse-shoe magnet. If the conductor is moved perpendicular to the magnetic lines of force, the galvanometer’s pointer deflects, indicating an induced current flow (iind) resulting from the induced voltage, vind. conductor

S

N

Induced current

Galvano meter

φ

81

ELECTROMAGNETIC INDUCTION – THE TRANSFORMER The transformer is essentially just two (or more) inductors, sharing a common magnetic path. Any two inductors placed reasonably close to each other will work as a transformer, and the more closely they are coupled magnetically, the more efficient they become. When a changing magnetic field is in the vicinity of a coil of wire (an inductor), a voltage is induced into the coil which is in sympathy with the applied magnetic field. A static magnetic field has no effect, and Magnetostatic Fields ENEL2FT FIELD THEORY 82 generates no output. Many of the same principles apply to generators, alternators,

ENEL2FT

FIELD THEORY

Magnetostatic Fields

ELECTROMAGNETIC INDUCTION – THE TRANSFORMER The figure shows the basics of all transformers. A coil (the primary) is connected to an AC voltage source  - typically the mains for power transformers. The flux induced into the core is coupled through to the secondary, a voltage is induced into the winding, and a current is produced through the load.

83

ELECTROMAGNETIC INDUCTION – THE TRANSFORMER Note that the time-varying current ip flowing in the primary coil, generates a time-varying flux φp in the space surrounding the coil. Part of this flux, φps, will link with the secondary coil. This part is called the mutual flux. Another part of this flux, φpp, will not link with the secondary coil. This last part is called the leakage flux of the primary coil.

dφ p

The primary and the induced secondary voltages, vs and vp, vp = N p dt are:

vs = N s Magnetostatic Fields

ENEL2FT

dφ ps dt

FIELD THEORY

84

ENEL2FT

FIELD THEORY

Magnetostatic Fields

ELECTROMAGNETIC INDUCTION – THE TRANSFORMER Note that the flux is proportional to the current producing it. Thus:

φp = N p Rp =

ip

Rp

lp

µAp

; φs = N p

; R ps =

ip

R ps

ls µAps

Here, Rp is the reluctance of the magnetic path of φp, and Rps is the reluctance of the magnetic path of φps. Thus the expressions dφ pfor vp and i p  N p2 di p d  vs become:

Np =  dt dt  R p  R p dt dφ ps i p  N p N s di p di p d   vs = N s = Ns Np = =M   dt dt  R ps  R ps dt dt vp = N p

= Np

85

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FIELD THEORY

Magnetostatic Fields

ELECTROMAGNETIC INDUCTION – THE TRANSFORMER The constants of proportionalities are the inductance, Lp N 2p the mutual N p N s inductance, M, of the primary coil, and Lp = ; M= between the two coils: Rp R ps

Both M and Lp have the same physical units (the Henry), and both are constants, depending on the physical parameters and dimensions of the magnetic flux paths, related through the coupling coefficient, k. By exciting the secondary coil and determining the induced voltage in the primary N p N s and N pcurrent Ns N s2 Ls = ; M= = side, we find that:

Rs

R ps

Rsp

φ ps φ ps + φss ∴ M = k L p Ls ; k = = ⇒ 0 ≤ k ≤1 φp φp 86

ELECTROMAGNETIC INDUCTION – THE TRANSFORMER The first and most important characteristic of an ideal transformer is that its primary and secondary fluxes have no leakage components. Thus all the flux produced due to the flow of ip links with the secondary coil, while all the flux produced due to the flow of is links with the φ pprimary = φ ps , φs coil. = φsp ; That is:

φ ps φsp ∴k = = =1 φ p φs

∴ M = k L p Ls = L p Ls vp = N p Magnetostatic Fields

dφ p dt

; vs = N s ENEL2FT

dφ ps dt

= Ns

dφ p dt

FIELD THEORY

⇒

vp vs

=

Np Ns 87

ELECTROMAGNETIC INDUCTION – THE TRANSFORMER From KVL, we have, for a signal with frequency ω:

v p = jωL p I p ; vs = jωLs I s ∴

Ip Is

( v p / L p ) N p Ls = = ( vs / Ls )

N s Lp

N 2p

lp N s2 l Lp = ; Ls = ; Rp = ; Rs = s ⇒ R p = Rs Rp Rs µAp µAs N p  Ls  N p  N s2  N s  =  = ∴ = 2 I s N s  L p  N s  N p  N p ∴V p I p = Vs I s Ip

Magnetostatic Fields

ENEL2FT

FIELD THEORY

88

ENEL2FT

FIELD THEORY

Magnetostatic Fields

Consider vector A. Then the divergence of vector A at a point P is the outward flux per unit volume as the volume about P. That is:  shrinks  ∫ A.dS    S divA = ∇. A = lim ∆v →0 ∆v The divergence theorem states that, for a closed surface S, which closes a volume v:     A . d S = ∇ ∫ ∫v . Adv S

The curl of vector A is defined as the circulation per unit area. That is:    ∫ A.dl     aˆ curlA = ∇xA = lim  L ∆s → 0  ∆s     Here, the area ∆S is bounded by the curve L, and a is the unit vector normal to the surface ∆S. Then Stoke’s theorem states that:      ∫ A.dl = ∫S ( ∇xA).dS

L

89

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FIELD THEORY

Magnetostatic Fields

1. Conservation of Electrostatic Field: Consider an electrostatic field which produces a field E. Then,      ( E . d l = ∇ ∫ ∫S xE ).dS

L

But, from definition of potential difference between point 1 and point 2, we have:   ∫ E.dl = V2 − V1

L2 L1

For a closed loop, the potential difference would be V2-V2=0 Thus:      ( E . d l = ∇ ∫ ∫S xE ).dS = V1 − V1 = 0 L   ∴ ∇xE = 0

This is Maxwell’s curl equation for an electrostatic filed. 90

ENEL2FT

FIELD THEORY

Magnetostatic Fields

2. Maxwell’s Curl Equation for Magnetostatic Field Similarly, from Ampere’s circuital law:

       I encl = ∫L H .dl = ∫S ( ∇xH ).dS = ∫S J .dS    ∴ ∇xH = J

3. Divergence Theorem: Maxwell’s Scalar equation for D. From Gauss’s law, we have:   Q = ∫S D.dS If we assume that Q is the total charge in a volume enclosed by surface S, with volume charge density ρv; and applying the divergence theorem, we have:     ( Q = ∫S D.dS = ∫V ∇.D ) dv But Q = ∫V ρ v dv   ∴ ∇.D = ρ v

91

ENEL2FT

FIELD THEORY

Magnetostatic Fields

4. Divergence Theorem: Maxwell’s Scalar equation for B For the magnetic field intensity, the total flux enclosed by a surface S is given by:   Ψ = ∫S B.dS   B = µH Applying the divergence theorem, and noting that there is no magnetic charge density in a given volume – that is, it is not possible to have isolated magnetic poles, since magnetic flux lines always close upon themselves, we have:     ( B . d S = ∇ ∫S ∫V .B ) dv = 0   ∴ ∇.B = 0 This is Maxwell’s fourth equation for static fields.

92