5. MAGNETOSTATICS Applied EM by Ulaby, Michielssen and Ravaioli

5. MAGNETOSTATICS Applied EM by Ulaby, Michielssen and Ravaioli

Chapter Outline

Maxwell’s Equations Magnetic Forces and Torques The total electromagnetic force, known as Lorentz

force Biot- Savart’s law Gauss’s law for magnetism Ampere’s law for magnetism Magnetic Field and Flux Vector magnetic potential Properties of 3 different types of material Boundary conditions between two different media Self inductance and mutual inductance Magnetic energy

Chapter 5 Overview

Course Outcome 3 (CO3)

Ability to analyze the concept of electric current density and boundary conditions, magnetic flux and magnetic flux density in a steady magnetic field and the basic laws of magnetic fields.

Maxwell’s equations

Where;

E = electric field intensityD = electric flux densityρv = electric charge density

per unit volumeH = magnetic field intensityB = magnetic flux densityt

t

DHH

0B

BE

D v

Maxwell’s equations:

Maxwell’s equations

For static case, ∂/∂t = 0. Maxwell’s equations is reduced to:

Electrostatics Magnetostatics

0

E

D vJH

B

0

Electric vs Magnetic Comparison

Electric & Magnetic Forces

Electromagnetic (Lorentz) force

Magnetic force

Magnetic Force on a Current Element

Differential force dFm on a differential current I dl:

I

Magnetic Force

qq

q

N m BuF q

Magnetic Forces and Torques The electric force Fe per unit charge acting

on a test charge placed at a point in space with electric field E.

When a charged particle moving with a velocity u passing through that point in space, the magnetic force Fm is exerted on that charged particle.

where B = magnetic flux density (Cm/s or Tesla T)

N m BuF q

Magnetic Forces and Torques If a charged particle is in the presence of

both an electric field E and magnetic field B, the total electromagnetic force acting on it is:

q BuEF

me FFF

;EF qe BuF qm

BuEFFF qqme

force)(Lorentz

Magnetic Force on a Current- Carrying Conductor

For closed circuit of contour C carrying I , total magnetic force Fm is:

In a uniform magnetic field, Fm is zero for a closed circuit.

C

m dI N BlF N m BuF q

Magnetic Force on a Current- Carrying Conductor

On a line segment, Fm is proportional to the vector between the end points.

BF Im

Example 1

The semicircular conductor shown carries a current I. The closed circuit is exposed to a uniform magnetic field .

Determine (a) the magnetic force F1 on the straight section of the wire and

(b) the force F2 on the curved section.

0yBB

C

m dI N BlF

BF Im

Solution to Example 1

a) the magnetic force F1 on the straight section of the wire

N 2ˆˆ2ˆ 001 IrBBIr zyxF

0ˆ

2ˆ

, Using

B

r

Im

yB

x

BF

N 2ˆ

)11(ˆ

cosˆsin

:direction ˆ- in the is d ofproduct the

section. curved on the F2 force the F2 b)

02

0

00

0

0

0

2

IrB

IrB

IrBBdrI

dI

zF

z

z

BlF

zBl

))(sin(

ˆ

0

rdl

ByB o

Torque

d = moment armF = forceT = torque

Magnetic Torque on Current Loop

No forces on arms 2 and 4 ( because I and B are parallel, or anti-parallel)

Magnetic torque:

Area of Loop

Inclined Loop

For a loop with N turns and whose surface normal is at angle theta relative to B direction:

The Biot–Savart’s Law

The Biot–Savart law is used to compute the magnetic field generated by a steady current, i.e. a continual flow of charges, for example through a wire

Biot–Savart’s law states that:

where:

dH = differential magnetic field dl = differential length

A/m R

dd

2Rl

4

1H

Biot-Savart Law

Magnetic field induced by a differential current:

For the entire length:

Magnetic Field due to Current Densities

Example 2

Determine the magnetic field at the apex O of the pie-shaped loop as shown. Ignore the contributions to the field due to the current in the small arcs near O.

O A

C O

A C

= dl

= -dl

0

?

• For segment AC, dl is in φ direction,

• Using Biot- Savart’s law:

add zRl aR

radians in is wherea

H

a

ad

a

ad

4

1z

z

4

1z

4

1H

22

A/m ˆ

4

12

l R

d RlH

Example 5-2: Magnetic Field of Linear Conductor

Cont.

Example 5-2: Magnetic Field of Linear Conductor

Magnetic Field of Long Conductor

Example 5-3: Magnetic Field of a Loop

Cont.

dH is in the r–z plane , and therefore it hascomponents dHr and dHz

z-components of the magnetic fields due to dl and dl’ add because they are in the same direction, but their r-components cancel

Hence for element dl:

Magnitude of field due to dl is

Example 5-3:Magnetic Field of a Loop (cont.)

For the entire loop:

Magnetic Dipole

Because a circular loop exhibits a magnetic field pattern similar to the electric field of an electric dipole, it is called a magnetic dipole

Forces on Parallel Conductors

Parallel wires attract if their currents are in the same direction, and repel if currents are in opposite directions

Gauss’s Law for Magnetism Gauss’s law for magnetism states

that:

Magnetic field lines always form continuous closed loops.

form)(integral0form)ial(different0 S

dsBB

0 BsD

Ampère’s Law

Ampere’s law states that: true for an infinite length of conductor

C

law sAmpere' IdlH

Ampere’s law for magnetism

true for an infinite length of conductor

I, +az

HdlC, +aø

r

Internal Magnetic Field of Long ConductorFor r < a

Cont.

External Magnetic Field of Long Conductor

For r > a

Magnetic Field of Toroid

Applying Ampere’s law over contour C:

The magnetic field outside the toroid is zero. Why?

Ampere’s law states that the line integral of H around a closed contour C is equal to the current traversing the surface bounded by the contour.

Magnetic Flux

The amount of magnetic flux, φ in Webers from magnetic field passing through a surface is found in a manner analogous to finding electric flux:

SB d

a

2H

r

I

a

2B 0 r

I

HB 0

Example 4

An infinite length coaxial cable with inner

conductor radius of 0.01m and outer conductor

radius of 0.05m carrying a current of 2.5A

exists along the z axis in the + az direction.

Find the flux passing through the region between

two conductors with height of 2 m in free space.

Solution to Example 4

1) inner conductor radius = r1 0.01m

2) outer conductor radius = r2 0.05m

3) current of 2.5A (in the +az direction)

4) Flux radius = 2m

z

xy

a

2HB 00 r

I

Sd

r

I

a20

Iaz=2.5A

r1 r2

Flux,z

SB d

aø

Solution to Example 4

where dS is in the aø direction.

So,

Therefore,

aS drdzd

Wb I

adrdzar

I

SdB

z r

60

2

0

05.0

01.0

0

1061.101.0

05.0ln

2

2

2

Magnetic Vector Potential A

Electrostatics

Magnetostatics

Vector Magnetic Potential

For any vector of vector magnetic potential A:

We are able to derive: .

Vector Poisson’s equation is given as:

where

0A

2Wb/mA B

J2 A

Wb/m ''4 '

dvR

JA

v

Magnetic Properties of Materials Magnetization in a material is associated

with atomic current loops generated by two principal mechanisms: Orbital motions of the electrons around the

nucleus, i.e orbital magnetic moment, mo

Electron spin about its own axis, i.e spin magnetic moment, ms

Magnetic Properties of Materials

Magnetization vector M is defined as

where = magnetic susceptibility (dimensionless)

Magnetic permeability is defined as:

and to define the magnetic properties in term of relative permeability is defined as:

Magnetic Permeability

m

HM m

H/m 10 m mH 104 where 70

mr 10

HB

metals have a very weak and negative susceptibility ( ) to magnetic field

slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed

Most elements in the periodic table, including copper, silver, and gold, are diamagnetic.

Magnetic Materials - Diamagnetic

m

Magnetic Materials - Paramagnetic

Paramagnetic materials have a small and positive susceptibilities to magnetic fields.

slightly attracted by a magnetic field and the material does not retain the magnetic properties when the external field is removed.

Paramagnetic materials include magnesium, molybdenum, lithium, and tantalum.

)( m

However, the absolute susceptibilities value of both materials is in the order 10-5. Thus, can be ignored. Hence, we have Magnetic permeability:

Diamagnetic and paramagnetic materials include dielectric materials and most metals.

Magnetic Materials – Diamagnetic, Paramagnetic

m

0or 1 r

Magnetic Materials – Ferromagnetic Materials Ferromagnetic materials is characterized

by magnetized domain - a microscopic region within which the magnetic moments of all its atoms are aligned parallel to each other.

Hysteresis – “to lag behind”. It determines how easy/hard for a magnetic material to be magnetized and demagnetized.

Process of Magnetic Hysteresis

material is magnetizedand can serve as permanent magnet!

material is demagnetize

B

Magnetic Hysteresis of Ferromagnetic Materials Comparison of hysteresis curves for (a) a hard

and (b) a soft ferromagnetic material is shown.

Hard magnetic material- cannot be easily magnetized & demagnetized by an

external magnetic field.

Soft magnetic material – easily magnetized & demagnetized.

Magnetic Hysteresis

Boundary Conditions

Magnetic boundary conditions Boundary condition related to normal

components of the electric field;

By analogy, application of Gauss’s law for magnetism, we get first boundary condition:

Since , For linear, isotropic media, the first

boundary condition which is related to H;

SnnSDDQd 21 sD

nnSBBd 21 0 sB

nn HH 2211

HB

z

xy

stt JHH 12

By applying Ampere’s law

C

IdlH

Magnetic boundary conditions

The result is generalized to a vector form:

Where However, surface currents can exist only

on the surfaces of perfect conductors and perfect superconductors (infinite conductivities).

Hence, at the interface between media with finite conductivities, Js=0. Thus:

sJHHn 212ˆ

2 medium fromaway pointing vector normal theis ˆ 2n

tt HH 21

Example

tt HH 21

))(( 0 r

xy (plane)

• Solution:1)H1t = H2t thus, H2t = 6ax + 2ay

2)Hn1 = 3az,

but, Hn2 = ??

6000μ0(3az) = 3000 μ0(Hn2)

Hn2 = 6az

thus, H2 =6ax + 2ay + 6az

))(( 0 r

μr1 = 6000 ; μr2 = 3000 ;

Inductance

An inductor is the magnetic analogue of an electrical capacitor.

Capacitor can store electric energy in the electric field present in the medium between its conducting surfaces.

Inductor can store magnetic energy in the volume comprising the inductors.

)(1

1212 H

IL

INDUCTANCEstore magnetic

energy

H I

L

32 J/m 2

1H

v

Ww m

m

Solenoid

Inside the solenoid:

Inductance

Example of an inductor is a solenoid - a coil consisting of multiple turns of wire wound in a helical geometry around a cylindrical core.

Magnetic Field in a Solenoid For one cross section of

solenoid,

When l >a, θ1≈−90° and θ2≈90°,

Where, N=nl =total number of turns

over the length l

12 sinsin2

ˆ

nIzB

l

NInI

al with solenoidlong For

zzB

1/

Self Inductance

The self-inductance of a circuit is used to

describe the reaction of the circuit to a

changing current in the circuit,

(The ratio of the magnetic flux to the

current)

The self-inductance of a circuit is used to

describe the reaction of the circuit to a

changing current in the circuit,

(The ratio of the magnetic flux to the

current)

Self Inductance

Self-inductance of any conducting structure is the ratio of the magnetic flux linkage, Λ to the current I flowing through the structure.

Magnetic flux linkage, Λ is the total magnetic flux linking a given conducting structure.

H I

L

(Wb) 2

IS

l

NN

Self Inductance

Magnetic flux, linking a surface S is given by:

In a solenoid with uniform magnetic field, the flux linking a single loop is:

Wb

S

dsB

loop the of area sectional-cross Swhere

ISl

N

dsIl

NS

zz

Wb d flux, Magnetic•

S sB

ISl

N

dsIl

NS

zz

l

NIzB

Sd

S s

Self Inductance – magnetic flux in solenoid

Self Inductance

Magnetic flux, linking a surface S is given by:

In a solenoid with uniform magnetic field, the flux linking a single loop is:

Wb S

dsB

loop the of area sectional-cross Swhere

ISl

N

dsIl

NS

zz

Self Inductance

For a solenoid:

For two conductor configuration:

solenoid S2

l

NL

S

dIII

L sB 1

Self Inductance for a solenoid

H I

L

ISl

NNN

(Wb) IS

l

N

2

Thus,

HI

ISl

N

I

L

2

Sl

NL

2

Mutual Inductance

Mutual inductance – produced by magnetic coupling between two different conducting structures.

Mutual Inductance Magnetic field B1 generated by current I1 results

in a flux Φ12 through loop 2:

If loop 2 consists of N2 turns all coupled by B1 in exactly the same way, the total magnetic flux linkage through loop 2 due to B1 is:

2 112

SdSB

2 1212212

SdNN SB

Mutual Inductance

Hence, the mutual inductance:

H dI

N

IL

s

2

sB11

2

1

1212

2 112

SdSB

2 1212212

SdNN SB

Inductance

Magnetic Flux

Flux Linkage

Inductance

Solenoid

Magnetic Energy

Consider an inductor with an inductance L connected to a current source.

The current I flowing through the inductor is increased from zero to a final value I.

The energy expended in building up the current in the inductor:

i.e the magnetic energy stored in the inductor

2

02

1LIidiLivdtpdtW

l

m

Magnetic Energy

Magnetic energy density (for solenoid):

i.e magnetic energy per unit volume

Magnetic energy in magnetic field:

322

1J/m H

v

Ww m

m

32

1J/m BdvHW

vm

The magnetic field in the region S between the two conductors is approximately

Example 5-7: Inductance of Coaxial Cable

Total magnetic flux through S:

Inductance per unit length:

Magnetic Energy Density

Magnetic field in the insulating material is

The magnetic energy stored in thecoaxial cable is

Summary