M.1

Magnetostatics

Magnetic Forces

Biot-Savart Law

Gauss’s Law for Magnetism

Ampere’s Law

Magnetic Properties of Materials

Inductance

BuFm ×= q

2

ˆ

4 RdId RIH ×

=π

0=⋅∇ B

JH =×∇

HB µ=

M.2

Magnetic Forces

The electric field E at a point in space has been defined as the electric force Fe per unit charge acting on a test charge when place at that point.The magnetic flux density B at a point in space is defined in terms of the magnetic force Fm that would be exerted on a charged particle moving with a velocity u.

BuFm ×= q

M.3

Magnetic Forces on Current-Carrying Conductor

A current flowing through a conducting wire consists of charged particles drifting through the material of the wire. Therefore, when a current-carrying wire is place in a magnetic field, it will experience a force.

Demonstration: M5.1 Q1-Q3, M5.2

∫ ×=⇒

×=

cdI

Idd

BlF

BlF

m

m

M.4

Biot-Savart Law

Used for calculating the magnetic field intensity dH due to a current I flowing in an element dl.

The resultant magnetic field intensity H is found by integratingdH for all current elements

mAR

dId /ˆ

4 2

RIH ×=

π

∫×

=l RdI

2

ˆ

4RIH

π

M.5

Gauss’s Law for Magnetism

For electric fields, the net outward flux of the electric flux density D through a closed surface enclosing a net charge Q is equal to Q. The magnetic analogue to a point charge is a magnetic pole, but magnetic poles do not exist in isolation. ---they occur in pairs.

may be called “law of nonexistence of isolated monoploes” or “the law of conservation of magnetic flux” and “Gauss’s law for magnetism”

∫ =⋅⇔

=⋅∇

sd 0

0

sB

B

M.6

Ampere’s Law

The line integral of H around any closed path C is equal to the current I enclosed by the path (in a right-handed sense).

Idc

=⋅⇔

=×∇

∫ lH

JHI

C

dl

Not enclosed

M.7

Example: Magnetic Field of a Long Wire

(a) ar ≤Choose the Amperian contour C1 to be a circular path of radius r1

∫ =⋅1

111CId lH

( )11

2

0 1111

2

ˆˆ1

Hr

drHdC

π

φφφπ

=

⋅=⋅∫ ∫lH

LHS:

The current flowing through the area enclosed by C1 is equal to

IarI

arI

=

= 2

21

2

21

1 ππ

M.8

Example: Magnetic Field of a Long Wire

(b)

21

11 2ˆˆ

aIrH

πφφ ==∴ H

ar >Choose the Amperian contour C2 which encloses all the current I

( )IHr

drHdC

==

⋅=⋅∫ ∫22

2

0 2222

2

ˆˆ2

π

φφφπ

lH

222 2

ˆˆrIHπ

φφ ==∴ H

M.9

Example: Magnetic Field of a Long Wire

21 2ˆ

arIπ

φ=HrIπ

φ2

ˆ2 =H

M.10

Example: Magnetic Field inside a Toroidal Coil

A toroidal coil (also called torus or toroid) is a doughnut-shaped structure with closely spaced turns of wire wrapped around it.

M.11

Example: Magnetic Field inside a Toroidal Coil

Applying the Ampere’s law along the contour C:

( ) ( )

brar

NIH

NIrH

rdHdC

<<−=−=⇒

−=−=

⋅−=⋅∫ ∫

πφφ

π

φφφπ

2ˆˆ

2

ˆˆ2

0

H

lH

M.12

Magnetic Properties of Materials

Magnetization in a material is associated with atomic current loops generated by – Orbital motions of the electrons around the nucleus– Electron spin

The magnetization vector M of a material is defined as the vector sum of the magnetic dipole moments of the atoms contained in a unit volume of the material

( )( )

HHMH

MHMHB

ro

mmo

o

oo

µµχχµ

µµµ

==+=

+=+=

Q1

Magnetic susceptibility

Relative permeability

M.13

Ferromagnetic Materials

Ferromagnetic materials, which include iron, nickel, and cobalt,exhibit strong magnetic properties due to the fact that their magnetic moments tend to align readily along the direction ofthe external magnetic field.

such material remain partially magnetized even after the removal of the external field.

M.14

Ferromagnetic Materials

Unmagnetized domains Magnetized domains

M.15

Ferromagnetic Materials

Typical hysteresis curve for a ferromagnetic material

SaturationResidual flux density

M.16

Summary

Curl equation

Divergence equation(Gauss’s law)

CurrentChargesSource

Static Magnetic fieldStatic Electric field

vρ=⋅∇ D 0=⋅∇ B

JH =×∇0=×∇ E

Qs

=⋅∫ dsD 0=⋅∫s dsB

0=⋅∫c dlE Ic

=⋅∫ dlH

M.17

Inductance

Lij= flux linking due to current in Sicurrent in Si

L11 is called the self-inductance; L12 is called the mutual-inductance

∫ ⋅=Φ2

12 s 21 dsB

ijΦ

1

1212 IL Φ

=

M.18

Example: Inductance of a Coaxial Transmission Line

Due to the current I in the inner conductor, the magnetic field generated inside the coaxial transmission line is given by

rIrIH

IrdH

Idc

πµφ

π

φφφ

φ

π

φ

2ˆ

2

ˆˆ2

0

=⇒

=⇒

=⋅⇒

=⋅

∫∫

B

lH

C

M.19

Example: Inductance of a Coaxial Transmission Line

Over the planar surface S, B is perpendicular to the surface. Hence, the flux through S is

The inductance per unit length is

)/ln(2

2

abIl

drrIl

Bdrl

b

a

b

a

πµ

πµ

=

=

=Φ

∫

∫

)/ln(2

/

ab

lIL

πµ

=

Φ=