chapter 2

MAGNETOSTATIC

CHAPTER 2

ELECTROMAGNETIC
FIELDS & WAVES
BEV 20303

•What appear in
your mind when talk
about “MAGNET”?

Why magnetic bar has
magnetic field surround

it ?

•A magnetic bar can have a magnetic field because of the alignment of its
atomic or molecular magnetic dipoles.

•In materials like iron, nickel, and cobalt, the atoms have magnetic
moments due to the alignment of their electron spins.

•When these materials are magnetized, the magnetic moments of
individual atoms align in the same direction, creating a net magnetic field
around the bar.

•This alignment can persist even after the external magnetic field is
removed, making the bar act as a magnet.

Magnetic Field

Magnetic fields can be produced by permanent magnets
and steady electric currents (DC Current) as illustrated in
figure below:

Magnetic Field Sources

Current in a wire

Solenoid

Magnetic Field

Magnetic fields can be produced by permanent magnets
and steady electric currents as illustrated in figure
below:

Magnetic Field Sources

Why Earth is a big
magnet?

• The movement of molten
iron and nickel located in the
outer core of the Earth is
responsible for generating the
Earth's magnetic field. This
process is known as the
geodynamo

Application of Electromagnets

Magnetic Relay is a switch to turn a circuit ON or OFF

A magnetic relay uses an electromagnet to switch
electrical circuits on or off. When current passes
through the relay's coil, it creates a magnetic field that
moves a switch (armature) to either open or close the
circuit, controlling the flow of electricity.

Application of Electromagnets

Doorbell

Magnetic Levitation
Application of Electromagnets

The train basically floats at a height of 1 or more centimeters above the track, which is made
possible by magnetic levitation. The train carries superconducting electromagnets that
induce currents in coils built into the guide rails alongside the train. The magnetic interaction
between the train’s superconducting electromagnets and the guide-rail coils serves not only
to levitate the train but also to propel it along the track.

Introduction

• An electrostatic field is

produced by static (stationary)
charges.

• The magnetostatic field or static

magnetic field is produced by a
constant current flow (direct
current).

• Two major laws governing

magnetostatic fields are;

• Biot-Savart’s law
• Ampere’s circuit law

A permanent magnet such as bar magnet has a magnetic field

surrounding it that consists of lines of forces or flux lines.

Iron filings in the
presence of a magnet

 The direction of the

magnetic

field

near

the poles of a magnet
is revealed by placing
compasses nearby.

 The

magnetic

field

points

towards

a

magnet's south pole
and

away

from

its

north pole.

Magnetic Field

Magnetic Field

• Attraction and Repulsion:

• Like poles repel each other

• Opposite poles attract each

other

Magnetic

Field

Magnetic Flux
Density, Tesla(T) or
Webers per square
meter (Wb/m²) or
volt-seconds per
square meter (V s/m²)

Magnetic Field
Intensity, Amperes
per meter (A/m).

Relationship

B

H

B=μH

Biot-Savarts’s Law states that the differential magnetic field
intensity, dH produced at a point P by the differential current
element I dl is proportional to the product I dl 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.











2

3

2

4

ˆ

4

ˆ

4

sin

R

Id

or

R

Id

R

Idl
d









R

R

l
H

l
H

The

direction

of

dH

can

be

determined by using:
 The right hand rule: where the

right hand thumb pointing in the
direction of the current and the
rest of right hand fingers encircling
the wire in the direction of dH.

 The

right-handed

screw

rule:

where the screw placed along the
wire and pointed in the direction
of current flow. The direction of
advance

of

the

screw

is

the

direction of dH.

The direction of dH can be represented by using a small
circle:
 with a dot sign to indicate the dH is outward and
 with a cross sign to indicate the dH is inward.

•

dH is out
dH is in

The magnetic field intensity, H for different current distributions:

where K is surface current density (A/m)

J is volume current density (A/m2)

I

dl

K
KdS

J
Jdv






L

R

Id

2

4

ˆ



R

l
H

Line current

Surface current

Volume current





v

R

dv

2

4

ˆ



R

J
H






S

R

dS

2

4

ˆ



R

K
H








































ˆ
cos

cos
4

ˆ

in
4

ˆ

cosec
cosec

4

cosec

cot

ˆ

)

(4

ˆ

zˆ

rˆ

zˆ
4

1

2

2

1

2

1

3

3

2

2

2

2/32

2

3



































r

I
H

d

s
I
H

d
r
r

I
H

d

r

dz

r

z

z

r

Irdz
H

rdz

R

ld

z

r

R

dz

ld

R

R

lId
Hd






















r

Exercise 1

Exercise 1

A current filament of 10 A is oriented in the positive y
direction, starting at (0,2,0) to (0,8,0). Determine the
magnetic field intensity at (0, 6, 1).

Exercise 2

Exercise 2

A filament is carrying 5 A of current from (0, 0 -5) to
(0, 0, 4). Compute the magnetic field intensity at (1, 3,
7).

Determine the magnetic field at (3, 0, 1) when there is a
2A of current element oriented from origin to (6, 0, 0).

Exercise 3

Exercise 3

Ans = 1.48 A/m

Ans = 0.011 -0.033

Ans :

Ans = 1.48 A/m

Ans = 0.011 -0.033

Ans :

Check Your Answer!

Check Your Answer!

Exercise 3

Exercise 3

Exercise 2

Exercise 2

Exercise 1

Exercise 1










ˆ
cos

cos
4

1

2


r
I
H


Semi infinite line

Infinite line

X

To infinity

X

To infinity

from infinity








ˆ

4

0

90

2

1

r
I
H

o

o















ˆ

2

0

180

2

1

r
I
H

o

o









Exercise 1

Exercise 1

A infinite current line of 3 A is laying along the x axis, find the intensity
magnetic field at (1,2,3).

Exercise 2

Exercise 2

An infinitely long conducting filament is placed along the x-axis and
carries current 10mA in the direction. Find H at (2,3,4).

Exercise 3

Exercise 3

A semi infinite current filament begin at (0, 6, 0) and extend to –
infinity in - direction. If it carries 4 A of current, Find the H at (-
2,8,1).

The z- and x- axis respectively carry filamentary
currents of 20A along and 30A along . Calculate

at (6,8,-6).

Exercise 4

Exercise 4

Biot-Savart’s Law Biot-Savart’s Law

Ans = 0.073

Ans =

A/m

Ans =

A/m

Ans :

Ans = 0.073

Ans =

A/m

Ans =

A/m

Ans :

Check Your Answer!

Check Your Answer!

Exercise 3

Exercise 3

Exercise 2

Exercise 2

Exercise 1

Exercise 1

Exercise 4

Exercise 4

A semi infinite line with current of 4 A begins at the + infinity of z axis, it is
bent toward the positive x axis at the origin and stop at (4,0,0). Determine
the magnetic field at (-3,4,0)

Homework 1

Homework 1

Homework 2

Homework 2

A square conducting loop 3 cm on each side carries a current of 10A. Calculate
the magnetic field intensity at the center of the loop.

Ampere’s Law

• Why need this law?
• When to apply this law

Ampere’s law states that the line integral of the
tangential components of H around a closed path is the
same as the net current Ienc enclosed by the path.
 The integral form of Ampere’s Circuit Law

 Ampere’s Circuit Law is used when we want to determine H

when the current distribution is symmetrical.





enc
I

dl

H

Ampere’s Law




enc
I

dl

H

 By applying Stokes’s theorem to integral form of Ampere’s

Circuit Law. We obtain

 But

 Therefore, the differential or point form of Ampere’s Circuit

Law is given by;

**Third Maxwell’s Equation: Ampere’s law in point form

S

H

l

H

d

d

I

S

L
enc













)

(







S
enc
d

I

S

J

J

H 



Infinite Line Current:

To

determine

H

at

an

observation point P, we

allow a closed path pass

through P known as an

Amperian path (analogous

to Gaussian surface)

Infinite Line Current:

 Since the amperian path encloses the whole current I,

according to Ampere’s law:

 Thus




ˆ

2 r

I


H






enc
I

dl

H

r

H

d

Hr

rd

H

I













2
ˆ

ˆ














Infinite Sheet of current:
 Consider an infinite current sheet in the z = 0 plane with a

uniform current density K = Ky
A/m

Z

x

Z

x

y

a

b

1

2

4

3

x

Infinite Sheet of current (solution):

Plane z = 0 carry current density (k) of -10

A/m, determine the magnetic

field at (1, 1, 1) and (0, -3, -10) respectively.

Exercise 1

Exercise 1

Restructing of Ideas

Restructing of Ideas

Exercise 2

Exercise 2

Additional plane z= 4 that carrying current density K=10

is added into

the Exercise 1. Find the total magnetic field at the same locations

Ampere’s Law Ampere’s Law

Exercise 3

Exercise 3

Infinite line y=1, and z=4 carries filamentary current 50

along

direction, while the z=0 plane carries 20mA/m along . Find

at (3, 4, 5)

A

Ans :

,

Ans :

A

Ans :

,

Ans :

Check Your Answer!

Check Your Answer!

Exercise 2

Exercise 2

Exercise 1

Exercise 1

Restructing of Ideas

Restructing of Ideas

Ampere’s Law Ampere’s Law

Exercise 3

Exercise 3

Infinitely long coaxial transmission line:

Infinitely long coaxial transmission line:

t

b

a

conductor

Dielectric

Infinitely long coaxial transmission line:

t

b

a

conductor

Dielectric

 Consider an infinitely long transmission

line consisting of two concentric
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.

 Since the current distribution is

symmetrical, we apply Ampere's law
for the Amperian path for each of the
four possible regions: 0 ≤ r ≤ a,
a ≤ r ≤ b, b ≤ r ≤ b + t, a n d r ≥ b + t .

Infinitely long coaxial transmission line:

Generation Of Ideas

Generation Of Ideas

Ampere’s Law Ampere’s Law

t

b

a

conductor

Dielectric

0 ≤

≤ a,

Infinitely long coaxial transmission line:

Generation Of Ideas

Generation Of Ideas

Ampere’s Law Ampere’s Law

t

b

a

conductor

Dielectric

a ≤

≤ b,

Infinitely long coaxial transmission line:

Generation Of Ideas

Generation Of Ideas

Ampere’s Law Ampere’s Law

t

b

a

conductor

Dielectric

b ≤ ≤ b + t,

Infinitely long coaxial transmission line:

Generation Of Ideas

Generation Of Ideas

Ampere’s Law Ampere’s Law

t

b

a

conductor

Dielectric

b + t ≤

Magnetic flux through a surface S is
given by:

Unit is webers (Wb) and

Unit of B is webers/square meter
(Wb/m2)

Magnetic flux density, B is given by:

Magnetic Flux Lines

I
S

N

S

(H/m)

10

4

7

0




 


is permeability of free space.

H

B
0








S
dS

B



Magnetic flux lines always close upon themselves -

NOT POSSIBLE to have isolated magnetic poles (or

magnetic charges)

An isolated magnetic charge does not exist

Thus the total flux through a closed surface is zero

0




S

B d

Law of conservation of magnetic flux
or Gauss’ Law for magnetostatic field

Magnetic Flux Density Magnetic Flux Density

By applying divergence theorem;

It suggest that magnetic field lines are always continuous

0

B

B

S

B
















0

S

v
dv

d

Fourth Maxwell's equation

Force Generated by Magnetic Field

Force Generated by Magnetic Field

• Force acting on a moving charged particle in a B

field.

Force Generated by Magnetic Field

• Force on a current element in an external B field.

Orientation

Orientation

Force Generated by Magnetic Field

• Force between two current elements.

I

I

Orientation

Orientation

Force Generated by Magnetic Field

• Force between two current elements.

I

I

Orientation

Orientation

Force Generated by Magnetic Field

• Force between two current elements.

I

I
F

Force Generated by Magnetic Field

• Force acting on a moving charged particle in a B

field.

Generation Of Ideas

Generation Of Ideas

Force acting on a moving charged

particle in a B field.

Generation Of Ideas

Generation Of Ideas

 The electric force, Fe acting on a charge q within an electric

field, E is,

Fe = qE

(N)

 The magnetic force, Fm acting on the individual charges, q

moving with constant velocity, u in a magnetic field, B is

Fm = q u X B (N)  Vector cross product

 The magnitude of magnetic force, Fm is

Fm = q u B sin (N)

where  is the angle between u and B

 If q is negative charge, the direction of Fm is reversed.

Force acting on a moving charged

particle in a B field.

Generation Of Ideas

Generation Of Ideas

Fm = q u B sin (N)

Force acting on a moving charged

particle in a B field.

Generation Of Ideas

Generation Of Ideas

 For a moving charge in the presence of both E and B fields, the

total electromagnetic on the charge is given by Lorentz force
equation as

F = Fe + Fm = qE+ qu X B = q(E+ u X B) (N)

 If the mass of the charged particle moving in E and B fields is m,

by Newton's second law of motion;

B)

u

E
u
a

F











(q
dt
d
m

m

A comparison between the electric force

and the magnetic force:

Magnetic Force, Fm
Electric Force, Fe

Fm is perpendicular to both u and B.
Fe and E have the same
direction.

Fm depends on the charge velocity
(charge in motion).

Fe is independent of the
velocity of the charge.

Fm cannot perform work when particle is
displaced because it is at right angles to
the direction of motion of the charge.

Fe expends energy (work
done) in displacing a
charged particle.

Fm does not cause an increase in kinetic
energy of the charge.

Fe change its kinetic
energy.

The magnitude of Fm is generally small
compared to Fe except at high velocities.

An negative charge of -50

is moving in the positive x-direction at the constant

velocity of 0.03m/s. It is perpendicular to a magnetic flux density of
Calculate the magnetic force act on this negative charge.

Exercise 1

Exercise 1

Exercise 2

Exercise 2

A proton Q = 30

moving with a speed of 2 x 106

m/s through a magnetic field

with magnetic flux density of 2.5

Tesla. Calculate the magnetic force, Fm

Exercise 3

Exercise 3

A proton, Q = 5

is moving with a speed of 1 x 106m/s through a magnetic field

with magnetic flux density of 5 Tesla. It experiences a magnetic force of magnitude
0.014 N. What is the angle between the magnetic field and the moving direction of
proton?

Exercise 4

Exercise 4

A charged particle with velocity u is moving in a medium containing uniform fields

and

. What should the velocity of the particle be so that the

particle experiences no net force on it?

Ans :

Ans : 150

Ans : 34.05 degree

Ans :

Ans :

Ans : 150

Ans : 34.05 degree

Ans :

Check Your Answer!

Check Your Answer!

Exercise 1

Exercise 1

Exercise 3

Exercise 3

Exercise 2

Exercise 2

Exercise 4

Exercise 4

Force Generated by Magnetic Field

• Force on a current element in an external B field.

Force on a current element in an

external B field

 Now we wish to find force, Fm on a line conducting current in

the presence of a magnetic field, B.

 The differential force, dFm on a small segment dq of a charge

moving with velocity u is

 The velocity of charge can be written as

 Then,

Force on a current element in an

external B field

• The total force on any closed current loop in a uniform magnetic

field is zero because the integral of the displacement vector dl
over a closed contour is zero.

***where dq/dt corresponds to the current I in the line and the B

field is external to the current element I dl

Force on a current element in an

external B field

No Current, no F

�

Exercise 1

Exercise 1

A finite current element is oriented from (8, 0, 0) and stop at the
origin. It is carrying 5 A of in the -x direction. This current
element is exposed to magnetic field flux density,

of 5 Tesla.

Solve for the force which is experienced by the current element.
Sketch a diagram and show the direction of the force.

Exercise 2

Exercise 2

2A of current element flows from (0, 3, 0) to (0, 3, 10). It is
placed in a environment which contains magnetic field flux
density,

of 10 Tesla. Predict the direction and magnitude of

the force exerted on this current element.

Exercise 3

Exercise 3

The semicircular conductor shown below lies in the y-x plane and carries a
current I. The closed circuit is exposed to a uniform magnetic field
�.

Determine
(a) the magnetic force,
�on the straight section of the wire,

(b) the force,
�on the curved section,

(c) the total force on the closed loop.

Force between two current

elements.

I1

I2
F

Force between two current

elements.

I1

I2

Ampere’s Law:

I1

I2

Force between two current

elements.

Ampere’s Law:

Exercise 1

Exercise 1

Two infinite and parallel filamentary current are separated by a
distance d = 4 m and carrying I = 6 A in opposite direction along
z-axis as shown in figure below. Determine :

y

z

x

I2
I1

d

a. The magnetic flux density

produced by current I1 at I2.

b. The magnetic flux density

produced by current I2 at I1.

c. Find the force per unit length

exerted on both current filament.

Exercise 2

Exercise 2

Two infinite length of wires are located as shown in figure below.
Determine the force per unit length on both of the wires.

y

z

x

I2 = 2 mA

I1= 15 A

(3, 0, 0)

Exercise 3

Exercise 3

A square loop of wire in the z = 0 plane carrying current 2 mA
in the field of an infinite filament on the y-axis as shown below.
Determine the total force on the loop.
Ans :

y

z

x

I2 = 2 mA

I1= 15 A

(1, 2, 0)

(3, 0, 0)

(1, 0, 0)

1

2

3

4