ENE 325 Electromagnetic Fields and Waves Lecture 4 Magnetostatics.

ENE 325ENE 325Electromagnetic Electromagnetic Fields and WavesFields and Waves

Lecture 4 Lecture 4 MagnetostaticsMagnetostatics

IntroductionIntroduction (1)(1) source of the steady magnetic field may be a per

manent magnet, and electric field changing linear ly with time or a direct current.

a schematic view of a bar magnet showing the m agnetic field. Magnetic flux lines begin and termi

nate at the same location, more like circulation.

IntroductionIntroduction (2)(2) Magnetic north and south poles are alwa

ys together.

N

S

N

S

N

S

N

S

N

S

N

S

IntroductionIntroduction (3)(3) Oersted’s experiment shows th at current produ

ces magnetic fields that loop around the conducto r. The field grows weaker as one compass moves a

way from the source of the current.

- Bi o Savart l aw- Bi o Savart l aw(1)(1) - The law of Bio Savart states that at any point P the magni

tude of the magnetic field intensity produced by the differ ential element is proportional to the product of the current

, the magnitude of the differential length, and the sine of t he angle lying between the filament and a line connecting the filament to the point P at which the filed is desired.

The magnitude of the magnetic field intensity is inversely proportional to the square of the distance from the differe

ntial element to the point P.

- Bi o Savart l aw- Bi o Savart l aw(2)(2) The direction of the magnetic field intensity is n

ormal to the plane containing the differential fila ment and the line drawn from the filament to the point P.

- Bio Savart law is a method to determine the mag netic field intensity. It is an analogy to Coulomb’s

law of Electrostatics.

- Bi o Savart l aw- Bi o Savart l aw(3)(3)

2 34 4rId L a Id L R

dHr r

��

��

from this picture:

1 121

2 2124

I d L adH

R

��

��

Total field A/m

24rId L a

Hr

��

Magnetic field intensity resulting fr Magnetic field intensity resulting fr om om an infinite length line of current an infinite length line of current

(1)(1)

Pick an observation point P located on axis.

The current

The vector from the source to the test point is

zId L Idza

��

R zRa za a

a unit vector

2 2

zR

za aa

z

Magnetic field intensity resulting fr Magnetic field intensity resulting fr om om an infinite length line of current an infinite length line of current

(2)(2)

then

From a table of Integral,

3/ 22 24

z zIdza za aH

z

��

3/ 22 2.

4

�� I a dzH

z

3/ 2 2 2 22 2

dx x

a x ax a

then

2 2 24

I a zH

z

�� /

2

I a

A m

Magnetic field intensity resulti Magnetic field intensity resulti ng from ng from a ring of current a ring of current (1)(1)

A ring is located on z = 0 plane with the radius a. The observation point is at z = h.


Id L Iad a��

R zRa ha aa

2 2

zR

ha aaa

h a

A unit vector


Consider a symmetry 2zId L R had a a d a

��

components are cancelled out due to symmetry of two segments on the opposite sides of the ring.

Therefore from

24rId L a

Hr

��

we have

2

3/ 22 20 4

zIad a ha aaH

h a

��


then

2 2

3/ 22 2 0.

4

zIa aH d

h a

��

We finally get

2

3/ 22 22

��z

IaH a

h a

Magnetic field intensity resulti Magnetic field intensity resulti ng from ng from a rectangular loop of c a rectangular loop of c

urrenturrent (1)(1)

Find the magnetic field intensity at the origin. By symmetry, there will be equal magnetic field intensity at each half width (w/2).


urrenturrent (2)(2)Consider 0 x w/2, y = -w/2

xId L Idxa

��

( / 2)R x yRa xa w a

A unit vector

2 2

( / 2)

( / 2)

x yR

xa w aa

x w

We have

3/ 22 2

( / 2)

4 ( / 2)

x x yIdxa xa w adH

x w

��


urrenturrent (3)(3)

3/ 22 2

( / 2)

4 ( / 2)

zIdx w a

x w

Then the total magnetic field at the origin is

/ 2

3 / 22 20

( / 2)8

4 ( / 2)

wzIdx w a

Hx w

��

Look up the table of integral, we find 3/ 2 2 2 22 2

dx x

a x ax a

then A/m. 2 2z

IH a

W

��

Bio-Savart law in different Bio-Savart law in different formsforms

We can express - Bio Savart law in terms of surface and volume current densities by re placing with and :

Id L��

KdS��

Jdv��

24rKdS a

Hr

��

where K = surface current density (A/m) I = K x width of the current sheet

and 24rJdv a

Hr

��

Ri ght handrul e Ri ght handrul e

The method to determine the result of cros s product (x).

F qv B ��

AmpAmpéé re’s circuital law re’s circuital law

Analogy to Gauss’s law Use for magnetostatic’s problems with sufficient symme

try. Ampere’s circuital law – the integration of around any c

losed path is equal to the net current enclosed by that p ath.

To find , choose the proper Amperian path that is every where either tangential or normal to and over which is

constant.

encH dL I��

A

Use Ampere’s circuital law to d Use Ampere’s circuital law to d etermine etermine

from the infinite line of current from the infinite line of current . .H��

From encH dL I��

H H a��

dL d a ��

then 2

0.

�� encH dL H a d a I

�� IH a

2A/m.

Magneti c fi el dof Magneti c fi el dof the the uni foruni for msheet of current msheet of current (1)(1)

xK Ka

��

Create path a-b-c-d and perform the integration along the path.

Magnetic field of Magnetic field of the the uniform s uniform s heet of current heet of current (2)(2)


( ) ( ) ( ) ( ) ,y z y z xH w H h H w H h K w

divide the sheet into small line segments along x-axis, by symmetry Hz is cancelled.

y x2H K .

Because of the symmetry, the magnetic field intensity on one side of the current sheet is the negative of that on the other.

.xyK

H2

where is a unit vector normal to the current sheet.

Magnetic field of Magnetic field of the the uniform s uniform s heet of current heet of current (3)(3)

.

Above the sheet,

y1 x1

H K2

(z > 0)

and y2 x1

H K2

(z < 0)

or we can write A/m n

1H K a

2

��

na

Magnetic field inside the sol Magnetic field inside the sol enoid enoid

a

b c

d

x

y z

h

w


Magnetic field inside the Magnetic field inside the toroid that has a circular cross toroid that has a circular cross section (1)section (1)

Magnetic field inside the Magnetic field inside the toroid that has a circular cross toroid that has a circular cross section (2)section (2) From encH dL I

��

Ex1Ex1 Determine Determine at point P (0.01, at point P (0.01, 0, 0) m from two current filaments 0, 0) m from two current filaments as shown. as shown.

H��

2Ex2Ex Determine for the coaxial c Determine for the coaxial c able that has a inner radius able that has a inner radius a a = 3 m = 3 m

m, m, bb 9= mm, and 9= mm, and cc 12= mm. Giv 12= mm. Giv en en II AA08 AA08

H��

a) at < a

b) at a < < b

c) at b < < c

d) at > c

33 Determine at point (10, 0, 0) Determine at point (10, 0, 0) mm resulted from three current sh mm resulted from three current sh

eets: eets: KK11

15 15 A/m at A/m at x x A AAA 6 A AAA 6 KK22

A A -3-3 A/m at A/m at xx A A A A AAA 9 A A A A AAA 9 KK

33 A A1 A A1

55 A/m at A/m at xx AAA12 AAA12

H��

ya

yaya

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ENE 325 Electromagnetic Fields and Waves Lecture 4 Magnetostatics.

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