Chapter 7 book for engineering mathematics, Exams of Electromagnetic Engineering

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Magnetostatic Fields

Dr. Talal Skaik

Islamic University of Gaza – Palestine

2

Magnet and Magnetic Field

A compass needle is deflected by the

direct current flowing in a conductor

Magnetic Field Around

Current Carrying Wires

The iron filings form circles around the wire along the magnetic field

Biot-Savart’s law states that the magnetic field intensity dH

produced at a point P by a 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 and its direction can be obtained by right handed screw rule.

Biot-Savart’s Law

7

2

2

2

sin

= sin , 1/ 4

= sin 4

dH I dl R or dH kI dl k R dH I dl R



 (^) 

 





8

2

2 3

R

sin

4

H l^ l^ R 4 4 where =|R| and a =R/

R

I dl dH R

d I d^ a^ I d R R R R

 

 

 ^  

The direction of dH can be

determined by the right-

hand rule or right-handed

screw rule.

For different current distributions:

10

2

2

2

l (line current)

K S (surface current) (K: surface current density)

J (volume current) (J: volume current density)

R L R S R v

H I d^ a

R

H d^ a

R

H dv^ a

R













Biot-Savart’s Law

Current distributions: (a) line current, (b) surface current, (c) volume current.

Consider conductor of finite length AB, carrying current from point A

to point B.

Magnetic Field of straight Conductor

11

3 z

2 2 3/ 2 2 2 3

Consider the contribution H at

due to l at (0,0,z): H l^ R ,

But l and R= a a , ,

l R a , Hence H= a

Letting z= cot , cosec ,

H cosec

z

d P

d d I d

R

d dz a z so

d dz I^ dz

z

dz d

I d

  

               



2 2 1 3 1

2 1

sin

cosec 4

H= cos cos

a I a d

or I a

     



   

  

 

tan / cot

z z

   

  ^ 1+cot x=cosec x^2

H= (^4) ^ I^  cos  2 cos 1  a 

α 1 :outside, α 2 : inside 13

The conducting triangular loop in the figure carries a current of 10 A.

Find H at (0,0,5) due to side 1 of the loop.

14

Example 7.

(a) conducting triangular loop,

(b) side 1 of the loop.

16

Example 7.

Find H at (-3,4,0) due to the current filament shown in the figure.

(a) current filament along semi-infinite x - and z -axes, a ℓ and a  for H 2 only; (b) determining a  for H 2.

1 2 1 2

2 2 1 1/

1 o^2 o

Let H=H +H at P(-3,4,0), where H is due to current filament along x-axis, and H is due to current filament along z-axis. H cos cos 4 At P(-3,4,0), =(9+16) 5, =90 , = l

I (^) a

a a a



 

   

 

 



  , but , and 3 4 5 5 3 4 4 3 5 5 5 5

l z x y

z x y x y

a a a a a

a a a a a a





    

   ^        17

Example 7.2 Solution

A circular loop located at X^2 +y^2 = 9 , z= 0 carries a direct current of 10 A

along aФ. Determine H at ( 0 , 0 , 4 ) and ( 0 , 0 ,- 4 ).

19

Example 7.

(a) circular current loop, (b) flux lines due to the current loop.

3

The magnetic field intensity dH at point P(0,0,h) contributed by current element Idl is :

H l^ R 4 l , R=(0,0, ) ( , ,0) R , and

l R 0 0

z z

d I d R where d d a h x y a ha a a a d d

   

   

 

 

2

3/2^2 2 2

Hence,

H= 4

By symmetry, the contributions along add up to zero H 0

z

z z z

h d a d a h

d I h d a d a dH a dH a h a



  

 

    

     

Example 7.3 - solution