1. CHAPTER 3
MAGNETOSTATICS

2. MAGNETOSTATICS 3.1 BIOT-SAVART’S LAW 3.2 AMPERE’S CIRCUITAL LAW
3.3 MAGNETIC FLUX DENSITY
3.4 MAGNETIC FORCES
3.5 MAGNETIC MATERIALS

3. INTRODUCTION
Magnetism and electricity were considered distinct phenomena until 1820 when Hans Christian Oersted introduced an experiment that showed a compass needle deflecting when in proximity to current carrying wire.

4. INTRODUCTION (Cont’d)
He used compass to show that current produces magnetic fields that loop around the conductor. The field grows weaker as it moves away from the source of current.
A  represents current coming out of paper.
A represents current heading into the paper. 

5. INTRODUCTION (Cont’d)
The principle of magnetism is widely used in many applications:
Magnetic memory
Motors and generators
Microphones and speakers
Magnetically levitated high-speed vehicle.

6. INTRODUCTION (Cont’d)
Magnetic fields can be easily visualized by sprinkling iron filings on a piece of paper suspended over a bar magnet.

7. INTRODUCTION (Cont’d)
The field lines are in terms of the magnetic field intensity, H in units of amps per meter.
This is analogous to the volts per meter units for electric field intensity, E.
Magnetic field will be introduced in a manner paralleling our treatment to electric fields.

8. 3.1 BIOT-SAVART’S LAW
Jean Baptiste Biot and Felix Savart arrived a mathematical relation between the field and current.

9. BIOT-SAVART’S LAW (Cont’d)
To get the total field resulting from a current, sum the contributions from each segment by integrating:

10. BIOT-SAVART’S LAW (Cont’d)
Due to continuous current distributions:
Line current
Surface current
Volume current

11. BIOT-SAVART’S LAW (Cont’d)
In terms of distributed current sources, the Biot-Savart’s Law becomes:
Line current
Surface current
Volume current

12. DERIVATION Let’s apply
to determine the magnetic field, H everywhere due to straight current carrying filamentary conductor of a finite length AB .

13. DERIVATION (Cont’d)

14. DERIVATION (Cont’d)
We assume that the conductor is along the z-axis with its upper and lower ends respectively subtending angles and at point P where H is to be determined.
The field will be independent of z and φ and only dependant on ρ.

15. DERIVATION (Cont’d)
The term dL is simply and the vector from the source to the test point P is:
Where the magnitude is:
And the unit vector:

16. DERIVATION (Cont’d)
Combining these terms to have:

17. DERIVATION (Cont’d)
Cross product of :
This yields to:

18. DERIVATION (Cont’d) Trigonometry from figure, So,
Differentiate to get:

19. DERIVATION (Cont’d)
Remember!

20. DERIVATION (Cont’d)
Simplify the equation to become:

21. DERIVATION 1
Therefore,
This expression generally applicable for any straight filamentary conductor of finite length.

22. DERIVATION 2
As a special case, when the conductor is semifinite with respect to P,
The angle become:
So that,

23. DERIVATION 3
Another special case, when the conductor is infinite with respect to P,
The angle become:
So that,

24. HOW TO FIND UNIT VECTOR aφ ?
From previous example, the vector H is in direction of aφ, where it needs to be determine by simple approach:
Where,
unit vector along the line current
unit vector perpendicular from the line current to the field point

25. EXAMPLE 1
The conducting triangular loop carries of 10A. Find H at (0,0,5) due to side 1 of the loop.

26. SOLUTION TO EXAMPLE 1
Side 1 lies on the x-y plane and treated as a straight conductor.
Join the point of interest (0,0,5) to the beginning and end of the line current.

27. SOLUTION TO EXAMPLE 1 (Cont’d)
This will show how is applied for any straight, thin, current carrying conductor.
From figure, we know that
and from trigonometry and

28. SOLUTION TO EXAMPLE 1 (Cont’d)
To determine by simple approach:
and so that,

29. EXAMPLE 2
A ring of current with radius a lying in the x-y plane with a current I in the direction. Find an expression for the field at arbitrary point a height h on z axis.

30. SOLUTION TO EXAMPLE 2 Can we use ?
Solve for each term in the Biot-Savart’s Law

31. SOLUTION TO EXAMPLE 2 (Cont’d)
We could find:

32. SOLUTION TO EXAMPLE 2 (Cont’d)
It leads to:
The differential current element will give a field with:
from
from

33. SOLUTION TO EXAMPLE 2 (Cont’d)
However, consider the symmetry of the problem:
The radial components cancel but the components adds, so:

34. SOLUTION TO EXAMPLE 2 (Cont’d)
This can be easily solved to get:
At h=0 where at the center of the loop, this equation reduces to:

35. BIOT-SAVART’S LAW (Cont’d)
For many problems involving surface current densities and volume current densities, solving for the magnetic field using Biot-Savart’s Law can be quite cumbersome and require numerical integration.
There will be sufficient symmetry to be able to solve for the fields using Ampere’s Circuital Law.

36. SUMMARY (5) Material permeability µ can be written as:
and the free space permeability is:
The amount of magnetic flux Φ in webers through a surface is:
Since magnetic flux forms closed loops, we have Gauss’s Law for static magnetic fields:

37. SUMMARY (6)
The total force vector F acting on a charge q moving through magnetic and electric fields with velocity u is given by Lorentz Force equation:
The force F12 from a magnetic field B1 on a current carrying line I2 is:

38. SUMMARY (7)
The magnetic fields at the boundary between different materials are given by:
Where a21 is unit vector normal from medium to medium 1, and:

39. VERY IMPORTANT!
From electrostatics and magnetostatics, we can now present all four of Maxwell’s equation for static fields:
Integral Form
Differential Form