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CHAPTER 3 MAGNETOSTATICS 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 2 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. 3 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. Figure 3-7 (p. 102) Oersted’s experiment with a compass placed in several positions in close proximity to a current-carrying wire. The inset shows used to represent the cross section for current coming out of the paper: this represents the head of an arrow. A 4 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. 5 INTRODUCTION (Cont’d) Magnetic fields can be easily visualized by sprinkling iron filings on a piece of paper suspended over a bar magnet. (a) (b) 6 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. 7 3.1 BIOT-SAVART’S LAW Jean Baptiste Biot and Felix Savart arrived a mathematical relation between the field and current. dH I1dL1 a12 4 R12 2 Figure 3-8 (p. 103) Illustration of the law of Biot–Savart showing magnetic field arising from a 8 BIOT-SAVART’S LAW (Cont’d) To get the total field resulting from a current, sum the contributions from each segment by integrating: IdL a R H 2 4R 9 BIOT-SAVART’S LAW (Cont’d) Due to continuous current distributions: Line current Surface current Volume current 10 BIOT-SAVART’S LAW (Cont’d) In terms of distributed current sources, the Biot-Savart’s Law becomes: IdL a R H 4R 2 KdS a R H 4R 2 JdV a R H 4R 2 Line current Surface current Volume current 11 DERIVATION Let’s apply IdL a R H 2 4R to determine the magnetic field, H everywhere due to straight current carrying filamentary conductor of a finite length AB . 12 DERIVATION (Cont’d) DERIVATION (Cont’d) We assume that the conductor is along the zaxis with its upper and lower ends respectively subtending angles 1 and 2 at point P where H is to be determined. The field will be independent of z and φ and only dependant on ρ. 14 DERIVATION (Cont’d) The term dL is simply dza z and the vector from the source to the test point P is: R Ra R za z a Where the magnitude is: R z2 2 And the unit vector: aR za z a z2 2 15 DERIVATION (Cont’d) Combining these terms to have: IdL a R IdL R H 2 3 4R 4R B Idza za a z z 2 2 32 A 4 z 16 DERIVATION (Cont’d) Cross product of dza z za z a : a dL R 0 a 0 0 az dz dza z This yields to: B H A 4 z Idz 2 2 32 a 17 DERIVATION (Cont’d) Trigonometry from figure, tan So, z cot z Differentiate to get: I H 4 2 1 dz cos ec 2d 2 cos ec 2d 2 cot 2 2 32 a 18 DERIVATION (Cont’d) Remember! u 2 cot(u ) cos ec (u ) x x 2 2 1 cot (u ) cos ec (u ) 19 DERIVATION (Cont’d) Simplify the equation to become: I H 4 I 2 2 cos ec 2d 1 3 cos ec 3 a 2 sin d a 4 1 I 4 cos 2 cos 1 a 20 DERIVATION 1 Therefore, H I 4 cos 2 cos1 a This expression generally applicable for any straight filamentary conductor of finite length. 21 DERIVATION 2 As a special case, when the conductor is semifinite with respect to P, A at 0,0,0 B at 0,0, or 0,0, The angle become: So that, H I 4 1 900 , 2 00 a 22 DERIVATION 3 Another special case, when the conductor is infinite with respect to P, A at 0,0, B at 0,0, The angle become: So that, H I 2 1 180 0 , 2 00 a 23 HOW TO FIND UNIT VECTOR aaφ ? From previous example, the vector H is in direction of aφ, where it needs to be determine by simple approach: a al a Where, al unit vector along the line current a unit vector perpendicular from the line current to the field point 24 EXAMPLE 1 The conducting triangular loop carries of 10A. Find H at (0,0,5) due to side 1 of the loop. 25 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. 26 SOLUTION TO EXAMPLE 1 (Cont’d) This will show how H I 4 cos 2 cos1 a is applied for any straight, thin, current carrying conductor. 1 900 cos 1 0 2 and 5 cos 2 29 From figure, we know that and from trigonometry 27 SOLUTION TO EXAMPLE 1 (Cont’d) To determine a by simple approach: al a x and a az so that, a al a a x a z a y H I 4 cos 2 cos1 a 10 2 0 a y 59.1 a y m A m 4 5 29 28 EXAMPLE 2 A ring of current with radius a lying in the x-y plane with a current I in the a direction. Find an expression for the field at arbitrary point a height h on z axis. 29 SOLUTION TO EXAMPLE 2 Can we use H I 4 cos 2 cos1 a ? Solve for each term in the Biot-Savart’s Law 30 SOLUTION TO EXAMPLE 2 (Cont’d) We could find: dL ada R Ra R ha z aa R h a 2 a R 2 ha z aa h a 2 2 31 SOLUTION TO EXAMPLE 2 (Cont’d) It leads to: IdL a R IdL R H 2 3 4R 4R 2 Iada ha aa z 2 2 32 0 4 h a The differential current element will give a field with: a from a a z az from a a 32 a) SOLUTION TO EXAMPLE 2 (Cont’d) (b) the problem: However, consider the symmetry of The radial components cancel but the a z components adds, so: 3-10a (p. 105) nt to find H a height h 2above a ring 2 a z (b) The entered in the x – Ia y plane. H d 3 2 values are shown for use in the 2 2 0 4 h a t equation. (c) The radial s of H cancel by symmetry. Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth Copyright © 2005 by John Wiley & Sons. All rights reserved. 33 SOLUTION TO EXAMPLE 2 (Cont’d) This can be easily solved to get: H Ia 2 2 h2 a 2 a 32 z At h=0 where at the center of the loop, this equation reduces to: I H az 2a 34 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. 35 SUMMARY (5) Material permeability µ can be written as: and the free space permeability is: 0 r 0 4 10 7 H m • The amount of magnetic flux Φ in webers through a surface is: B dS Since magnetic flux forms closed loops, we have Gauss’s Law for static magnetic fields: B dS 0 36 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: F qE u B The force F12 from a magnetic field B1 on a current carrying line I2 is: F12 I 2dL2 B1 37 SUMMARY (7) • The magnetic fields at the boundary between different materials are given by: a 21 H1 H 2 K Where a21 is unit vector normal from medium 2 to medium 1, and: BN BN 1 2 38 VERY IMPORTANT! From electrostatics and magnetostatics, we can now present all four of Maxwell’s equation for static fields: D dS Qenc B dS 0 E dL 0 H dL I enc D v Integral Form Differential Form B 0 E 0 H J 39