Magnetostatic Fields Dr. Talal Skaik Islamic University of Gaza Palestine 01
Introduction In chapters 4 to 6, static electric fields characterized by E or D (D=εE) were discussed. This chapter considers static magnetic fields, characterised by H or B (B=μH). As we have noticed, a distribution of static or stationary charges produces static electric field. If the charges move at a constant rate (direct current- DC), a static magnetic field is produced (magnetostatic field). Static magnetic field are also produced by stationary permanent magnets. 3
Magnet and Magnetic Field
The iron filings form circles around the wire along the magnetic field Magnetic Field Around Current Carrying Wires A compass needle is deflected by the direct current flowing in a conductor
Electrostatic Fields have dual equations for magnetostatic fields 6
Biot-Savart s Law 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. I dl sin dh R ki dl sin or dh =, k 1/ 4 R I dl sin dh = 4 R 7
I dl sin dh = 4 R I dl ar I dl R dh 3 4R 4R where R= R and a =R/ R R The direction of dh can be determined by the righthand rule or right-handed screw rule. 8
Biot-Savart s Law The direction of magnetic field intensity H (or current I) can be represented by a small circle with a dot or cross sign depending on whether H (or I) is out of, or into the page. 9
For different current distributions: Biot-Savart s Law Current distributions: (a) line current, (b) surface current, (c) volume current. H H H L S v I dl a 4 R K ds a 4 R J dv a 4 R R R R (line current) (surface current) (K: surface current density) (volume current) (J: volume current density) 10
Magnetic Field of straight Conductor Consider conductor of finite length AB, carrying current from point A to point B. Consider the contribution dh at P IdlR due to dl at (0,0,z): dh, 3 4 R But dl dz a and R= a za, so, I dz dlr dz a, Hence H= a 3/ 4 z Letting z= cot, dz cosec d, I H 4 tan / z z cot 1+cot x=cosec x z 3 3 cosec 4 1 1 or z cosec d I a a sind I H= cos cos1 4 a 11
I H= cos cos1 4 When the conductor is semi-infinite, so that point A is now at O(0,0,0) o o while B is at (0,0, ), 1=90, =0 : H= o o B is at (0,0, ), 1=180, =0 : I a 4 When the conductor is of infinite length, point A is at (0,0, ) while H= A simple approach to determine a : I a where a is is a unit vector along the line current, and a is a unit vector l along the perpendicular line from the line current to the field point. 1 l a a a a
I H= cos cos1 4 a α 1 :outside, α : inside 13
Example 7.1 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. (a) conducting triangular loop, (b) side 1 of the loop. 14
I Applying H= cos cos 1 a, that is valid for any straight, 4 thin, current carrying conductor.,, and a are found as follows: 1 o cos1 cos 90 0, cos, =5 9 a a a, but a a, and a a 1 l l x so a a a a x z y I H1 cos cos1 4, Hence 10 H1 0 ( ay ) 4 (5) 9 H 59.1 a ma/m y Example 7.1 Solution a z 15
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 only; (b) determining a for H. 16
Let H=H +H at P(-3,4,0), where H is due 1 1 to current filament along x-axis, and H is due to current filament along z-axis. I H cos cos1 4 1/ At P(-3,4,0), =(9+16) 5, =90, =0 o 1 a a a, l but a a, l z o Example 7. Solution a 3 4 and a ax ay 5 5 3 4 4 3 a az ax ay ax ay 5 5 5 5 17
or alternatively, from figure b, 4 3 a sina cosa a a 5 5 x y x y 4a 3 x ay 3 Thus, H 1 0 4 (5) 5 H 38.a 8.65 a ma/m a x y in this case is the negative of the regular 3 cylindrical coordinates: H 1 0 ( a 4 (5) To find H at P: 1 o 1 1 a of ) 47.75 a ma/m =4, 0, cos 3 / 5, and a a a a a a. Hence, 3 3 H1 1 az 3.88 az ma/m 4 (4) 5 Thus H=H +H 38.a 8.65a 3.88 a ma/m x y z or H= 47.75a 3.88 ma/m l x y z a 18 z
Example 7.3 A circular loop located at X +y =9, z=0 carries a direct current of 10 A along a Ф. Determine H at (0,0,4) and (0,0,-4). (a) circular current loop, (b) flux lines due to the current loop. 19
The magnetic field intensity dh at point P(0,0,h) contributed by current element Idl is : IdlR dh 3 4 R where dl da, R=(0,0, h) ( x, y,0) R a a a dl R 0 d 0 0 h Hence, I Example 7.3 - solution a z ha z, and h d a d az dh= h d a 3/ d az dh a dh zaz 4 h By symmetry, the contributions along a add up to zero H 0 0
Thus, H= dh a z z I d az I a z 3/ 3/ 0 4 h 4 h I or H= a 3/ z h (a) Substituting I=10 A, =3, h=4 gives 10(3) az H0, 0, 4 = 0.36 a A/m 3/ z 9 16 (b) replacing h by - h, dl R h d a d az (z-component remains the same) Example 7.3 - solution Hence, H 0, 0, 4 H 0, 0, 4 0.36 a A/m 1 z
AMPERE'S CIRCUIT LAW - MAXWELL'S EQUATION Ampere's circuit law states that the line integral of the tangential component of H around a closed path is the same as the net current I enc enclosed by the path. In other words, the circulation of H equals I enc ; that is, H dl= I (integral form of 3rd Maxwell's equation) enc Apply Stoke's theorem, I H dl H ds enc L But, I J ds enc S S H J (differential form of third Maxwell's equation)
AMPERE'S CIRCUIT LAW - MAXWELL'S EQUATION Hdl= I enc H dl apply to Amperian path ( L). L I enc enc current enclosed by amperian path. I I d is found if J(A/m ) is known: enc J S S or if K(A/m) is known for surface current density. 3
APPLICATIONS OF AMPER'S LAW A. Infinite Line Current Consider infinite filamentary current I along the z-axis. To determine H at point P, allow Amperian path passing through P, such that H is constant provided ρ is constant. The whole current I is enclosed by the path, according to Amper's law: I H a d a I = H d H ( ) 0 I or H= a Hdl= I enc 4
B. Infinite Sheet of Current Consider an infinite current sheet in the z=0 plane with uniform current desnity K= K y a A/m. Applying Ampere's law to the y rectangular closed 1--3-4-1 path gives: H dl I K b Consider the sheet as comprising of filaments Consider dh above and below the sheet due to a pair of filamentary currents. enc y 5
The resultant dh has only an x-component. H on one side is the negative of that on the other side. H0ax z 0 H= H0ax z 0 Evaluating the line integral of H along the closed path: 3 4 1 H. dl H. dl= Ienc K yb 1 3 4 1 0( a)+( H0)( b) 0( a) H0( b)=h0b K yb H 0 K 1 K yax z 0 H= 1 K yax z 0 In general, for an infinite sheet of current density K A/m, 1 H= K a n where a n is a unit normal vector directed from the current sheet to the point of interest y 6
Magnetic field of Infinite Sheet of Current 1 H= K a n 7
Infinitely Long Coaxial Transmission Line 8
Infinitely Long Coaxial Transmission Line Consider infinitely long coaxial transmission line of two concentric cylinders The inner conductor has radius a and carries current I, the outer conductor has inner radius b and thickness t and carries return current I. To determine H everywhere, apply Ampere's law in four possible regions: 0 a, a b, b b t, b t 9
For region 0 a : apply Ampere's law to path L L 1 H dl I J ds enc Since the current is uniformly distributed over I the cross section, J= a, S= z d d d a a I I I J ds d d enc a a 0 0 1 I H dl H or H a a L For region a b : Apply Ampere's law to path L L H dl I I (since the whole current is enclosed by L ) enc I H I or H (same as infinite straight filamentary current) I z I a 1 30
For region b b t : Apply Ampere's law to path L L 3 H dl H I where I Ienc I J ds, J= a b t b enc t 3 I b Thus, Ienc I d d I 1 b t b t bt 0 b I H 1 For region bt: Apply Ampere's law to path L 4 L 4 b bt H. dl I I 0 or H 0 z 31
Infinitely Long Coaxial Transmission Line Putting all equations together: I a 0 a a I a a b H= I b H 1 a b b t t bt 0 bt 3
Infinitely Long Coaxial Transmission Line 33
Example 7.5 Planes z=0 and z=4 carry current K= -10 a x A/m and K=10 a x A/m, respectively. Determine H at (a) (1,1,1) (b) (0,-3,10) 34
Example 7.5 Let H=H +H, where H +H are 0 4 0 4 the contributions due to current sheets z=0 and z=4, (a) At (1,1,1), which is between the plates (0 ( z 1) 4), H (1/ )K a (1/ )( 10a ) a 5a A/m 0 n 1 n Hence, H=10a A/m x x z y H (1/ )K a (1/ )(10a ) ( a ) y z 5a A/m y (b) At (0,-3,10), which is above the sheets ( z 10 4 0), H (1/ )K a (1/ )( 10a ) a 5a A/m 0 n H (1/ )K a (1/ )(10a ) a 5a A/m 1 n Hence, H= 0 A/m x z y x z y 35
Example 7.6 A toroid whose dimensions are shown in the figure has N turns and carries current I. Determine H inside and outside the toroid. 36
The net current enclosed by the Amperian path is H dl= I H ( ) NI or H= where 0 enc NI, for a a 0 0 is is the mean radius of the toroid. NI. Hence, Outside the toroid, the current enclosed by an Amperian path is NI - NI 0 and Hence H =0 Example 7.6 37
Magnetic Field of a Toroid 38
Electric flux density and Electric field intensity are related by D=ε 0 E in free space. Similarly, the magnetic flux density B is related to the magnetic field intensity H by: Magnetic Flux Density Where μ 0 is known as the permeability of free space. The magnetic flux through a surface is given by B H 0 7 0 4 10 H/m = BdS S Where Ψ is in webers (Wb), B is in (Wb/m ) or teslas (T). 39
Magnetic Flux Lines Magnetic flux line is a path to which B is tangential at every point on the line. Each flux line is closed and has no beginning or end. 40
In an electrostatic field the flux crossing a closed surface is the same as the charge enclosed. DdS Q S So it is possible to have an isolated electric charge and the flux lines produced by it need not be closed. Unlike electric flux lines, magnetic flux lines always close upon themselves. This is because it is not possible to have isolated magnetic poles (or magnetic charges). BdS 0 S 41
It is not possible to isolate the north and south poles of a magnet. 4
Broken Magnet 43
Gauss s Law for magnetostatic fields The total flux through a closed surface in a magnetic field is zero. BdS 0 Applying divergence theorem, Or S BdS B dv 0 v B 0 This is the Maxwell s fourth equation. It states that magnetostatic fields have no sources or sinks. 44
Maxwell s Equations for Static Fields 45
Magnetic Scalar and Vector Potentials *In electrostatics electric field intensity and potential are related by: E V *Similar to this we can relate magnetic field intensity with two magnetic potentials: Magnetic scalar potential (V m ) Magnetic vector potential (A) *Magnetic scalar potential V m is related to H by the relation: J= H= ( V ) 0 (since for any scalar, ( V) 0) so the magnetic scalar potential V is only defined in the region where J=0. V m satisfies Laplace's equation m H V m if J=0 m V m 0 (J=0) 46
Since for a magnetostatic field B=0 For any vector ( A)=0 we can define magnetic vector potential A such that B= A In many EM problems it is more convenient to first find A and then find B from it. A 0 L Idl 4 R Magnetic Vector Potential for line current A S KdS 0 4 R for surface current A v Jdv 0 4 R for volume current 47
Magnetic flux from vector potential The magnetic fluc through a given area can be found from = B ds Applying Stoke's theorem, we obtain, = B ds ( A) ds A dl S S L S A dl L 48
Example 7.7 Given the magnetic vector potential A=-ρ /4 a z Wb/m, calculate the total magnetic flux crossing the surface Ф=π/, 1 ρ m, 0 z 5 m. Example 7.7 Solution Method 1 Az B A= a a, ds= d dz a Hence, 5 B ds= d dz (5) 1 1 15 4 4 3.75 Wb z0 1 1 49
Method A dl= L 1 3 1 3 4 where L is the path bounding surface S. Since A has only z-component, Example 7.7 Solution Continued 0 1 4 1 dz () 4 1 15 (1 4)(5) 3.75 Wb 4 4 5 0 0 5 dz A=-ρ /4 a z 50