Electromagnetics Wave and Fields Lecture 7 AIUB Dr. M. Tanseer Ali EMWF Lec 7 /1 Engineering Electromagnetics Chapter 8 The Steady Magnetic Field AIUB Dr. M. Tanseer Ali EMWF Lec 7 /2 Chapter 8 The Steady Magnetic Field Steady Magnetic Field This chapter discusses the magnetic field with definition of the magnetic field itself and show how it arises from a current distribution. The source of magnetic field may be a permanent magnet, an electric field changing linearly with time or a direct current. Biot-Savart’s Law describes the relation between the magnetic field produced by the differential DC element in free space. AIUB Dr. M. Tanseer Ali EMWF Lec 7 /3 Chapter 8 The Steady Magnetic Field Biot-Savart Law When current I is flowing through the differential length dL of cylindrical conductor of circular cross section as the radius approaches to zero; then in vector notation the Biot-Savart Law can be given as: dH IdL aR IdL R 2 4R 4R 3 Where H is the magnetic field intensity, R is the distance at which the magnetic field is measured and aR is the unit vector along radial distance of the point from the filament. dH AIUB I dL sin aR 4R 2 Dr. M. Tanseer Ali θ dH 2 I1dL1 aR12 4R122 EMWF Lec 7 /4 Chapter 8 The Steady Magnetic Field Biot-Savart Law Integral form of Biot-Savart Law H IdL aR 4R 2 Current can be expressed with surface density (K) or volume density (J) I dL K dS J dV K aR dS J aR dV vol 4R 2 S 4R 2 H AIUB Dr. M. Tanseer Ali EMWF Lec 7 /5 Chapter 8 The Steady Magnetic Field Biot-Savart Law Let us find the magnetic field intensity produced by an infinite long filament carrying a current I. The filament lies on the z axis in free space, flowing to az direction. R a z ' a z and 4 2 z '2 I 4 AIUB 3/ 2 4 2 z '2 z 2 z '2 I1dz ' a z a z ' a z Considering cylindrical coordinate system, no variation with z or Φ can exist. a R12 I1dz ' a z a z ' a z dH 2 H a z' a z z I a 3/ 2 dz ' a 2 z' 2 3/ 2 z' 2 2 2 z '2 Dr. M. Tanseer Ali z I 2 a EMWF Lec 7 /6 Chapter 8 The Steady Magnetic Field Biot-Savart Law The direction of the magnetic-field-intensity vector is circumferential. The streamlines are therfore circles about the filament. H AIUB Dr. M. Tanseer Ali I 2 a EMWF Lec 7 /7 Chapter 8 The Steady Magnetic Field Biot-Savart Law For a finite length of wire, H at any point can be expressed in terms of α1 and α2 H I 4 sin 2 sin1 a If one or both ends are below point 2, then α1 , or both α1 and α2 are negative. This equation can be used to find the magnetic field intensity caused by current filaments arranged as sequence of straight line segments. AIUB Dr. M. Tanseer Ali EMWF Lec 7 /8 Chapter 8 The Steady Magnetic Field Example Determine H at P2(0.4,0.3,0) in the field of an 8-A filament current directed inwards from infinity to the origin on the positive x axis and outward to infinity along y axis. x 0.3 8 sin 53.1 1a H 2 x 4 0.3 12 a A/m The unit vector aϕ on z = 0 plane is same downward direction along az with respect to x axis. 1x 90 2x 0.4 tan 53.1 0.3 -1 H 2 x 12 a z A/m aΦ or -az AIUB Dr. M. Tanseer Ali EMWF Lec 7 /9 Chapter 8 The Steady Magnetic Field Example Determine H at P2(0.4,0.3,0) in the field of an 8-A filament current directed inwards from infinity to the origin on the positive x axis and outward to infinity along y axis. y 0.4 8 1 sin 36.9 a z H 2 y 4 0.4 8 a z A/m H 2 H 2 x H 2 y 2 y 90 0.3 1y tan 36.9 0.4 -1 aΦ or -az AIUB Dr. M. Tanseer Ali 12 20 az 8 a z A/m a z 6.37a z A/m EMWF Lec 7 /10 Chapter 8 The Steady Magnetic Field Ampere’s Circuital Law In solving electrostatic problems, whenever a high degree of symmetry is present, we found that they could be solved much more easily by using Gauss’s law compared to Coulomb’s law. Again, an analogous procedure exists in magnetic field. Here, the law that helps solving problems more easily is known as Ampere’s circuital law. The derivation of this law will waits until several subsection ahead. For the present we accept Ampere’s circuital law as another law capable of experimental proof. Ampere’s circuital law states that the line integral of magnetic field intensity H about any closed path is exactly equal to the direct current enclosed by that path, H dL I AIUB Dr. M. Tanseer Ali EMWF Lec 7 /11 Chapter 8 The Steady Magnetic Field Ampere’s Circuital Law • The line integral of H about the closed path a and b is equal to I • The integral around path c is less than I. The application of Ampere’s circuital law involves finding the total current enclosed by a closed path. AIUB Dr. M. Tanseer Ali EMWF Lec 7 /12 Chapter 8 The Steady Magnetic Field Ampere’s Circuital Law Let us again find the magnetic field intensity produced by an infinite long filament carrying a current I. The filament lies on the z axis in free space, flowing to az direction. We choose a convenient path to any section of which H is either perpendicular or tangential and along which the magnitude H is constant. The path must be a circle of radius ρ, and Ampere’s circuital law can be written as 2 H dL 0 AIUB 2 H d H d H 2 I 0 Dr. M. Tanseer Ali H I 2 EMWF Lec 7 /13 Chapter 8 The Steady Magnetic Field Ampere’s Circuital Law As a second example, consider an infinitely long coaxial transmission line, carrying a uniformly distributed total current I in the center conductor and –I in the outer conductor. A circular path of radius ρ, where ρ is larger than the radius of the inner conductor a but less than the inner radius of the outer conductor b, leads immediately to I H ( a b) 2 If ρ < a, the current enclosed is I encl 2 H I 2 a2 Resulting H I ( a) 2 2 a AIUB Dr. M. Tanseer Ali EMWF Lec 7 /14 Chapter 8 The Steady Magnetic Field Ampere’s Circuital Law If the radius ρ is larger than the outer radius of the outer conductor, no current is enclosed and H 0 ( c) Finally, if the path lies within the outer conductor, we have 2 b2 2 H I I 2 2 c b I c2 2 H 2 c 2 b2 (b c) • ρ components cancel, z component is zero. • Only φ component of H does exist. AIUB Dr. M. Tanseer Ali EMWF Lec 7 /15 Chapter 8 The Steady Magnetic Field Ampere’s Circuital Law The magnetic-field-strength variation with radius is shown below for a coaxial cable in which b = 3a, c = 4a. It should be noted that the magnetic field intensity H is continuous at all the conductor boundaries The value of Hφ does not show sudden jumps. Outside the coaxial cable, a complete cancellation of magnetic field occurs. Such coaxial cable would not produce any noticeable effect to the surroundings (“shielding”) AIUB Dr. M. Tanseer Ali EMWF Lec 7 /16 Chapter 8 The Steady Magnetic Field Ampere’s Circuital Law As final example, consider a sheet of current flowing in the positive y direction and located in the z = 0 plane, with uniform surface current density K = Ky ay. Due to symmetry, H cannot vary with x and y. If the sheet is subdivided into a number of filaments, it is evident that no filament can produce an Hy component. Moreover, the Biot-Savart law shows that the contributions to Hz produced by a symmetrically located pair of filaments cancel each other. Hz is zero also. Thus, only Hx component is present. AIUB Dr. M. Tanseer Ali EMWF Lec 7 /17 Chapter 8 The Steady Magnetic Field Ampere’s Circuital Law We therefore choose the path 1-1’-2’-2-1 composed of straightline segments which are either parallel or perpendicular to Hx and enclose the current sheet. Ampere's circuital law gives H x1L H x 2 ( L) K y L H x1 H x 2 K y If we choose a new path 3-3’-2’-2-3, the same current is enclosed, giving H x3 H x 2 K y and therefore H x3 H x1 AIUB Dr. M. Tanseer Ali EMWF Lec 7 /18 Chapter 8 The Steady Magnetic Field Ampere’s Circuital Law Because of the symmetry, then, the magnetic field intensity on one side of the current sheet is the negative of that on the other side. Above the sheet H x 12 K y ( z 0) while below it H x 12 K y ( z 0) Letting aN be a unit vector normal (outward) to the current sheet, this result may be written in a form correct for all z as H 12 K a N AIUB Dr. M. Tanseer Ali EMWF Lec 7 /19 Chapter 8 The Steady Magnetic Field Ampere’s Circuital Law If a second sheet of current flowing in the opposite direction, K = –Ky ay, is placed at z = h, then the field in the region between the current sheets is H K aN (0 z h) and is zero elsewhere H0 AIUB ( z 0, z h) Dr. M. Tanseer Ali EMWF Lec 7 /20 Chapter 8 The Steady Magnetic Field Ampere’s Circuital Law The difficult part of the application of Ampere’s circuital law is the determination of the components of the field which are present. The surest method is the logical application of the Biot-Savart law and a knowledge of the magnetic fields of simple form (line, sheet of current, “volume of current”). • Solenoid AIUB • Toroid Dr. M. Tanseer Ali EMWF Lec 7 /21 Chapter 8 The Steady Magnetic Field Ampere’s Circuital Law For an infinitely long solenoid of radius a and uniform current density Ka aφ, the result is H Kaa z ( a) H0 ( a) AIUB If the solenoid has a finite length d and consists of N closely wound turns of a filament that carries a current I, then NI well within H a z (the solenoid) d Dr. M. Tanseer Ali EMWF Lec 7 /22 Chapter 8 The Steady Magnetic Field Ampere’s Circuital Law For a toroid with ideal case 0 a inside H Ka a (toroid ) (outside H0 toroid) AIUB For the N-turn toroid, we have the good approximations Dr. M. Tanseer Ali NI H a 2 H0 (inside toroid) (outside toroid) EMWF Lec 7 /23 Chapter 8 The Steady Magnetic Field Curl For an incremental closed path of sides ∆x and ∆y which produces a reference value for H at the center. H L12 H y,12y H 0 H x 0a x H y 0 a y H z 0 a z 1 x 2 1 x 2 1 y 2 1 y 2 H y 1 H y ,12 H y 0 x x 2 H H L12 H y 0 y 1 x y x 2 H L23 H x 0 H x 1 y x H L34 y 2 H y 1 H y 0 x y x 2 H L41 H x 0 H x 1 y x AIUB Dr. M. Tanseer Ali y 2 EMWF Lec 7 /24 Chapter 8 The Steady Magnetic Field Curl H y H x H dL x y xy H dL I Combining all the paths, By amperes circuital law, Here, I J z xy H 0 H x 0a x H y 0 a y H z 0 a z ∆I H y H x H dL x y xy J z xy or, H dL H xy lim xy 0 AIUB Dr. M. Tanseer Ali x y H x Jz y H dL H xy x y H x Jz y EMWF Lec 7 /25 Chapter 8 The Steady Magnetic Field Curl The Curl of any vector is a vector, and any component of the curl is given by the limit of the quotient of the closed line integral of the vector, about a small path in a plane normal to that component desired and the area enclosed. By amperes circuital law, H 0 H x 0a x H y 0 a y H z 0 a z curl H N lim S N 0 H dL H S N ∆I H AIUB Dr. M. Tanseer Ali lim S N 0 H dL J S N z EMWF Lec 7 /26 Chapter 8 The Steady Magnetic Field Curl Properties of Curl: lim xy 0 lim yz 0 lim zx 0 H dL H xy x H dL H yz H x Jz y H y Jx y z H dL H zx y z z x ax H J x Hz ay y Hz az z Hz H z Jy x H z H y H y H x H x H z a x a z J H a y z x y z y x AIUB Dr. M. Tanseer Ali EMWF Lec 7 /27 Chapter 8 The Steady Magnetic Field Curl Rectangular Coordinate: ax H J x Hx ay y Hy az z Hz Cylindrical Coordinate: 1 H z H H 1 H 1 H H z a a a z H J z z Spherical Coordinate: 1 H sin H H rsin AIUB 1 1 H r rH 1 rH H r a r a a r sin r r r Dr. M. Tanseer Ali EMWF Lec 7 /28 Chapter 8 The Steady Magnetic Field Curl Example 5.2: Calculate H dL about a square path with side d centered at (0, 0, z1) in the y = 0 plane where z1 > 2d. 2 H 0 . 2 z a x for z > 0 and .H 0 elsewhere. Where 2 2 1 1 H dL 0 . 2 z d d 0 0 . 2 z d d 0 1 1 2 2 0.4 z1d 2 H lim d 0 ax H J x 0.2 z 2 AIUB Dr. M. Tanseer Ali H dL d2 ay y 0 lim 0.4 z1d 2 0.4 z1 2 d 0 d az 0.2 z 2 a y 0.4 z a y z z 0 EMWF Lec 7 /29 Chapter 8 The Steady Magnetic Field Stokes Theorem From Curl of H H dL S H dL S S H N H N S H dL S S H dS Stokes’ Theorem AIUB Dr. M. Tanseer Ali EMWF Lec 7 /30 Chapter 8 The Steady Magnetic Field Stokes Theorem Example Consider the portion of sphere shown in the figure. The surface is specified by r 4, 0 0.1 , 0 0.3 and the closed path forming its parameter is composed of three circular arcs. Given,H 6r sin a r 18r sin cos a . Evaluate each side of Stokes’ Theorem dL dr a r rd a r sin d a H dL S H rd H r sin d H rd 1 0.3 0 2 3 184sin 0.1 cos 4sin 0.1 d 288 sin 2 0.1 sin0.3 22.2 A AIUB Dr. M. Tanseer Ali H 0 EMWF Lec 7 /31 Chapter 8 The Steady Magnetic Field Stokes Theorem H 1 36r sin cos cos a r 1 1 6rcos 36r sin cos a r sin r sin Here, the surface dS r 2sin d d a r 0.3 0.1 0 0 H dS 36 cos cos 16 sin d d S 0.3 0 0.1 1 576 sin 2 cos d 2 0 288 sin 2 0.1 sin0.3 22.2 A AIUB Dr. M. Tanseer Ali EMWF Lec 7 /32 Chapter 8 The Steady Magnetic Field Stokes Theorem Stokes theorem relates surface integral to a closed line integral. While divergence theorem relates a volume integral to a closed surface integral Let consider A to be any vector, then A T Is T is vector or scalar? vol A dv Tdv vol S A dS volTdv Divergence theorem, SB dS vol B dv dS approaches to zero (from curl conditions); hence closed path also approaches to zero and stokes theorem gives S A dS 0 volTdv 0 T 0 A 0 AIUB Dr. M. Tanseer Ali This is a useful identity in vector calculus EMWF Lec 7 /33 Chapter 8 The Steady Magnetic Field Stokes Theorem In case of magnetic field, H 0 J 0 Ampere’s circuital law from stokes theorem H J H dS J dS S S or, H dL J dS I S AIUB Dr. M. Tanseer Ali EMWF Lec 7 /34 Chapter 8 The Steady Magnetic Field Magnetic Flux and Magnetic Flux Density In free space, the magnetic flux density B 0 H Unit of magnetic flux density is webers per meter sq (Wb/m2) or tesla (T) Magnetic flux, B dS Wb S Gauss’s law, states D dS Q S Divergence theorem, SB dS vol B dv The charge Q is the source of the lines of electric flux and these lines begins & ends on positive and negative charge. While the magnetic flux lines are closed and do not terminate on a “magnetic charge”. Hence Gauss’s law for the magnetic field is B dS 0 S AIUB B 0 Dr. M. Tanseer Ali This is Maxwell’s fourth Equations EMWF Lec 7 /35 Chapter 8 The Steady Magnetic Field Magnetic Flux and Magnetic Flux Density Maxwell’s Equations, D v E 0 H J B 0 Electric Field Equations, SD dS Q vol v dv E dL 0 Magnetic Field Equations, SH dS I B dS 0 AIUB Dr. M. Tanseer Ali J dv vol EMWF Lec 7 /36 Chapter 8 The Steady Magnetic Field Magnetic Flux and Magnetic Flux Density Example: consider an infinitely long coaxial transmission line, carrying a uniformly distributed total current I in the center conductor and –I in the outer conductor. A circular path of radius ρ, where ρ is larger than the radius of the inner conductor a but less than the inner radius of the outer conductor b, leads immediately to I H ( a b) 2 0 I B 0 H a 2 d b 0 I B dS a d dz a S 0 a 2 0 Id b ln 2 a AIUB Dr. M. Tanseer Ali EMWF Lec 7 /37 Chapter 8 The Steady Magnetic Field Scalar and Vector Magnetic Potentials Similar to electric potential, scalar magnetic potential H Vm Taking Curl of both sides, H J Vm Curl of the gradient of any scalar is identically zero. Vm 0 Hence, H Vm is true only if J 0 Laplace’s Equation B 0 H 0 0 Vm 0 2Vm 0 AIUB Dr. M. Tanseer Ali J 0 EMWF Lec 7 /38 Chapter 8 The Steady Magnetic Field Scalar and Vector Magnetic Potentials Example: consider an infinitely long coaxial transmission line, carrying a uniformly distributed total current I in the center conductor and –I in the outer conductor. I H a for a b, J 0 2 I 1 Vm Vm 2 Vm I 2 I Thus, Vm 2 1 2n 4 n 0,1,2, 1 Vm I n 8 n 0,1,2, I Vm 2 AIUB Dr. M. Tanseer Ali EMWF Lec 7 /39 Chapter 8 The Steady Magnetic Field Scalar and Vector Magnetic Potentials Properties of Scalar Magnetic Potential Vm is not single-valued function of position Vm is not a conservative field a Vm,ab H dL b (specified path) Vector magnetic potential A B A H 1 0 A H J AIUB 1 0 Divergence of the Curl of any vector field is zero Curl of the curl of a vector field is not zero A Dr. M. Tanseer Ali EMWF Lec 7 /40 Chapter 8 The Steady Magnetic Field Scalar and Vector Magnetic Potentials Vector magnetic potential A defined by Biot-Savart Law 0 IdL A 4R 0 IdL dA 4R For a filament at the origin in free space and allowing it to extend in the positive z direction so that dL=dz az dA dA z 0 Idza z dA 0 4 2 z 2 0 Idza z 4 2 z 2 dA 0 dA 0 AIUB Dr. M. Tanseer Ali EMWF Lec 7 /41 Chapter 8 The Steady Magnetic Field Scalar and Vector Magnetic Potentials Vector magnetic potential A defined by Biot-Savart Law 1 dAz a dH dA 0 0 1 Idz dH 4 2 z 2 IdL KdS 0KdS A S 4R AIUB 3/ 2 a IdL Jdv 0 Jdv A vol 4R Dr. M. Tanseer Ali EMWF Lec 7 /42 Chapter 8 The Steady Magnetic Field Assignment 4 Problem 1. Two filamentary current of 8 A each are directed inward from infinity to the origin on the positive x axis, and outward to infinity along the y axis. Determine H at P2(0.4, 0.3, 0). Problem 2. A current filament carrying 15 A in the az direction lies along the entire z axis. Find H in rectangular coordinates at: (a) PA(√20, 0, 4); (b) PB(2,−4, 4). Problem 3. A coaxial cable is excited from a current source of 20 A. The radius of inner conductor is 4 m. Radius of inner surface of Outer conductor is 4.25 m. The Outer conductor having thickness of 0.75 m. Find the magnitude of magnetic field intensity at the region: (i) 2.5 m, (ii) 4.8 m, (iii) 4.15 m, (iv) 5.75 m. Problem 4. Suppose that H = 0.2 z2 ax for z > 0, and H = 0 elsewhere. Calculate about a square path with side d, centered at (0, 0, z1) in the y = 0 plane where z1 > d/2. Problem 5. Calculate the value of the vector current density: (a) in rectangular coordinates at PA(2, 3, 4) if H = x2z ay − y2x az; (b) in cylindrical coordinates at PB(1.5, 90◦, 0.5) if H 2 cos0.2 a ; (c) in spherical coordinates at PC(2, 30◦, 20◦) if H 1 a sin AIUB Dr. M. Tanseer Ali EMWF Lec 7 /43 Chapter 8 The Steady Magnetic Field Assignment 4 Problem 6. Consider the portion of a sphere where the surface is specified by r = 4, 0 ≤ θ ≤ 0.1π, 0 ≤ φ ≤ 0.3π, and the closed path forming its perimeter is composed of three circular arcs. Evaluate each side of Stokes’ theorem if H = 6r sin φ ar +18r sin θ cos φ aφ. Problem 7. Evaluate both sides of Stokes’ theorem for the field H = 6xy ax − 3y2 ay A/m and the rectangular path around the region, 2 ≤ x ≤ 5, −1 ≤ y ≤ 1, z = 0. Let the positive direction of dS be az. (a) Find H in cartesian components at P(2, 3, 4) if there is a current filament on the z axis carrying 8 mA in the az direction. (b) Repeat if the filament is located at x = −1, y = 2. (c) Find H if both filaments are present. Problem 8. Let a filamentary current of 5 mA be directed from infinity to the origin on the positive z axis and then back out to infinity on the positive x axis. Find H at P(0, 1, 0). Problem 9. When x, y, and z are positive and less than 5, a certain magnetic field intensity may be expressed as H = [x2yz/(y + 1)] ax + 3x2z2 ay − [xyz2/(y + 1)] az. Find the total current in the ax direction that crosses the strip x = 2, 1 ≤ y ≤ 4, 3 ≤ z ≤ 4, by a method utilizing: (a) a surface integral; (b) a closed line integral. Problem 10. Evaluate both sides of Stokes’ theorem for the field H = 6xy ax − 3y2 ay A/m and the rectangular path around the region, 2 ≤ x ≤ 5, −1 ≤ y ≤ 1, z = 0. Let the positive direction of dS be az. Problem 11. Given H = (3r 2/ sin θ)aθ + 54r cos θ aφ A/m in free space: (a) Find the total current in the aθ direction through the conical surface θ = 20◦, 0 ≤ φ ≤ 2π, 0 ≤ r ≤ 5, by whatever side of Stokes’ theorem you like the best. (b) Check the result by using the other side of Stokes’ theorem. AIUB Dr. M. Tanseer Ali EMWF Lec 7 /44