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3.2 AMPERE’S CIRCUITAL LAW In magnetostatic problems with sufficient symmetry, we can employ Ampere’s Circuital Law more easily that the law of Biot-Savart. The law says that the integration of H around any closed path is equal to the net current enclosed by that path. i.e. H dL Ienc 1 AMPERE’S CIRCUITAL LAW (Cont’d) • The line integral of H around the path is termed the circulation of H. • To solve for H in given symmetrical current distribution, it is important to make a careful selection of an Amperian Path (analogous to gaussian surface) that is everywhere either tangential or normal to H. • The direction of the circulation is chosen such that the right hand rule is satisfied. 2 DERIVATION 4 Find the magnetic field intensity everywhere resulting from an infinite length line of current situated on the z-axis using Ampere’s Circuital Law. 3 DERIVATION 4 (Cont’d) Select the best Amperian path, Figure 3-15 (p. 113) Two possible Amperian paths around an infinite length line of current. where here are two possible Amperian paths around an infinite length line of current. Choose path b which has a constant value of Hφ around the circle specified by the radius ρ Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth Copyright © 2005 by John Wiley & Sons. All rights reserved. 4 DERIVATION 4 (Cont’d) Using Ampere’s circuital law: H dL Ienc We could find: H H a dL da So, 2 H dL I enc H a da I 0 5 DERIVATION 4 (Cont’d) Solving for Hφ: H I 2 Where we find that the field resulting from an infinite length line of current is the expected result: H I 2 a Same as applying Biot-Savart’s Law! 6 DERIVATION 5 Use Ampere’s Circuital Law to find the magnetic field intensity resulting from an infinite extent sheet of current with current sheet K K xa x in the x-y plane. 7 DERIVATION 5 (Cont’d) Rectangular amperian path of height Δh and width Figure 3-16 (p. 113) Calculating H resulting from a current K = K a hand in the x–y plane. Δw. According to sheet right rule, perform the x x circulation in order of a b c d a Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth Copyright © 2005 by John Wiley & Sons. All rights reserved. 8 DERIVATION 5 (Cont’d) We have: b c d a a b c d H dL I enc H dL H dL H dL H dL From symmetry argument, there’s only Hy component exists. So, Hz will be zero and thus the expression reduces to: b d a c H dL I enc H dL H dL 9 DERIVATION 5 (Cont’d) So, we have: b d a 0 c H dL H dL H dL w H y a y dya y H ya y dya y w 0 2 H y w 10 DERIVATION 5 (Cont’d) The current enclosed by the path, I KdS w This will give: H dL Ienc 2H y w K x w Kx Hy 2 K x dy K x w 0 Or generally, 1 H K aN 2 11 EXAMPLE 3 An infinite sheet of current with K 6a A z m exists on the x-z plane at y = 0. Find H at P (3,2,5). 12 SOLUTION TO EXAMPLE 3 Use previous expression, that is: 1 H K aN 2 is a normal vector from the sheet to the test point P (3,4,5), where: aN aN a y So, and K 6a z 1 H 6a z a y 3a x A m 2 13 EXAMPLE 4 Consider the infinite length cylindrical conductor carrying a radially dependent current J J 0 a z Find H everywhere. 14 SOLUTION TO EXAMPLE 4 What components of H will be present? Finding the field at some point P, the field has both a and a components. (a) 15 SOLUTION TO EXAMPLE 4 (Cont’d) The field from the second line current results in a cancellation of the a components (b) 16 SOLUTION TO EXAMPLE 4 (Cont’d) To calculate H everywhere, two amperian paths are required: Path #1 is for a Path #2 is for a 17 SOLUTION TO EXAMPLE 4 (Cont’d) Evaluating the left side of Ampere’s law: 2 H dL H a da 2H 0 This is true for both amperian path. The current enclosed for the path #1: I J dS J 0 a z dda z 2 3 2 J 0 J 0 2 d d 3 0 0 18 SOLUTION TO EXAMPLE 4 (Cont’d) Solving to get Hφ: J0 2 H 3 J0 2 H a for a 3 Or The current enclosed for the path #2: 2 3 2 J a 2 0 I J dS J 0 dd 3 0 0 a Solving to get Hφ: J 0a3 H a for 3 a 19 EXAMPLE 5 Find H everywhere for coaxial cable as shown. 20 (a) SOLUTION TO EXAMPLE 5 Even current distributions are assumed in the inner and outer conductor. Consider four amperian paths. (a) (b) 21 SOLUTION TO EXAMPLE 5 (Cont’d) It will be four amperian paths: a a b b c c Therefore, the magnetic field intensity, H will be determined for each amperian paths. 22 SOLUTION TO EXAMPLE 5 (Cont’d) As previous example, only Hφ component is present, and we have the left side of ampere’s circuital law: 2 H dL H a da 2H 0 For the path #1: Ienc J dS 23 SOLUTION TO EXAMPLE 5 (Cont’d) We need to find current density, J for inner conductor because the problem assumes an event current distribution (ρ<a is a solid volume where current distributed uniformly). Where, I J az dS dS dd , S 2 a 2 d d a 0 0 24 SOLUTION TO EXAMPLE 5 (Cont’d) So, J I a I a z 2 z dS a We therefore have: 2 I I enc J dS a dda z 2 z 0 0 a I 2 2 a 25 SOLUTION TO EXAMPLE 5 (Cont’d) Equating both sides to get: I 2 I H 2 a 2 2a 2 for a For the path #2: The current enclosed is just I, I enc I Therefore: H dL 2H I I H I 2 for enc a b 26 SOLUTION TO EXAMPLE 5 (Cont’d) For the path #3: For total current enclosed by path 3, we need to find the current density, J in the outer conductor because the problem assumes an event current distribution (a<ρ<b is a solid volume where current distributed uniformly) given by: I I a z 2 2 a z J dS c b 27 SOLUTION TO EXAMPLE 5 (Cont’d) We therefore have (for AP#3): 2 I J dS c 2 b2 a z dda z 0 b I 2 b2 c2 b2 But, the total current enclosed is: I enc I J dS 2 2 2 2 b c I I I 2 2 2 2 c b c b 28 SOLUTION TO EXAMPLE 5 (Cont’d) So we can solve for path #3: H dL 2H I enc c2 2 I 2 c b2 I c 2 2 for b c H 2 c 2 b 2 For the path #4, the total current is zero. So, H 0 for c This shows the shielding ability by coaxial cable!! 29 SOLUTION TO EXAMPLE 5 (Cont’d) Summarize the results to have: I a 2 2a I a 2 H I c2 2 a c2 b2 2 0 a a b b c c 30