ECE 3318 Applied Electricity and Magnetism. Spring Prof. David R. Jackson ECE Dept. Notes 28

Transcription

1 EE 3318 Applied Electricit and Magnetism Spring 217 Prof. David R. Jackson EE Dept. Notes 28 1

2 Magnetic Field S ε r N v q S N Note: Flu lines come out of north poles! Lorent force Law: F = qv B This eperimental law defines B. B is the magnetic flu densit vector. n general, (with both E and B present): F = q( E+ v B) The units of B are Webers/m 2 or Tesla [T]. 2

3 Magnetic Field (cont.) F = q( v B) Beam of electrons moving in a circle, due to the presence of a magnetic field. Purple light is emitted along the electron path, due to the electrons colliding with gas molecules in the bulb. (From Wikipedia) A stable orbit is a circular path. 2 mv R = qv B q > F R B = B ˆ v = ˆ φ v F = ρˆ q v ( qv B ) 3

4 Magnetic Field (cont.) The most general stable path is a heli, with the heli ais aligned with the magnetic field. B F = qv B There is no force in the ais direction (hence a constant velocit in this direction). Magnetic field lines thus guide charged particles. 4

5 Magnetic Field (cont.) The earth's magnetic field protects us from charged particles from the sun. The particles spiral along the directions of the magnetic field, and are thus directed towards the poles. 5

6 Magnetic Field (cont.) This also eplains the auroras seen near the north pole (aurora borealis) and the south pole (aurora australis). The particles from the sun that reach the earth are directed towards the poles. 6

7 Magnetic Gauss Law S (closed surface) S B nˆ ds = N B Magnetic pole (not possible)! N S S Note: Magnetic flu lines come out of a north pole and go into a south pole. No net magnetic flu out! 7

8 Magnetic Gauss Law: Differential Form S B nˆ ds = From the definition of divergence we then have 1 B lim B nˆ ds V V S Hence B = 8

9 Ampere s Law ron filings Eperimental law: ˆ B = φ µ 2πρ µ = π [ ] H/m Note: A right-hand rule predicts the direction of the magnetic field. (This is an eact value: please see net slide.) 9

10 Ampere s Law (cont.) Note: The definition of the Amp* is as follows: 1 [A] current produces: B φ = [ ] [ ] T at ρ = 1 m * This is one form of the definition. Another form involves the force between two infinite wires. Hence B φ = µ 2πρ [ ] = T [ ] [ ] 1A ( ) 2π 1m µ so µ π [ ] 7 = 4 1 H/m 1

11 Ampere s Law (cont.) Define: H 1 µ B (This is the definition in free space.) Hence B = µ H H is called the magnetic field. The units of H are [A/m]. H ˆ = φ 2πρ [ A/m] (for single infinite wire) 11

12 Ampere s Law (cont.) Wire inside path A current (wire) is inside a closed path. We integrate counterclockwise. ( ˆ φ ) ( ˆ ˆ ˆ ) φ φρ φ ρ ρ H dr = H d + d + d = H 2π = ρdφ 2πρ 2π = dφ 2π = 2π = φ ρdφ ( 2π) 12

13 Ampere s Law (cont.) Wire outside path H dr = = 2 dφ π A current (wire) is outside a closed path. Note: The angle φ smoothl goes from ero back to ero as we go around the path, starting on the ais. 13

14 Ampere s Law (cont.) Hence H dr = encl Ampere s Law (D currents) encl Although the law was derived for an infinite wire of current, the assumption is made that it holds for an shape current. This is now an eperimental law. Right-Hand Rule Right-hand rule for Ampere s law: The fingers of the right hand are in the direction of the path, and the thumb gives the reference direction for the current that is enclosed b the path. (The contour goes counterclockwise if the reference direction for current is pointing up.) Note: The same D current encl goes through an surface that is attached to. 14

15 Amperes Law: Differential Form S ˆn S is a small planar surface. ˆn is constant S S ( ) H nˆ ds = ( ) H nˆds = J nˆds ( ) H dr = S encl encl H nˆ S J nˆ S (from Stokes s theorem) ( H) nˆ = J nˆ Since the unit normal is arbitrar (it could be an of the three unit vectors), we have Let S H = J 15

16 Mawell s Equations (Statics) D = ρ v Electric Gauss law B = Magnetic Gauss law E = H = J Farada s law Ampere s law 16

17 Mawell s Equations (Dnamics) D = ρ v Electric Gauss law B = Magnetic Gauss law B E = t D H = J + t Farada s law Ampere s law 17

18 Ampere s Law: Finding H H dr = encl Rules: 1) The Amperian path must be a closed path. 2) The sign of encl is from the right-hand rule. 3) Pick in the direction of H (to the etent possible). 18

19 Eample alculate H 1) First solve for H φ. Note : H = ( ) H ρ ρ r Amperian path An infinite line current along the ais 19

20 Eample (cont.) ( ) ( ˆ φρ φ) 2π H dr = ( 2π) encl H d = = H H ρdφ = φ φ ρ = encl ρ r Amperian path H φ = 2πρ 2

21 Eample (cont.) 2) H = H ρ dρ cancels H = at ρ h ρ = H dr = = encl ( ρ )( ) + ( )( ) = H h H h 3) H ρ = Magnetic Gauss law: ( ρ ) H = h B S ρ B nˆ ds = ( πρh) 2 = S 21

22 Eample (cont.) Summar ρ H ˆ = φ 2πρ [ A/m] r Note: There is a right-hand rule for predicting the direction of the magnetic field from a wire. (The thumb is in the direction of the current, and the fingers are in the direction of the field.) 22

23 oaial cable Eample D current a b c a c b ρ r This inner wire is solid. The outer shield (jacket) of the coa has a thickness of t = c b. ρ < a J π a A 2 = J = A/m 2 Note: The permittivit of the material inside the coa does not matter here. b < ρ < c J πb B 2 = J = A/m 2 2 πc Note: At D, the current densit is uniform, since the electric field is uniform (due to the fact that the voltage drop is path independent). 23

24 Eample (cont.) H = ˆ φ H φ The other components are ero, as in the wire eample. H dr = encl ( ˆ φρ φ) H d = 2π ρ φ = encl Hφ d encl 2πρ Hφ = encl a c b ρ r H φ = encl 2πρ This formula holds for an radius, as long as we get encl correct. 24

25 Eample (cont.) ρ < a a <ρ < b b <ρ < c encl J ( πρ ) π a A 2 2 = = πρ 2 encl encl = B ( 2 2 πρ π ) = + J b = πc πb ( 2 2 πρ πb ) a c b ρ r ρ > c = + ( ) = encl Note: There is no magnetic field outside of the coa (a perfect shielding propert ). 25

26 Eample (cont.) Summar ρ < a H 2 ˆ ρ φ πρ 2 = 2 a [A/m] a b a <ρ < b H ˆ = φ 2 πρ [A/m] c ρ b <ρ < c H 2 2 ˆ πρ πb = φ 1 [A/m] πρ π c π b ρ > c H = [A/m] 26

27 Eample Solenoid n = # turns/meter deal solenoid: n alculate H a µ = µµ r Find H H ρ dρ cancels ρ < a h H is ero at H dr = encl encl ( ) H h = = nh H = n 27

28 Eample (cont.) ρ > a Hh = encl H ρ dρ cancels Hence h ( ) [ ] H = n A/m, ρ < a H = at H =, ρ > a so encl Hh = = H = 28

29 Eample (cont.) The other components of the magnetic field are ero: 1) H φ = since encl = Hφ πρ = 2 encl 2) H ρ = from S B nˆ ds = S Bρ 2πρ h= 29

30 Eample (cont.) Summar n = # turns/meter Note: A right-hand rule can be used to determine the direction of the magnetic field (the same rule as for a straight wire). a µ = µµ r ( ) [ ] H = ˆ n A/m, ρ < a =, ρ > a B= µµ H ρ< a r, Note: We will sa more about relative permeabilit later. Note: A larger relative permeabilit will give a larger magnetic flu densit. This results is a stronger magnet, or a larger inductance. 3

31 Eample alculate H J s = J ˆ s [ A/m] - side nfinite sheet of current Top view + side 31

32 Eample (cont.) H H + ( ) H = H ˆ H = (superposition with line currents) H = ( ) = ( + ) H H (magnetic Gauss Law) Also, b smmetr: ( ) = ( + ) H H ( ) 2H + = 32

33 Eample (cont.) w - side ( ) H = H ˆ H + ( ) + side H dr = encl Note: There is no contribution from the left and right edges (the edges are perpendicular to the field). front back w/2 H dr = H d = H w w/2 w/2 w/2 + H dr = H d = H w 33

34 Eample (cont.) H w+ H w= J w + s H H = J + + s H + = 1 2 J s J H = > 2 s ˆ [ A/m], J H = + < 2 s ˆ [ A/m], Note: We can use a right hand-rule to quickl determine the direction of the magnetic field: Put our thumb is in the direction of the current, and our fingers will give the overall direction of the magnetic field. Note: The magnetic field does not depend on. 34

35 Eample Find H everwhere Parallel-plate transmission line h Assume w >> h w (We neglect fringing here.) J J bot s top s = ˆ w = ˆ w [ A/m] [ A/m] 35

36 Eample (cont.) h Two parallel sheets (plates) of opposite surface current J J bot s top s = w = w [ A/m] [ A/m] 36

37 Eample (cont.) bot top h Magnetic field due to bottom plate Magnetic field due to top plate bot H = H + H top 37

38 Eample (cont.) We then have H bot J 2 ˆ s, < < h = 2, otherwise Recall that J bot s = w [ A/m] Hence H = ˆ [A/m], < < h w H =, otherwise 38

39 Eample (cont.) We could also appl Ampere s law directl: H = Note: t is convenient to put one side (top) of the Amperian path in the region where the magnetic field is ero. h H = H ˆ H dr = encl top H = Js = w H = H = / w Hence ˆ H = [A/m], < < h w 39

40 Low Frequenc alculations At low frequenc, the D formulas should be accurate, as long as we account for the time variation in the results. Eample (line current in time domain) ρ << λ = c/ f ( ) = ( ω + φ ) i t cos t i(t) ρ r ( ) H t ( ) ˆ i t φ 2πρ [ A/m] 4

41 Low Frequenc alculations (cont.) We can also used the D formulas in the phasor domain. Eample (line current in the phasor domain) ρ << λ = c/ f i(t) ( ) = ( ω + φ ) i t cos t = phasor current j e φ ( ) ρ r H ˆ φ 2πρ [ A/m] (phasor H field) 41

42 Low Frequenc alculations (cont.) Eample (cont.) onverting from phasor domain to the time domain: H ˆ φ 2πρ (phasor H field) 1 H t He e e 2πρ 2πρ jωt ( ) Re( ) Re ˆ jωt ˆ jωt = φ = φ Re( ) ˆ 1 = φ Re 2πρ jφ jωt (( e ) e ) ( ) ˆ 1 H t φ cos ( ωt+ φ) [A/m] 2πρ 42