Notes 18 Faraday s Law

Transcription

1 EE 3318 Applied Electricity and Magnetism Spring 2018 Prof. David R. Jackson Dept. of EE Notes 18 Faraday s Law 1

2 Example (cont.) Find curl of E from a static point charge q y E q = rˆ 2 4πε0r x ( E sinθ ) ( re ) ( re ) 1 φ E 1 θ 1 E r φ ˆ 1 θ E r E = rˆ + θ + φ r sinθ θ φ r sinθ φ r r r θ = 0 2

3 Example (cont.) Note: If the curl of the electric field is ero for the field from a static point charge, then by superposition it must be ero for the field from any static charge density. This gives us Faraday s law: E = 0 (in statics) 3

4 Faraday s Law in Statics (Integral Form) ˆn Stokes's theorem: S ( ) ˆ 0 E d r = E n ds = Here S is any surface that is attached to. Hence E dr = 0 4

5 Faraday s Law in Statics (Differential Form) We show here how the integral form also implies the differential form. Assume E dr = 0 ( for any path ) S url is calculated here S x We then have: S y y 1 xˆ curl E lim E dr = 0 s 0 S x 1 yˆ curl E lim E dr = 0 s 0 S y 1 ˆ curl E lim E dr = 0 s 0 S x y x x y Hence E = 0 5

6 Faraday s Law in Statics (Summary) E = 0 Differential (point) form of Faraday s law Stokes s theorem Definition of curl E dr = 0 Integral form of Faraday s law 6

7 Path Independence and Faraday s Law The integral form of Faraday s law is equivalent to path independence of the voltage drop calculation in statics. E dr = Proof: 0 (in statics) A 2 B Also, E dr = E dr E dr Hence, E dr = 0 E dr = E dr 1 2 7

8 Summary of Path Independence Path independence for V AB Equivalent E dr = 0 E = 0 Equivalent properties of an electrostatic field 8

9 Summary of Electrostatics Here is a summary of the important equations related to the electric field in statics. D = ρ v Electric Gauss law E = 0 Faraday s law D = ε E 0 onstitutive equation 9

10 Faraday s Law: Dynamics Experimental Law (dynamics): B E = t This is the general Faraday s law in dynamics. 2 B magnetic flux density Webers/m = Michael Faraday* (from Wikipedia) *Ernest Rutherford stated: "When we consider the magnitude and extent of his discoveries and their influence on the progress of science and of industry, there is no honour too great to pay to the memory of Faraday, one of the greatest scientific discoverers of all time". 10

11 Faraday s Law: Dynamics (cont.) B E = t B ˆ ( E) = < 0 t Assume a B field that increases with time: ( ) B = B ˆ t db dt > 0 y The changing magnetic field produces an electric field. Experiment x Magnetic field B (increasing with time) Electric field E 11

12 Eddy urrents B ˆ ( E) = < 0 t y Transformer core ( ρ φ ) E E B = ρ ρ φ t 1 ρ x Assume no φ variation J = σ E ψ ψ ψ Magnetic field B (increasing with time) Electric field E 12

13 Eddy urrents (cont.) Laminated cores are used to reduce eddy currents. J φ ψ Solid core ψ ψ The laminated core consists of layers of iron material coated with electrical insulation. Laminated core 13

14 Faraday s Law: Integral Form B E = t Integrate both sides over an arbitrary open surface (bowl) S: S B E n ds n ds t ( ) ˆ = ˆ S ˆn S (open) Apply Stokes s theorem for the LHS: B E d r = nˆ ds t S (closed) Note: The right-hand rule determines the direction of the unit normal, from the direction along. Faraday's law in integral form 14

15 Faraday s Law: Integral Form (cont.) B E d r = nˆ ds t S Assume that the surface and the path are not changing with time: d E d r = B nˆ ds dt Define magnetic flux through the surface S: We then have E dr = dψ dt S ψ S B nˆ ds Note: The right-hand rule determines the direction of the unit normal in the flux calculation, from the direction along. 15

16 Faraday s Law: Summary Summary E dr = dψ dt ˆn S (open) where ψ B nˆ ds (closed) S The voltage drop around a closed path is equal to the rate of change of magnetic flux through the path. 16

17 Faraday s Law for a Loop We measure a voltage across a loop due to a changing magnetic field inside the loop. y + ( ) v t - Open-circuited loop x Magnetic field B (B is changing with time) 17

18 Faraday s Law for a Loop (cont.) B = AB = = A v v E dr E dr Also E dr = dψ dt S y A + - B x v( t) v So we have dψ = dt ψ = ( B ) ds = S ( nˆ = ˆ ) ψ Note: The voltage drop along the PE wire (from B to A inside the wire) is ero. 18

19 Faraday s Law for a Loop (cont.) Final Result v = dψ dt y + - v( t) ψ B ds S x 19

20 Faraday s Law for a Loop (cont.) Assume a uniform magnetic field for simplicity (at least uniform over the loop area). Assume ψ = BA y + - v( t) so v = db A dt x A = area of loop 20

21 Faraday s Law for a Loop (cont.) Summary v = Assume dψ dt General form y + v( t) - v db A dt = Uniform field x A = area of loop ψ B ds S ψ = magnetic flux through loop in direction 21

22 Len s Law This is a simple rule to tell us the polarity of the output voltage (without having to do any calculation). Assume The voltage polarity is such that it would set up a current flow that would oppose the change in flux in the loop. y A R + B v - ( t ) > 0 I x Note: A right-hand rule tells us the direction of the magnetic field due to a wire carrying a current. (A wire carrying a current in the direction produces a magnetic field in the positive φ direction.) A = area of loop B is increasing with time. db v= A > 0 dt 22

23 Example: Magnetic Field Probe A small loop can be used to measure the magnetic field (for A). Assume B v= A t y + - ( ) v t Assume B = B t+ cos 0 ( ω φ) x Then we have ω = 2π f A = area of loop v= AωB ωt+ φ 0 sin ( ) A = π a 2 At a given frequency, the output voltage is proportional to the strength of the magnetic field. 23

24 Applications of Faraday s Law Faraday s law explains: How A generators work How transformers work Output voltage of generator v = dψ dt Output voltage on secondary of transformer Note: For N turns in a loop we have v = dψ N dt 24

25 The world's first electric generator! (invented by Michael Faraday) A magnet is slid in and out of the coil, resulting in a voltage output. (Faraday Museum, London) 25

26 The world's first transformer! (invented by Michael Faraday) The primary and secondary coils are wound together on an iron core. (Faraday Museum, London) 26

27 A Generators Diagram of a simple alternator (A generator) with a rotating magnetic core (rotor) and stationary wire (stator), also showing the output voltage induced in the stator by the rotating magnetic field of the rotor. v( t) + - ω = angular velocity of magnet B = B cos 0 γ γ = angle between north pole and the axis (Magnetic flux comes out of the north pole.) A = area of loop γ = ωt + γ

28 A Generators (cont.) v( t) + - ( ω γ ) B = B cos t+ 0 0 db NA dt ( ) = v t so ( ) = ( ω) sin ( ω + γ ) v t NAB t 0 0 or ( ) = ( ω + φ ) v t V cos t 0 A = area of loop ( V NAB ω, φ γ + π /2)

29 A Generators (cont.) v( t) + - Summary v ( t ) = V cos ( ωt+ φ ) 0 If the magnet rotates at a fixed speed, a sinusoidal voltage output is produced. A = area of loop Note: The angular velocity of the magnet is the same as the radian frequency of the output voltage. 29

30 A Generators (cont.) Thévenin Equivalent ircuit ( ) = ( ω + φ ) v t V cos t 0 V NAB ω, φ γ + π / v( t) + - R Th V Th + - N turns VTh = j Ve φ 0 A = area of loop R Th = resistance of wire in coil 30

31 A Generators (cont.) Generators at Hoover Dam 31

32 A Generators (cont.) Generators at Hoover Dam 32

33 Transformers A transformer changes an A signal from one voltage to another. 33

34 Transformers (cont.) High voltages are used for transmitting power over long distances (less current means less conductor loss). Low voltages are used inside homes for convenience and safety. 34

35 Transformers (cont.) µ r ψ dψ v t = N dt p ( ) p ψ ψ dψ v t = N dt s ( ) s Hence v ( ) s t N = v t N p ( ) Ideal transformer (no flux leakage): s p (time domain) V V s p = N N s p (phasor domain) µ r 35

36 Transformers (cont.) Ideal transformer (no losses): ( ) ( ) = ( ) ( ) v t i t v t i t p p s s Hence so ( ) ( ) ( t) ( ) ( ) ( ) is t v v t ip t vs t vp t p s = = 1 ( ) ( ) is t N ip t N s = p (time domain) 1 I I s p N s = N p 1 (phasor domain) 36

37 Transformers (cont.) Impedance transformation (phasor domain): Z in = V I p p Z out = V I s s ψ ψ ψ Hence so Z Z in out 1 1 N s N s I p Z V N in p I V s p p N p Np = = = = Z I V I N N N out p s p s s s Vp N p N p N p = Ns

38 Transformers (cont.) Impedance transformation Z in N p = N s 2 Z L N :1 This means N p N = s N Z in Z L 38

39 Transformers (cont.) Example: audio matching circuit The PA should see a matched load (50 [Ω]) for maximum power transfer to the load (speaker). N :1 N = 50 8 Audio output circuit Power amplifier (PA) [ ] Z = 50 Ω out Transformer Speaker Z L = 8 [ Ω] 39

40 Transformers (cont.) Isolation transformer An isolation transformer is used isolate the input and output circuits (no direct electrical connection between them). It can be used to connect a grounded circuit to an ungrounded one. The transformer is being used as a form of balun, which connects a balanced circuit (the two leads are at a +/- voltage with respect to ground) to an unbalanced circuit (where one lead is grounded). Ungrounded input Grounded output 40

41 Eddy urrents (Revisited) y J φ = σ E φ ψ ψ ψ ρ x Solid core dψ E dr = = dt d dt p 2 p E dr = jωπρ B 2 ( 2 ) = p p Eφ πρ jωπρ B ( 2 πρ B ) Phasors ( ) B = B0 cos ωt+ φ p j B = Be φ 0 p σωρ Jφ = j B 2 p 41

42 Measurement Error from Magnetic Field Find the voltage readout on the voltmeter. V m + - Voltmeter y B a x Perfectly conducting leads V Applied magnetic field: ( ) = ˆ cos( ω ) B t t Note: The voltmeter is assumed to have a very high internal resistance, so that negligible current flows in the circuit. (We can neglect any magnetic field coming from the current flowing in the loop.) 42

43 Measurement Error from Magnetic Field (cont.) Faraday s law: nˆ = ˆ E dr dψ = dt d = B nˆ ds dt d = dt S S B ds V m + - Voltmeter y a x B Perfectly conducting leads V d = dt B a db = dt ( 2 ) π ( 2 πa ) = ωsinωt π ( 2 a ) ( ) = ˆ cos( ω ) B t t db dt = ω ω sin ( t) 43

44 Measurement Error from Magnetic Field (cont.) From the last slide, ( 2 ) E dr =ωsinωt πa Therefore V m + - Voltmeter y B nˆ = a x ˆ V m 2 0 sin ( ) V V =ωπ a ωt or Perfectly conducting leads Note: There is no voltage drop along the perfectly conducting leads. m 2 0 sin ( ) V = V +ωπ a ωt 44

45 Measurement Error from Magnetic Field (cont.) Summary V m + - Voltmeter y B a x Perfectly conducting leads V m 2 0 sin ( ) V = V +ωπ a ωt Practical note: In such a measurement, it is good to keep the leads close together (or even better, twist them.) 45

46 Twisted Pair Transmission Line Twisted pair is used to reduced interference pickup in a transmission line, compared to twin lead transmission line. AT 5 cable (twisted pair) Twin lead oax Note: oaxial cable is perfectly shielded and has no interference. 46

47 Maxwell s Equations (Differential Form) D =ρ B E = t B = 0 D H = J + t v Electric Gauss law Faraday s law Magnetic Gauss law Ampere s law D= ε E B= µ H onstitutive equations

48 Maxwell s Equations (Integral Form) S S D nˆ ds = Q S B nˆ ds = 0 encl B E dr = t nˆ ds D H dr = i ˆ S + n ds t S Electric Gauss law Faraday s law Magnetic Gauss law Ampere s law i = J nˆ ds ( current through S) S S 48

49 Maxwell s Equations (Statics) In statics, Maxwell's equations decouple into two independent sets. D = ρ v E = 0 D =ρ B E = t B = 0 D H = J + t v B = 0 H = J Electrostatics Magnetostatics ρ E J B v 49

50 Maxwell s Equations (Dynamics) In dynamics, the electric and magnetic fields are coupled together. Each one, changing with time, produces the other one. B E = t D H = t D= ε E 0 B= µ H 0 From EE 3317: Example: A plane wave propagating through free space H E ρ = 0, J = 0 power flow E = xˆ cos( ωt k) 1 H = yˆ cos t k η0 v ( ω ) k = ω µε 0 0 η = µ / ε