# CHAPTER 21.pdf - CHAPTER 21 ELECTROMAGNETIC INDUCTION BASIC...

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Unformatted text preview: CHAPTER 21 ELECTROMAGNETIC INDUCTION BASIC CONCEPTS Faraday’s Law Lenz’s Law Motional emf 1 Electric Charge Produces an Electric Field Moving Electric Charge Produces a Magnetic Field Now Changing Magnetic Field Produces an εmf 2 a potential difference (voltage) What is changing and causing the emf is the magnetic flux. 3 We defined Electric Flux  =  Magnetic Flux is similar  =   Flux through an area 4 We will define a vector to represent area. The direction of the vector will be perpendicular to the surface. The length of the vector will be the area of the surface. The vector will be . The flux will be 5 First, flat area perpendicular to   =   0 =  6 For a surface parallel to   =   90 = 0 And for in between  =  Δ   =   7 If this flux changes an emf will be induced in the area. Consider a copper coil with one loop B A r 8  =   For N loops N loops B A r  =   9 FARADAY’S LAW The emf induced in a circuit is directly proportional to the time rate of change of the magnetic flux through the circuit. Or  =  10 If  changes with time an emf, , will be produced. Coil in magnetic field. 11 Flux through coil  =   Flux can change: 1. Magnitude of  changes. 2. Area of loop changes. 12 3. The angle  changes. 4. Any combination of the three. 13 Example: Consider a coil with 200 turns on a rectangular frame, 20 cm by 30 cm. The resistance of the coil is 2 Ω. 14 If  increases uniformly from zero to 0.5 "#/% in 0.8 s what is the current in the coil? = &0.2%(&0.3%( = 0.060% At  = 0  = 0 At  = 0.8  =  = &0.5"#/% (0.060% = 0.03"# 15  & = 0( −  & = 0.8 ( | | = 0.8 200&0.03"#( | | = 0.8 | | = 7.5/ The current will be  7.5/ 0= = = 3.75 1 2 Ω 16 Lenz’s Law A changing magnetic flux will cause an emf and if the emf is in a conductor there will be a current. What will be the direction of the current? Lenz’s Law will answer that question. 17 The polarity of the induced emf is such that it tends to produce a current that will create a magnetic flux to oppose the change in magnetic flux through the loop. 18 a. Magnet moves toward loop. Magnet field is down and increasing. Induced emf and thus current will be ccw to oppose the increase. 19 b. Magnet moves away from loop. Magnetic field is up and but decreasing. Induced current gives magnetic field to oppose the decrease. 20 Motional emf Δx 21 In time Δt the rod will move Δx. The area covered in Δt will be Δ = 3Δ4 Δ4 = 5Δ Δ = 35Δ Faraday’s Law 22 Δ = Δ Δ = Δ = 35Δ 35Δ = = 35 Δ A conducting rod moving in a magnetic field will have an emf Induced across it given by this equation. 23 Example: My airplane Luscombe 8E Built 1947 Wingspan 22 feet = 6.7m Speed 100 mph = 44.7 m/s 24 If I am flying it at a point on earth where the vertical 78 magnetic field is 5410 9 what is the voltage drop across the wings?  = :5  = &541078 9(&6.7%(&44.7%/ (  = 1.54107 / 25 Consider a loop around a long solenoid. 26 The current in the solenoid is increasing so the flux through the loop is increasing. An emf is induced in the loop by . 27 INDUCTANCE BASIC CONCEPTS Mutual Inductance Self Inductance Magnetic Field Energy Circuits with Inductors 28 Mutual Inductance I1 #1 ∙ = > If I1 changes with time t Δ0> ?@A Δ> BCCCD E = Δ Δ 29 Put second coil at P I1 ΔF Δ #1 #2 ?@A BCCCD  GH G: #2 30 ?@A  BCCCD 0 If I1 changes with time For example 0> = 0J GHK Then > = J GHK And  = − ΔLM Δ 31  = − J K K Or And  − J K = K 0 = 1 1 I1 #1 #2 I2 32  Φ> N= 0> >  PJ 0> : = 0> > PJ >  N =  PJ = : : Only depends on geometry, number of turns, area and length. 33 Self Inductance If the current is changing then there will be a changing magnetic flux through the coil. 34 A changing magnetic flux will induce an emf.  = − @L @ Therefore we define in analogy to the mutual inductance Φ 3= G This is Self-Inductance Solve for Φ 35 3G Φ =  Faraday’s Law  = − QLR Q = −3 Q Q Self-Induced emf ΔG  = −3 Δ 36 What is the inductance, 3, of the black coil? Area A ΦS 3= 0 > TUU@ = PJ 0> : > Φ = PJ 0> :   > PJ > 0>  > : 3= = PJ 0> : 37 Magnetic Field Energy Consider the coil [Type a quote from the document or the summary of an interesting point. You can position the text box anywhere in the document. Use the Drawing Tools tab to change the formatting of the pull quote text box.] Current starts at zero and increases to 0. 38  / = VW = 0  1 = = 0/ 1 Or = = 0 And Δ0  = −3 Δ Δ0 Δ0 |=| = 0 X3 Y = 03 Δ Δ Power is work per unit time. Δ" == Δ 39 Δ0 Δ" = =Δ = 03 Δ = 30Δ0 Δ How much work to bring current through inductor from zero to 0? To calculate the work integrate from zero current to current I We won’t do it but you may see how on the next page. 40 "=Z \]\ \]J \ 30[0 = 3 Z 0[0 J 1  " = 30 2 _UA U " ^ BCCCD ` W[ HWab 41 The energy stored in the magnetic field is 1  c = 30 2 Apply to solenoid 42 / :d%W = : >  = PJ 0 : So : 0=  PJ > >  3 = PJ : Put into equation for c  1  1 >  : c = 30 = PJ X Y 2 2 : PJ > 43 1 : c= 2 PJ 1  & : ( c= 2 PJ And / :d%W = : So 1  HWab 2 PJ & : ( = & : ( / :d%W 1  d= 2 PJ Energy stored in magnetic field  of coil. 44 R-L Circuit Consider this circuit 45 First consider that switch `> is closed and calculate how the current will increase with time. Second, then while there is a current in the circuit consider opening `> and closing ` and calculate how the current will decay. Close `> 46 At time  = 0 close switch. Kirchkoff’s loop starting at the switch and going ccw. 47 ΔG  − G1 − 3 = 0 Δ ΔG  − G1 = Δ 3 ΔG  1 = − G Δ 3 3 At t = 0 i=0 Then Δ0  X Y = Δ J 3 After long time Δ0 =0 Δ 48 And  0= 1 49 50 The L-C Circuit The capacitor has been charged to an initial charge e and then placed across the inductor. 51 The L-C Circuit Remember energy in Capacitor cf = > gh  i And Inductor > c = 3G   Then study 52 53 Transfromers 54 ...
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• Fall '19
• Magnetic Field, Faraday's law of induction
• • •  