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Exam 2 is coming ! Tues., March 21, 12:30 to 1:55 pm, in this room. Covering 5 chapters (27-31) 18 multiple-choice questions (just as before) 15 conceptual/numerical problems, 1 point each 3 questions are numerical (like homework problems) - 2 pts. I will pass out formula5 sheets at the exam. Please familiarize yourself with it. Any constants needed will be given. (Only the equations up to “AC Circuits” will come into play.) Personalized exams I will enter the grade on your Mastering Physics account (“Exam 2”). Recovery points Set 2 on sPH2435-04 opens March 22 (Weds., noon) and closes March 24 (Fri., 11:59 pm). Use the “old interface”. You need a 4-digit CAPA ID to access it. Will get this from on-class exam. You must try to recover everything, not just the ones you missed. You must get a higher grade on the recovery exam to get any points added to your class score. What can I bring to the exam? Pencil eraser calculator That’s all (no cell phones for example) Phys 2435: Chap. 27-31, Pg 1 Exam 2 coverage Chapter 27: Magnetic Field and Forces Magnets Magnetic Force Chapter 28: Sources of Magnetic Field Biot-Savart Law Ampere’s law Chapter 29: Electromagnetic Induction Induced EMF and applications General form of Faraday’s Law Chapter 30: Inductance Mutual and Self-inductance Energy storage and DC Circuits Chapter 31: Alternating Current AC Circuits Resonance and Transformers 7 lectures (including this one) 6 homework sets 6 quizzes 1 exam About 1/3 of the work Phys 2435: Chap. 27-31, Pg 2 Chapter 27: Magnetic Field and Forces Magnets and Magnetic Fields Magnetic Force (Lorentz Force) force on a moving charge force on a current in a wire torque on a current loop mass spectrometer Phys 2435: Chap. 27-31, Pg 3 Magnetism Natural magnets were observed by Greeks more than 2500 years ago in “Magnesia” (northern Greece) certain type of stone (lodestone) exert forces on similar stones Small lodestone suspended with a string aligns itself in a north-south direction due to Earth’s magnetic field! Direction of Magnetic Field tail out of in to page page Drawing vectors in head Phys 2435: Chap. 27-31, Pg 4 The Magnetic Force What happens if you put a charged particle in a magnetic field? it experiences a magnetic force! force (Lorentz force) Magnitude depends on Charge q Velocity v Field B Angle between v and B Fmagnetic = q v B sinθ Direction is “sideways” Fmag force is perpendicular to both v and B! Vector cross product ! ! ! F = qv ! B v B Phys 2435: Chap. 27-31, Pg 5 Direction of the Magnetic Force ! ! ! F = qv ! B Use the right-hand rule: point your fingers along the direction of velocity curl your fingers towards the magnetic field vector your thumb will then point in the direction of the force F v B Reverse direction if it’s a negative charge! Phys 2435: Chap. 27-31, Pg 6 Radius of Circular Orbit magnetic force: x x x x x x x x x x x x centripetal force: x x x x x x x x x x x vx B x x x x x x x x x x x x q v F F R ⇒ ⇒ This is an important result: It relates atomic quantities (m and q) to quantities we can measure (R, v, B) in a lab. Phys 2435: Chap. 27-31, Pg 7 Magnetic Force on a Current-Carrying Wire ! ! ! F = Il ! B The right-hand-rule Phys 2435: Chap. 27-31, Pg 8 Torque on a Current Loop • Define magnetic dipole moment ! " µ = NIA Then the torque can be written as a vector cross product ! ! ! " = µ!B The potential energy is ! ! U = "µ ! B Phys 2435: Chap. 27-31, Pg 9 Mass Spectrometer Velocity selector qE = qvB v = E/B Only those particles with v can pass through. Bending in a magnetic field qvB' = mv /r 2 m = qBB ' r / E Phys 2435: Chap. 27-31, Pg 10 Chapter 28: Sources of Magnetic Field Biot-Savart Law moving charge a straight wire force between parallel current loops wires Ampere’s Law straight wire solenoid toroidal solenoid Phys 2435: Chap. 27-31, Pg 11 Biot-Savart Law - Moving Charge µ 0 qv sin ! B= 2 4" r ! ! µ 0 qv ! rˆ B= 2 4" r µ0=4πx10-7 T.m/A is called permeability of free space where rˆ is the unit vector from the source point to the field point. Phys 2435: Chap. 27-31, Pg 12 Biot-Savart Law - Curent Segment Question: how to find B field (both direction and magnitude) due to a current segment ? ! ! µ 0 qv ! rˆ B= 2 4" r ! ! µ 0 Idl ! rˆ dB = 2 4" r The total magnetic field is an integral over the entire wire: ! ! µ 0 Idl " rˆ B= 2 ! 4# r Phys 2435: Chap. 27-31, Pg 13 Example: straight wire Set up the coordinate system as shown. The field at point P due to a small segment: µ0 I dy sin φ dB = 2 π 4 r The direction is the same from any segment, so the total is µ0 I B= 4π +a xdy x2 + y 2 ∫( −a ) 3/ 2 µ0 I 2a = 2π x x 2 + a 2 For a long wire (a ∞): µ0 I B= 2!x Phys 2435: Chap. 27-31, Pg 14 Force Between Two Parallel Current-Carrying Wires Field at wire 2 due to wire 1: µ0 I 1 B1 = 2πd Force on µ0 I 1 I 2 F2 = I 2 LB1 = L wire 2: 2πd Force per unit length: F2 = µ 0 I 1 I 2 L 2πd Phys 2435: Chap. 27-31, Pg 15 Example: a current loop Set up the coordinate system as shown. For a point on the axis µ0 I dl dB = 2 π 4 r By symmetry, total B⊥ is zero, so total B=B|| µ0 I a 2 a µ0 I a = B = ∫ dBx = ∫ dB cos θ = ∫ dB = dl ∫ r 4π r 3 2 r3 Phys 2435: Chap. 27-31, Pg 16 Ampere’s Law Question: Is there a general relation between a current in a wire of any shape and the magnetic field around it? ! ! B " d l = µ I 0 enclosed ! Ampere’s Law: The line integral of the magnetic field around any closed loop is equal to µ0 times the total current enclosed by the loop Phys 2435: Chap. 27-31, Pg 17 Example: a long straight wire Consider a circular path of radius r around the wire. The plane of the path is perpendicular to the wire. By symmetry, the B field has the same magnitude at every point along the path, with a direction tangential to the circle by the right-hand-rule. µ 0 I = ∫ B ⋅ dl = B ∫ dl =B (2πr ) Phys 2435: Chap. 27-31, Pg 18 Example: Solenoid Consider a rectangular path as shown By symmetry, the only non-zero contribution comes from the segment cd: d ∫ B ⋅ dl = ∫ B ⋅ dl = Bl c Bl = µ 0 NI Where n = N / l is the number loops per unit length Phys 2435: Chap. 27-31, Pg 19 Example: Toroidal Solenoid Inside: consider path 1 B(2πr ) = µ 0 NI Outside: consider path 2 The net current passing through is zero B(2πr ) = 0 Phys 2435: Chap. 27-31, Pg 20 Chapter 29: Electromagnetic Induction Magnetic Flux Induced EMF Faraday’s law Lenz’s law Motional EMF Applications of Induction generators motors Counter EMF Faraday Law (general form) Displacement Current and Maxwell Equations Phys 2435: Chap. 27-31, Pg 21 Magnetic Flux Consider the B field lines that pass through a surface define a quantity called the magnetic flux Φ Φ ≡ B A cosθ where θ is angle between magnetic field B and the normal to the plane. units of magnetic flux are T.m2 = Weber (Wb) Scalar product Phys 2435: Chap. 27-31, Pg 22 Faraday’s Law of Induction Lenz’s Law ε = −N ΔΦ Δ B t induced emf rate of change of flux with time minus sign comes from Lenz’s Law: The induced emf gives rise to a current whose magnetic field opposes the original change in flux. Phys 2435: Chap. 27-31, Pg 23 Motional EMF Consider a conducting rod moving on metal rails in an uniform magnetic field: ε ΔΦ B Δ (BA ) Δ (BLx ) Δx = = = = BL Δt Δt Δt Δt Current will flow counter-clockwise in this “circuit” x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x v L ε = BLv x Phys 2435: Chap. 27-31, Pg 24 Electric Generators Flux is changing in a sinusoidal manner: Φ = B A cos θ = B A cos (ω t) this leads to an alternating emf (AC generator) Φ ε = N d dt B = NBA d cos( ω t ) = NBA ω sin( ω t ) dt This is how most of our electricity is generated !! water or steam turns blades of a turbine which rotates a loop Phys 2435: Chap. 27-31, Pg 25 Motors Generators AC current + B field → rotation rotation + B field → AC current electrical ⇒ mechanical energy mechanical ⇒ electrical energy Phys 2435: Chap. 27-31, Pg 26 Counter EMF in a motor The armture windings of dc motor have a resistance of 5.0 Ω. The motor is connected to a 120-V line, and when the motor reaches full speed against its normal load, the counter emf is 108 V. Calculate the current into the motor when it is just starting up the current when it reaches full speed. Phys 2435: Chap. 27-31, Pg 27 Faraday’s Law (general form) d! B # =" dt d" B E $ dl = # ! dt # = ! E " dl The integral is taken around the loop through which the magnetic flux is changing. A changing magnetic flux produces an electric field. It’s a non-conservative field, because $ = # E " dl ! 0 Phys 2435: Chap. 27-31, Pg 28 Maxwell’s Displacement Current Can we understand why we must have a “displacement current”? • Consider applying Ampere’s Law to the current shown in the diagram. • If the surface is chosen as 1, 2 or 4, the enclosed current = I circuit • If the surface is chosen as 3, the enclosed current = 0! (ie there is no current between the plates of the capacitor) Big Idea: The added term is non-zero in this case, since the current I causes the charge Q on the capacitor to change in time which causes the Electric field in the region between the plates to change in time. The “displacement current” ID = ε0 (dφE/dt) in the region between the plates = the real current I in the wire. Phys 2435: Chap. 27-31, Pg 29 Maxwell’s Displacement Current In order to have for surface 2 to be equal to for surface 3, we want the displacement current in the region between the plates to be equal to the current in the wire. The Electric Field E between the plates of the capacitor is determined by the charge Q on the plate of area A: E = Q/(Aε0) What we want is a term that relates E to I without involving A. The answer: the time derivative of the electric flux!! Recall flux: Therefore, if we want ID = I, we need to identify: Phys 2435: Chap. 27-31, Pg 30 Maxwell’s Equations ! ! Q # E " dA = ! 0 Gauss’s law for electric field: electric charges produce electric fields. ! ! B ! d A = 0 " ! ! d! B $ E # dl = " dt J. C. Maxwell (1831 - 1879) Gauss’s law for magnetic field: but there’re no magnetic charges. Faraday’s law: changing B produces E. Ampere’s law as modified by Maxwell: ! ! d! E current or changing E # B " dl = µ0 I + µ0$ 0 dt electric produces B. All of electromagnetism is contained in this set of four equations. Phys 2435: Chap. 27-31, Pg 31 Chapter 30: Inductance Mutual Inductance Self-inductance Energy Stored in a Magnetic Field LR Circuits (DC) LC Circuits (DC) LRC Circuits (DC) Phys 2435: Chap. 27-31, Pg 32 Mutual Inductance The magnetic flux in coil 2 created by coil 1 is proportional to I1. Define N 2 ! 21 M 21 = I1 M21 is called the mutual inductance. It depends only on the geometric factors, NOT on the currents. Faraday’s law: The reverse situation is dI 2 " 1 = ! M 12 dt dI1 " 2 = ! M 21 dt M 12 = M 21 = M The SI unit for M is henry (H). 1 H = 1 V.s/A = 1 Ω.s Phys 2435: Chap. 27-31, Pg 33 Example: Solenoid and coil The magnetic field inside the solenoid is N1 B = µ0 I1 l The magnetic flux through the coil is N1 ! 21 = BA = µ 0 I1 A l Hence the mutual inductance is N 2 ! 21 M= I1 N1 N 2 A M = µ0 l Phys 2435: Chap. 27-31, Pg 34 Self-Inductance The total magnetic flux in the coil is proportional to the current I. Define N! B L= I L is called self-inductance. It depends only on the geometric factors, NOT on the current. Such coil is called an inductor. Use Faraday’s law: dI " = !L dt The SI unit for L is also henry (H). 1 H =1 Ω.s. Phys 2435: Chap. 27-31, Pg 35 Example: Solenoid inductance The magnetic field inside the solenoid is N B = µ0 I l The total magnetic flux through the coil is N ! B = BA = µ 0 IA l Hence the self-inductance is N! B L= or I N2A L = µ0 l For N=100, l=5 cm, A=0.3 cm2, L=4π x10-7x1002x0.3x10-4/0.05=7.5 µH. If filled with an iron core (µ=4000 µ0 ), L= 30 mH. Phys 2435: Chap. 27-31, Pg 36 Voltage across an inductor Inductor does not oppose current that flows through it. It opposes the change in the current. It’s a current stabilizer in the circuit. Phys 2435: Chap. 27-31, Pg 37 Energy Stored in a Magnetic Field When an inductor is carrying a current which is changing at a rate dI/dt, the energy is being supplied to the inductor at a rate dI P = I! = LI dt The work needed to increase the current from 0 to I is I 1 2 W = ! Pdt = ! LIdI = LI 2 0 By energy conservation, the energy stored in the inductor is 1 2 U = LI 2 Phys 2435: Chap. 27-31, Pg 38 Energy Stored in a Magnetic Field Question: Where exactly does the energy reside? 1 2 U= 2 LI Answer: It resides in the magnetic field. N2A Using L = µ 0 l N and B = µ 0 I l 2 1 & N A #& Bl # 1 B2 !! = !!$$ U = $$ µ 0 Al 2% l "% µ 0 N " 2 µ0 2 Or energy density The conclusion is valid for any region of space where a magnetic field exists. Compare with the electric case: 1 U = CV 2 2 1 B2 u= 2 µ0 1 u = !0E2 2 Phys 2435: Chap. 27-31, Pg 39 LR Circuits dI loop rule : V0 ! IR ! L = 0 dt Solve differential equation: I (t ) = V0 R (1 " e ) " t /! Where τ=L/R is called the time constant. Phys 2435: Chap. 27-31, Pg 40 LR Circuits dI loop rule : ! IR ! L = 0 dt Solve differential equation I (t ) = V0 R e " t /! time constant τ=L/R Summary: there is always some reaction time when a LR circuit is turned on or off. The situation is similar to RC circuits, except here the time constant is proportional to 1/R, not R. Phys 2435: Chap. 27-31, Pg 41 LC Circuits The capacitor is charged to Q0. At t=0, the circuit is closed. What will happen? Q dI loop rule : ! L = 0 C dt Using I=-dQ/dt, one gets d 2Q Q + =0 2 dt LC Solve differential equation where Q(t ) = Q0 cos("t + ! ) ! = 1 / LC Phys 2435: Chap. 27-31, Pg 42 LC Circuits Q(t ) = Q0 cos("t + ! ) 1 Q 2 Q02 UE = = cos 2 ("t + ! ) 2 C 2C 1 2 Q02 U B = LI = sin 2 ("t + ! ) 2 2C dQ I (t ) = = #Q0" sin ("t + ! ) dt Charge oscillates! So does the current and voltage. What about energy? The total energy is conserved. Phys 2435: Chap. 27-31, Pg 43 LRC Circuits Damped oscillations! 3 scenarios. A) Under-damped if R2<4L/C. B) Critical damping if R2=4L/C. C) Over-damped if R2>4L/C. For under-damping: Q(t ) = Q0 e T= $ R t 2L + - cos(" #t + ! ) 2# 2# LC = $" R 2C 1! 4L Phys 2435: Chap. 27-31, Pg 44 Summary of Various Direct Current Circuits RC circuit, time constant τ = RC (transient) LR circuit, time constant τ = L/R (transient) LC circuit, oscillation period T = 2π LC (oscillator) 2 R C = π − LRC circuit, damped oscillation period T 2 LC / 1 4L (damped oscillator) Phys 2435: Chap. 27-31, Pg 45 Chapter 31: Alternating Current AC Circuits (using phasors) AC Circuit, R only AC Circuit, L only AC Circuit, C only AC Circuit with LRC Resonance Transformers Phys 2435: Chap. 27-31, Pg 46 RMS current and voltage i = I cos !t T root-mean-square: 2 T I rms = i 2 2 1 2 I I i = ! i dt = (1 + cos 2"t )dt = ! T 0 2T 0 2 2 I rms I = 2 Vrms V = 2 Phys 2435: Chap. 27-31, Pg 47 AC Circuit Containing only Resistance The AC source is given as So the voltage across the resistor is i = I cos !t vR = iR = IR cos !t = VR cos !t The resistor voltage is in phase with the current. The power dissipated in the resistor is p=iv, or at an average rate 2 V 2 P = I rms R = rms R Phys 2435: Chap. 27-31, Pg 48 AC Circuit Containing only Inductor The AC source is given i = I cos !t di So the voltage across vL = L = ! IL" sin "t = VL cos("t + 900 ) the inductor is dt The inductor voltage leads the current by 90 degrees. Define reactance of inductor XL=ωL, then VL = I XL. Its unit is Ohm. Phys 2435: Chap. 27-31, Pg 49 AC Circuit Containing only Capacitor i = I cos !t The voltage across the capacitor is t q = " idt = 0 I I sin #t = cos(#t ! 900 ) # # v = q/C I 0 vC = cos("t ! 90 ) "C The capacitor voltage lags the current by 90 degrees. Define reactance of capacitor XC=1/ωC, then VC = I XC. Its unit is Ohm. Phys 2435: Chap. 27-31, Pg 50 AC-driven LRC Circuit in Series (Graphical Method) The goal is to cast the total voltage at any time into the form i = I cos !t v = vR + vL + vC = V cos("t + ! ) where φ is the phase angle with which v leads i. From the triangle: V = VR2 + (VL ! VC ) 2 = ( IR) 2 + ( IX L ! IX C ) 2 = I R 2 + ( X L ! X C )2 The impedance The phase difference V = IZ with Z = R 2 + ( X L ! X C ) 2 X L ! XC R tan " = or cos" = R Z Phys 2435: Chap. 27-31, Pg 51 The Impedance Triangle Z = R 2 + ( X L ! X C )2 X L = #L 1 XC = #C X L ! XC R tan " = or cos" = R Z General results, valid for any combination of L, R, C in series. For example, if LR circuit, set XC=0. Check special cases: R only : X L = 0, X C = 0, so Z = R, ! = 0 L only : R = 0, X C = 0, so Z = X L , ! = +90 0 C only : R = 0, X L = 0, so Z = X C , " = !90 0 Phys 2435: Chap. 27-31, Pg 52 Frequency Dependence i = I cos #t v = V cos(#t + $ ) V = IZ Z = R 2 + ( X L ! X C )2 X L = #L 1 #C X L ! XC R tan " = or cos" = R Z XC = Phys 2435: Chap. 27-31, Pg 53 The Power Factor The power is only dissipated in the resistor, so the average power is 2 P = I rms R But P=I 2 rms R = Z cos ! 1 Z cos ! = I rmsVrms cos ! = IV cos ! 2 The factor cosφ is called the power factor. For example: For a pure resistor (φ=0) , cosφ=1. For a pure inductor (φ=900) or capacitor (φ=-900), cosφ=0. Phys 2435: Chap. 27-31, Pg 54 Resonance in AC Circuit Recall: V V I= = 2 2 Z ( ) + − R X L XC For fixed R,C,L the current I will be a maximum at the resonant frequency ω0 which makes the impedance Z purely resistive. XL = XC Z is minimum when: So the frequency at which this condition is satisfied is given from: ωoL = 1 ω oC ⇒ ωo = 1 LC • Note that this resonant frequency is identical to the natural frequency of the LC circuit by itself! • At this frequency, the current and the driving voltage are in phase! R φ = = 1, so φ = 0 cos Phys 2435: Chap. 27-31, Pg 55 Z Resonance in AC Circuit At resonance, I and V are in phase. V, I and P are at their maximum. (V=100 V, L=2.0 H, C=0.5 µF) Phys 2435: Chap. 27-31, Pg 56 Transformers Transformers change alternating (AC) voltage to a bigger or smaller value Changing flux in secondary induces emf Vs Input AC voltage Vp in the primary produces a flux Same ΔΦ /Δt !! Transformer equation: Ns Vs = Vp Np Phys 2435: Chap. 27-31, Pg 57 Transformers Nothing comes for free, however! voltage increase comes at the cost of current output power cannot exceed input power power in = power out (assume no heat loss) I pVp = I sVs If voltage increases, then current decreases If voltage decreases, then current increases Phys 2435: Chap. 27-31, Pg 58