Physics 1502: Lecture 25 Today s Agenda

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Abraham Welch
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1 Physics 1502: Lecture 25 Today s Agenda Announcements: Midterm 2: NOT Nov. 6 Following week Homework 07: due Friday net week AC current esonances Electromagnetic Waves Mawell s Equations - evised Energy and Momentum in Waves C ε L ω 1

Phasors:LC Phasors:Tips This phasor diagram was drawn as a snapshot of

Sometimes, in working problems, it is easier to draw the diagram at a

y i m X L ε m φ i m i m X C Full Phasor Diagram From this diagram, we can

2 Phasors:LC Phasors:Tips This phasor diagram was drawn as a snapshot of time t=0 with the voltages being given as the projections along the y-ais. Sometimes, in working problems, it is easier to draw the diagram at a time when the current is along the -ais (when i=0). y i m X L ε m φ i m i m X C Full Phasor Diagram From this diagram, we can also create a triangle which allows us to calculate the impedance Z: Z X L -X C φ Impedance Triangle 2

3 esonance The current in an LC circuit depends on the values of the elements and on the driving frequency through the relation Z φ X L -X C Impedance Triangle Suppose you plot the current versus ω, the source voltage frequency, you would get: ε m / 0 i m = o 0 0 =2 o ω 1 2 2ω o Power and esonance in LC Power, as well as current, peaks at ω = ω 0. The sharpness of the resonance depends on the values of the components. Z ecall: X L -X C φ Therefore, We can write this in the following manner (which we won t try to prove): introducing the curious factors Q and 3

4 The Q factor Q also determines the sharpness of the resonance peaks in a graph of Power delivered by the source versus frequency. P av High Q Δω ω o Low Q ω Lecture 25, ACT 1 Consider the two circuits shown where C II = 2 C I. What is the relation between the quality factors, Q I and Q II, of the two circuits? I II (a) Q II < Q I (b) Q II = Q I (c) Q II > Q I 4

5 Lecture 25, ACT 2 Consider the two circuits shown where C II = 2 C I and L II = ½ L I. Which circuit has the narrowest width of the resonance peak? I II (a) I (b) II (c) Both the same Power Transmission How do we transport power from power stations to homes? At home, the AC voltage obtained from outlets in this country is 120V at 60Hz. Transmission of power is typically at very high voltages ( eg ~500 kv) (a high tension line) Transformers are used to raise the voltage for transmission and lower the voltage for use. We ll describe these net. But why? Calculate ohmic losses in the transmission lines: Define efficiency of transmission: Keep small Note for fied input power and line resistance, the inefficiency 1/V 2 Make V in big 5

coils. (primary) (secondary) The iron is used to maimize the mutual inductance. We assume that the entire flu produced by each turn of the primary is trapped in the iron.

6 Transformers AC voltages can be stepped up or stepped down by the use of transformers. iron The AC current in the primary circuit creates a time-varying magnetic field in the iron ε V 1 V 2 This induces an emf on the secondary windings due to the mutual inductance of the two sets of coils. (primary) (secondary) The iron is used to maimize the mutual inductance. We assume that the entire flu produced by each turn of the primary is trapped in the iron. Ideal Transformers (no load) No resistance losses All flu contained in iron Nothing connected on secondary The primary circuit is just an AC voltage source in series with an inductor. The change in flu produced in each turn is given by: The change in flu per turn in the secondary coil is the same as the change in flu per turn in the primary coil (ideal case). The induced voltage appearing across the secondary coil is given by: Therefore, N 2 > N 1 secondary V 2 is larger than primary V 1 (step-up) N 1 > N 2 secondary V 2 is smaller than primary V 1 (step-down) Note: no load means no current in secondary. The primary current, termed the magnetizing current is small! 6

7 Ideal Transformers What happens when we connect a resistive load to the secondary coil? Flu produced by primary coil induces an emf in secondary with a Load emf in secondary produces current i 2 This current produces a flu in the secondary coil N 2 i 2, which opposes the original flu -- Lenz s law This changing flu appears in the primary circuit as well; the sense of it is to reduce the emf in the primary... However, V 1 is a voltage source. Therefore, there must be an increased current i 1 (supplied by the voltage source) in the primary which produces a flu N 1 i 1 which eactly cancels the flu produced by i 2. Transformers with a Load With a resistive load in the secondary, the primary current is given by: It s time.. 7

8 Lecture 25, ACT 3 The primary coil of an ideal transformer is connected to a battery (V 1 = 12V) as shown. The secondary winding has a load of 2 Ω. There are 50 turns in the primary and 200 turns in the secondary. What is the current in the secondcary? ε V 1 N 1 (primary) iron V 2 N 2 (secondary) (a) 24 A (b) 1.5 A (c) 6 A (d) 0 A Lecture 25, ACT 4 The primary coil of an ideal transformer is connected to the wall (V 1 = 120V) as shown. There are 50 turns in the primary and 200 turns in the secondary. If 960 W are dissipated in the resistor, what is the current in the primary? (a) 8 A (b) 16 A (c) 32 A 8

Fields from Circuits? We have been focusing on what happens within the circuits we have been studying (eg currents, voltages, etc.) What s happening outside the circuits?

9 Fields from Circuits? We have been focusing on what happens within the circuits we have been studying (eg currents, voltages, etc.) What s happening outside the circuits?? We know that:» charges create electric fields and» moving charges (currents) create magnetic fields. Can we detect these fields? Demos:» We saw a bulb connected to a loop glow when the loop came near a solenoidal magnet.» Light spreads out and makes interference patterns. Do we understand this? f( f( ) z y 9

10 Mawell s Equations These equations describe all of Electricity and Magnetism. They are consistent with modern ideas such as relativity. They describe light! Mawell s Equations - evised In free space, outside the wires of a circuit, Mawell s equations reduce to the following. These can be solved (see notes) to give the following differential equations for E and B. These are wave equations. Just like for waves on a string. But here the field is changing instead of the displacement of the string. 10

11 Step 1 Plane Wave Derivation Assume we have a plane wave propagating in z (ie E, B not functions of or y) Eample: does this Step 2 Apply Faraday s Law to infinitesimal loop in -z plane E E Δ y B y z 1 z 2 ΔZ z Plane Wave Derivation Step 3 Apply Ampere s Law to an infinitesimal loop in the y-z plane: y B y E ΔZ z 1 z 2 z Δy B y Step 4 Combine results from steps 2 and 3 to eliminate B y!! 11

12 Plane Wave Derivation We derived the wave eqn for E : We could have also derived for B y : How are E and B y related in phase and magnitude? Consider the harmonic solution: (esult from step 2) where B y is in phase with E B 0 = E 0 / c eview of Waves from last semester The one-dimensional wave equation: has a general solution of the form: where h 1 represents a wave traveling in the + direction and h 2 represents a wave traveling in the - direction. A specific solution for harmonic waves traveling in the + direction is: h λ A A = amplitude λ = wavelength f = frequency v = speed k = wave number 12

E & B in Electromagnetic Wave Plane Harmonic Wave: where: y z Note: the direction of propagation is given by the cross product where are the unit vectors in the (E,B) directions.

13 E & B in Electromagnetic Wave Plane Harmonic Wave: where: y z Note: the direction of propagation is given by the cross product where are the unit vectors in the (E,B) directions. Nothing special about (E y,b z ); eg could have (E y,-b ) Note cyclical relation: Lecture 25, ACT 5 Suppose the electric field in an e-m wave is given by: 5A In what direction is this wave traveling? (a) + z direction (c) +y direction (b) -z direction (d) -y direction 13

(a) B = -(E o /c)cos(kz + ωt) (b) B = +(E o /c)cos(kz - ωt) (c) B = +(E o /c)sin(kz - ωt) Velocity of Electromagnetic Waves The wave equation

14 Lecture 25, ACT 5 Suppose the electric field in an e-m wave is given by: 5B Which of the following epressions describes the magnetic field associated with this wave? (a) B = -(E o /c)cos(kz + ωt) (b) B = +(E o /c)cos(kz - ωt) (c) B = +(E o /c)sin(kz - ωt) Velocity of Electromagnetic Waves The wave equation for E : (derived from Mawell s Eqn) Therefore, we now know the velocity of electromagnetic waves in free space: Putting in the measured values for µ 0 & ε 0, we get: This value is identical to the measured speed of light! We identify light as an electromagnetic wave. 14

15 The EM Spectrum These EM waves can take on any wavelength from angstroms to miles (and beyond). We give these waves different names depending on the wavelength. Gamma ays X ays Ultraviolet Visible Light Infrared Microwaves Short Wave adio TV and FM adio AM adio Wavelength [m] Long adio Waves Energy in EM Waves / review Electromagnetic waves contain energy which is stored in E and B fields: = Therefore, the total energy density in an e-m wave = u, where The Intensity of a wave is defined as the average power transmitted per unit area = average energy density times wave velocity: 15

RLC Circuit (3) We can then write the differential equation for charge on the capacitor. The solution of this differential equation is

RLC Circuit (3) We can then write the differential equation for charge on the capacitor The solution of this differential equation is (damped harmonic oscillation!), where 25 RLC Circuit (4) If we charge