1. Recall that a coil carrying a steady current I generates a uniform magnetic field within its enclosed area of                                                                                                                                                                                                                                                                                                                                      I B=oIB=oI B This circular loop encloses a total FLUX of  =BA =(  o I)  r 2

2. A wire coiled into two loops carries a current, I. It faces a single independent loop enclosing the same cross-sectional area. The flux through the secondary (red) loop must be 1)  =(  o I)  r 2. 2) ½ . 3) 2 . 4) 3 . 5) zero. Here assume  =BA =(  o I)  r 2

3. Compared to the voltage drop across coil A, the voltage across coil B is 1) less than 6 V 2) 6 V 3) greater than 6 V AB 6 V

4. What is the voltage across the light bulb? 1)60 V 2)120 V 3)240 V What is the current through the light bulb? 1)1/2 Amp 2)1 Amp 3)2 Amps What is the current through middle circuit? 1)1/2 Amp 2)1 Amp 3)2 Amps 240 V 120 V 120 Watt bulb

5. f  fofo R Z “ Resonance ” occurs when X C =X L (and they cancel!)

6. f  fofo R Z I=V/Z f  “ Resonance ” occurs when X C =X L (and they cancel!) fofo I=V/R

7. LC Oscillations Fully charged capacitor Energy: all E + + – – Capacitor charged Energy: all E Current starts to flow Energy: some E, some B Maximum current Energy: all B Current begins to die down  induced    C charges in opposite direction Energy: some E, some B Current flows ( opposite direction !) Energy: some E, some B + + – – + – II II + – + –

8. Fully charged capacitor Current starts to flow + + – – + – I I I I I Maximum current Maximum positive current

9. Capacitor fully charged Current flows ( opposite direction !) + + – – I + – I + – I Current begins to die down I I

10. Capacitor:Stores energy in E field Energy = ½ CV 2 Inductor : Stores energy in B field Energy = ½ LI 2 LC Circuits InductorCapacitor Two ways to store energy  Oscillations Frequency of oscillation = LC 1 2π 1 Adjusting the capacitance of an LC circuit is the basic was a radio tuner works. Question:What are the two ways to store energy in a pendulum? Answer:Gravitational potential and kinetic energy

11. A radio is tuned to 141 MHz. If the tuner is to be set for 100 MHz, the capacitance of the LC circuit in the radio must be A) quadrupled B) doubled C) not changed D) halved E) quartered

12. N S NSSN magnetic tape Thin, plastic tape with millions of tiny permanent magnets adhered to it AC current matching frequency of voice recording head (iron ring) Guides the magnetic field created by the current in the coil around to to the recording tape. coil Changing current in the coil creates a changing magnetic field in the iron Gap Gap in recording head lets the magnetic field out, so it can align the magnetic dipoles on the tape direction of tape movement magnetized areas of tape changing current produces changing B field

13. NSSN S N direction of tape movement Changing magnetic field inside the coil induces a current in the coil. changing current Tape induces a magnetic field in the recording head. (Just like holding a magnet near a nail makes it temporarily magnetic) Of course, there are amplifiers and other stereo components which, for clarity, are not shown. changing B field induces changing current

14. 3) 2 . The too black loops reinforce one another, combining to create a magnetic field B =  o I. 1) less than 6 V The voltage across B is zero Only a changing magnetic flux induces an EMF. Batteries provide DC current. 2)120 V Twice as many loops voltage doubles. THEN: half as many loops, voltage halves 2)1 Amp Supposed to be an easy one. P = IV, so 120 watts = I (120 volts). 1)1/2 Amp Twice the voltage, means half the current. 1 LC 2π 1 f = C  2C then f  (1/  2) f  2 = 1.41 141 MHz / 1.41 = 100 MHz B) doubled