1. Class Averages (3 exams w/bonus points and homework quizzes) average 70.4 +95 90 - 85 +80 75 - 70 +65 60 - 55 +50 45 - 40 These cutoffs will not be raised…you can use them to determine your guaranteed lowest grade. 35 40 45 50 55 60 65 70 75 80 85 90 95 100

2. + _ 1.up. 2. down. 3. left. 4. right. 5. into the screen. 6.out of the screen. 7. is zero. The B-field everywhere within this current-carrying loop points

3. +-

5. The SOLENOID pictured carries current in the direction indicated:

6. 1)A magnetic field is set up with a North Pole at the sphere’s top. 2)A magnetic field is set up with a South Pole at the sphere’s top. 3)The magnetic field lines form closed concentric circles centered on the spin axis. 4)Since the charge stays centered on a fixed position, no magnetic field is generated. A positively charged sphere is set spinning as shown at left.

7. N S SNSN

9. 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d Oxygen O 8 Iron Fe 26

10. Fe 3 O 4 [Magnetite] Or the simple cubic IRON crystal The slightly magnetic iron and oxygen atoms are locked into a crystal that fixes their orientation. the magnetic field of any atom may point in any of six different directions.

11. As molten iron solidifies, the magnetic orientation of any atom is totally random. So generally, any sample of iron is overall (on average) UNMAGNETIZED. By chance, neighboring atoms may share the same orientation, making microscopic, highly localized magnets.

12. The iron sample settles into microscopic regions that are locally magnetized. But overall, with no single dominant direction, the iron remains UNMAGNETIZED. These microscopic neighborhoods are called DOMAINS

13. In the presence of a strong external magnetic field magnetic flipping can help some domains to grow.

14. In the presence of a strong external magnetic field magnetic flipping can help some domains to grow and a sample of iron becomes magnetized!

15. An “induced” magnetic North pole.

16. What’s the resistance of a 60-Watt bulb (designed to be operated with 120 volt )? Ohm’s LawPower V = IRP = IV P = I 2 R P = V 2 /R A. R = 60/120 = 0.50  B. R = 120/60 = 2.00  C. R = 60 2 /120 = 30  D. R = 120 2 /60 = 240  E. R = 60  120 = 720  F. R = 60 2  120 = 432,000  What’s the resistance of a 100-Watt bulb (designed to be operated with 120 volt )? A. 1.20  B. 83  C. 144  D. 360 

17. Nichrome ( nickel/chromium alloy ) elements high resistance wire that can generate enough heat to glow red hot: 541 Watt toaster I = 4.5 A R = 26.6 ohm 1500 Watt hair dryer I = 12.5 A R = 9.6 ohm

18. Household wiring 12 AWG copper wire (0.08in diameter) 0.0052  /meter Power cords 18 AWG copper (0.04in diameter) 0.0210  /meter Appliance wiring “telephone wire” 24 AWG copper (0.02in diameter) 0.0842  /meter

19. Utility lines power transmission 6 AWG copper (0.162in diameter) 0.00130  /meter

20. Unlike electrical components (resistors, inductors, transformers, motors) we can think of the short copper wires and tiny gold-plated copper traces on a circuit board as being resistance-less.

21. Two bulbs are connected “in series” with a 120-volt backup regulator. Bulb B burns much brighter than Bulb A. The current through Bulb B is ___ Bulb A. 1. less than 2. equal to 3. greater than The voltage across Bulb B is ___ Bulb A. 1. less than 2. equal to 3. greater than The power used in Bulb B is ___ Bulb A. 1. less than 2. equal to 3. greater than

22. Two identical bulbs would share the energy, while doubling the resistance in the circuit. 120 V “60 Watt” 240  “60 Watt” 240  = 0.25 A total Notice: each would really be burning only = 15 Watts total

23. Each bulb only sees a fraction of the total available voltage across its ends: +120 V 240  = 60 V We say the voltage “drops” across each circuit element that uses energy. 0 V +60 V 0 V = 60 V Schematics include test point voltages for diagnosing parts of complicated circuits.

24. The resistance of a 100 ft (30.48 m) of gauge 18 copper wire is about 0.64 . A 120 volt generator supplies DC current through a 100 foot extension cord (gauge 18 copper wire) to a 60 Watt bulb. How much current would be drawn? How much power would the bulb really consume? How big would the voltage drop be across the extension cord? How much power is wasted in the cord?

25. A 120 volt generator supplies DC current through a 100 foot extension cord (gauge 18 copper wire) to a 60 Watt bulb. How much current would be drawn? = 0.49734748 A instead of 0.50 A How much power would the bulb really consume? = 59.365 Watts instead of 60 W The resistance of a 100 ft (30.48 m) of gauge 18 copper wire is about 0.64 .

26. The resistance of a 100 ft (30.48 m) of gauge 18 copper wire is about 0.64 . A 120 volt generator supplies DC current through a 100 foot extension cord (gauge 18 copper wire) to a 60 Watt bulb. I = 0.49734748 A Voltage drop across the extension cord? 59.365 Watts = 0.6238 volts How much power is wasted in the cord? = 0.304 Watts which is 0.5% of the total energy supplied

27. The resistance of a 100 ft (30.48 m) of gauge 18 copper wire is about 0.64 . What about running a 1500 Watt (9.6  ) space heater with that extension cord? 0.64  120-V 9.6  0.64  How much current would be drawn? = 11.294 A Voltage drop across the extension cord? = 14.45632 volts How much power is wasted in the cord? = 163.270 Watts close to 12% of the energy

28. QUESTION 2 5. into the screen. QUESTION 1 2) South Pole at top. Curl the fingers of your right hand in the direction the sphere spins. Since it is a charged sphere, current circulates that way. Your thumb reveals the North pole of this electromagnet! D. R = 120 2 /60 = 240  QUESTION 3 Given Power and Voltage, while asked for R, clearly the most direct approach is to simply use P=V 2 /R. Solving for R gives: R = V 2 /P. C. 144  QUESTION 4 The only difference now is P = 100 Watts. So R = V 2 /P means (120 volts) 2 /100 Watts = 14400/100 = 144. QUESTION 5 2. equal to As much current enters as leaves this circuit (returning to the battery). Since this is a series circuit, exactly as much current flows through bulb B as A. 3. greater than QUESTION 6 The brightness is evidence that B uses more energy! 3. greater than QUESTION 7 P=IV. If both have the same size current passing through them, the brighter (greater P) must have the larger V.