Understanding Ampere's Law: Magnetic Field Calculation

1. screen menu, calcs & info menu buttons # shows 0 volts for each channel autoset cursor control measure cursor triggering control usual channel controls

2. screen menu, calcs & info menu buttons # shows 0 volts for each channel autoset cursor control measure cursor triggering control usual channel controls

3. Ideal Solenoid B B Boutside=0 B Boutside=0 B uniform magnetic field inside

4. Use Ampere’s Law to Find Magnetic Field (Explain each step in your report.) BIN L Amperian Loop N solenoid loops enclosed, each with current I. where n is “loop density” N/L of solenoid.

5. voltage VR=RIR=RIL The current in the solenoid creates a magnetic field inside the solenoid due to Ampere’s Law. Bampere t L The changing magnetic field inside the solenoid causes a back EMF (voltage) due to Faraday’s Law. Notice that dI/dt causes a phase shift.

6. Inside the solenoid: velocity B S N R Direction of current inside the resistor?

7. velocity B S N Binduced R

8. Transmitter Receiver Transmitting magnetic fields reach inside coils. Oscillating transmitting magnetic fields. Oscillating transmitting voltage. Oscillating voltage received is measurable.

9. The LRC Circuit - AC Driven voltage VR VC VL t

10. The LRC Circuit - AC Driven: Source from Addition Vsource voltage VR VC VL t

11. A B C

12. 100  0.1 F 50 mH

13. Iamplitude Iamplitude Large R Small R fdrive fdrive fresonance fresonance

14. VR(t) VR(t) 45o VS(t) VS(t) in phase out of phase

15. System: Charged hollow sphere with inner radius a and outer radius b. Charges: nonuniform charge distribution in between (so not a conductor): Problem: The electric field is a radial vector field due to the symmetry of the system. Find the electric field magnitude in the radial direction at every distance from the origin. Required vector calculus knowledge: rI r2 r3 Problem solving strategy: 1) Draw non-physical Gaussian sphere at distance r where you want to find Er. 2) Use Gauss’s law to write equation for Er in terms of other parameters. 3) Solve for Er. In this case solve in 3 places, inside hollow region (rI), inside charged region (r2) and outside (r3).

16. System: Charged infinite cylinder with radius a. Required vector calculus knowledge: Charges: Nonuniform charge distribution inside cylinder (so not a conductor): Problem: The electric field is a radial vector field due to the symmetry of the system. Find the electric field magnitude in the radial direction at every distance from the origin. Try solving over a finite height zo: rI r2 Problem solving strategy: 1) Draw non-physical Gaussian cylinder at distance r where you want to find Er. 2) Use Gauss’s law to write equation for Er in terms of other parameters including an arbitrary height zo. 3) Solve for Er. In this case solve in 2 places, inside region (rI), and outside (r2). You will need to have the arbitrary height zo cancel in the end.

17. Another view of drawing a Gaussian cylinder of radius r and finite length zo around an infinite cylinder of charge (this one outside). zo r

18. System: Charged infinite slab of width w in x-y direction. Required vector calculus knowledge: Charges: Uniform slab of charge density : Try solving over a finite box xo and yo: Problem: The electric field is a vector field pointing perpendicular to the plane of the slab due to the symmetry of the system. Find the electric field magnitude in the perpendicular direction at a given distance from the middle of the slab. xo yo yo xo z2 z1 Set z=0 in middle of slab. Problem solving strategy: 1) Draw non-physical Gaussian rectangular prism from center of slab to height z where you want to find Ez. 2) Use Gauss’s law to write equation for Ez in terms of other parameters including arbitrary length and width xo and yo. 3) Solve for Er. In this case solve in 2 places, inside region (zI), and outside (z2). You will need to have the arbitrary xo and yo cancel in the end.

19. Three representations of the same circuit: V V RBULB + BATTERY

20. + BATTERY

21. + + BATTERY BATTERY V 3.0 (Note: bulb shape distorted.) 0 Circuit Position

22. a b a b (Note: bulb shape distorted.) c d d a b a b 3 V + + + + c d BATTERY BATTERY BATTERY BATTERY c 3 V c d

23. A. B. C. 1.5 V 1.5 V 1.5 V 1.5 V 1.5 V

24. A. B. D. C. 1.5 V 1.5 V 1.5 V 1.5 V 1.5 V 1.5 V 1.5 V

25. A. B. D. E. C. 1.5 V 1.5 V + + + + + 1.5 V 1.5 V 1.5 V 1.5 V V V V V V 1.5 V 1.5 V - - - - -

26. + + + + BATTERY BATTERY BATTERY BATTERY ITOTAL IA IB IC ID IE

27. 3 V + + + BATTERY BATTERY BATTERY

28. + BATTERY Measuring the voltage drop across a light bulb (DMM in parallel): Voltage VDC R V

29. Measuring the voltage drop across a light bulb (DMM in series): Amperes mA R A + BATTERY

30. Measuring the resistance of a light bulb (component disconnected): Ohms ()  R

31. V 1.5 0 Circuit Position + BATTERY

32. V 1.5 + Circuit Position BATTERY

34. Right-hand-wrap rule for finding direction of magnetic poles created by moving charges (current). Thumb points to North N Wrap fingers in direction of current. q If charge is negative, reverse poles. S

36. N S

37. N S N S

38. S S

39. SMAGNETIC N S NMAGNETIC

40. Excess negative charge on the surface of a cubic conductor. Excess positive charge on the surface of a spherical conductor. - + + - - - - + - - + - - - - + - + - - - - - - + - + - - - - + - - - + - - - - + - + - - + + (Excess charges always repel each other to the surface.)

41. Excess negative charge on the surface of a cubic conductor. Excess positive charge on the surface of a spherical conductor. - + + - - - - + - - + - - - - + - + - - - - - - + - + - - - - + - - - + - - - - + - + - - + + (Excess charges always repel each other to the surface.)

42. - + - - + - - Negatively charged attracts positive charge and repels negative charges on the surface of a conductor. + - + + - - + - - + - - + - -

43. + Positively charged object creates an electric field that rotates polar molecules in insulator. MAGNIFY rotate - - - + + + insulator/dielectric material

44. Electrons deposited on the surface of a balloon by rubbing it against your hair do not spread out. - - - - - - - - -