1. DC Electrical Circuits Chapter 28 (Continued) Circuits with Capacitors

2. Kirchhoff’s Laws 1. At any circuit junction, currents entering must equal currents leaving. The loop method is based on two laws devised by Kirchoff: 2. Sum of all  V’s across all circuit elements in a loop must be zero. I1I1 I2I2 I 3 = I 1 + I 2 + - r E R I E - Ir - IR = 0

3. RC Circuits So far we have considered simple circuits with either capacitors or resistors. Now we will consider more complicated circuits with both resistors and capacitors: RC Circuits. + E R C - The battery pushes current until the capacitor is fully charged. After this no current flows. (A small lie.) This problem is time dependent.

4. +++ - - - RC Circuits: Charging V R =IR V C =q/C When the switch closes, at first a high current flows: V R is big and V C is small. + R C openclosed E I + E R C - -

5. +++ - - - RC Circuits: Charging V R =IR V C =q/C When the switch closes, at first a high current flows: V R is big and V C is small. As q is stored in C, V C increases. This fights against the battery so I decreases. + R C openclosed E I + E R C - -

6. RC Circuits: Charging Now use dq/dt = I and rearrange: Apply the loop law: E – IR - q/C = 0 Take the derivative of this with respect to time: This is a differential equation for an unknown function I(t). It is solved subject to the initial condition I(0) = E / R. V R =IR V C =q/C + R C closed E I - - - - +++

7. RC Circuits: Charging And I(0) = I 0 = E / R  E  V R =IR V C =q/C + R C closed E I - - - - +++

8. RC Circuits: Charging From this we get: E E E q = V C C = E C (1 – e –t/RC ) V R =IR V C =q/C + R C closed E I - - - - +++ E

9. t/RC Potential Drop E E/ R Current VCVC VRVR Charging I

10. Discharging an RC Circuit R C q Current will flow through the resistor for a while. Eventually, the capacitor will lose all its charge, and the current will go to zero. Power P = IV = I 2 R will be dissipated in the resistor (as heat) while the current flows. -q Open circuit V C =V 0 R C q I -q After closing switch V R =IR V C =q/C

11. Discharging an RC Circuit R C +q+q I -q V R =IR V C =q/C Loop equation: q/C - IR = 0  I = q / (RC) Take d/dt  Here the current at t=0 is given by the initial voltage on the capacitor: I(0) = V 0 /R = q 0 /RC [Note that I = - dq/dt] This equation is solved very much like the other (charging case):

12. Discharging an RC Circuit R C q I -q The charge on the capacitor is given by: q/C - IR = 0 so q = C IR [q = C V] V R =IR V C =q/C

13. t/RC Potential Drop 0 E/ R Current VCVC VRVR Discharging

14. Example: A capacitor C discharges through a resistor R. (a) When does its charge fall to half its initial value ? R C Q I Charge on a capacitor varies as

15. Example: A capacitor C discharges through a resistor R. (a) When does its charge fall to half its initial value ? R Charge on a capacitor varies as Find the time for which Q=Q 0 /2 C Q I

16. Example: A capacitor C discharges through a resistor R. (a) When does its charge fall to half its initial value ? R Charge on a capacitor varies as Find the time for which Q=Q 0 /2 C Q I

17. Example: A capacitor C discharges through a resistor R. (a) When does its charge fall to half its initial value ? R Charge on a capacitor varies as Find the time for which Q=Q 0 /2 RC is the “time constant” C Q I

18. Example: A capacitor C discharges through a resistor R. (b) When does the energy drop to half its initial value? The energy stored in a capacitor is We seek the time for U to drop to U 0 /2:

19. Example: A capacitor C discharges through a resistor R. (b) When does the energy drop to half its initial value? The energy stored in a capacitor is We seek the time for U to drop to U 0 /2:

20. Magnetic Fields Chapter 29 Permanent Magnets & Magnetic Field Lines The Magnetic Force on Charges

21. Magnetism Our most familiar experience of magnetism is through permanent magnets. These are made of materials which exhibit a property called ferromagnetism - i.e., they can be magnetized. Depending on how we position two magnets, they will attract or repel, i.e. they exert forces on each other. Just as it was convenient to use electric fields instead of electric forces, here too it is useful to introduce the concept of the magnetic field B. There are useful analogies between electric and magnetic fields, but the analogy is not perfect: while there are magnetic dipoles in nature, there seem to be no isolated magnetic charges (called “magnetic monopoles”). And the force laws are different. We describe magnets as having two magnetic poles: North (N) and South (S). Like poles repel, opposite poles attract.

22. Field of a Permanent Magnet NS Shown here are field lines. The magnetic field B at any point is tangential to the field line there.

23. The south pole of the small bar magnet is attracted towards the north pole of the big magnet. Also, the small bar magnet (a magnetic dipole) wants to align with the B-field. The field attracts and exerts a torque on the small magnet. Field of a Permanent Magnet NS N S

24. Magnetism The origin of magnetism lies in moving electric charges. Moving (or rotating) charges generate magnetic fields. An electric current generates a magnetic field. A magnetic field will exert a force on a moving charge. A magnetic field will exert a force on a conductor that carries an electric current.

25. What Force Does a Magnetic Field Exert on Charges? If the charge is not moving with respect to the field (or if the charge moves parallel to the field), there is NO FORCE. q

26. What Force Does a Magnetic Field Exert on Charges? q q If the charge is moving, there is a force on the charge, perpendicular to both v and B. F = q v x B If the charge is not moving with respect to the field (or if the charge moves parallel to the field), there is NO FORCE.

27. Force on a Charge in a Magnetic Field v F B q m (Use “Right-Hand” Rule to determine direction of F)

28. Units of Magnetic Field Since Therefore the units of magnetic field are: (Note: 1 Tesla = 10,000 Gauss)

29. The Electric and Magnetic Forces are Different Whereas the electric force acts in the same direction as the field: The magnetic force acts in a direction orthogonal to the field:

30. Whereas the electric force acts in the same direction as the field: The magnetic force acts in a direction orthogonal to the field: (Use “Right-Hand” Rule to determine direction of F) The Electric and Magnetic Forces are Different

31. Whereas the electric force acts in the same direction as the field: The magnetic force acts in a direction orthogonal to the field: And – the charge must be moving. (Use “Right-Hand” Rule to determine direction of F) The Electric and Magnetic Forces are Different

32. Trajectory of Charged Particles in a Magnetic Field ++++++++++++++++++++++++++++++++++++++++ v B F (B field points into plane of paper.)

33. Trajectory of Charged Particles in a Magnetic Field ++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++++++++++++++++++ v v BB F F (B field points into plane of paper.)

34. Trajectory of Charged Particles in a Magnetic Field ++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++++++++++++++++++ v v BB F F (B field points into plane of paper.) Magnetic Force is a centripetal force

35. Rotational Motion r  s  = s / r  s =  r  ds/dt = d  /dt r  v =  r  atat arar a t = r  tangential acceleration a r = v 2 / r radial acceleration The radial acceleration changes the direction of motion, while the tangential acceleration changes the speed. Uniform Circular Motion  = constant  v and a r constant but direction changes a r = v 2 /r =  2 r F = ma r = mv 2 /r = m  2 r KE = (1/2) mv 2 = (1/2) m  2 r 2  = angle,  = angular speed,  = angular acceleration v  arar

36. Radius of a Charged Particle Orbit in a Magnetic Field ++++++++++++++++++++++++++++++++++++++++ v B F r Centripetal Magnetic Force Force =

37. Radius of a Charged Particle Orbit in a Magnetic Field Centripetal Magnetic Force Force = ++++++++++++++++++++++++++++++++++++++++ v B F r

38. Radius of a Charged Particle Orbit in a Magnetic Field Centripetal Magnetic Force Force = ++++++++++++++++++++++++++++++++++++++++ v B F r

39. ++++++++++++++++++++++++++++++++++++++++ Radius of a Charged Particle Orbit in a Magnetic Field v B F r Centripetal Magnetic Force Force = Note: as, the magnetic force does no work.

40. Cyclotron Frequency ++++++++++++++++++++++++++++++++++++++++ v B F r The time taken to complete one orbit is:

41. Cyclotron Frequency ++++++++++++++++++++++++++++++++++++++++ v B F r The time taken to complete one orbit is: Hence the orbit frequency, f

42. Cyclotron Frequency ++++++++++++++++++++++++++++++++++++++++ v B F r The time taken to complete one orbit is: Hence the orbit frequency, f - known as the “cyclotron frequency” T = 2  /  = 1/ƒ  ƒ =  /2 

43. The Electromagnetic Force If a magnetic field and an electric field are simultaneously present, their forces obey the superposition principle and must be added vectorially: + + + q The Lorentz force

44. Exercise B v v’v’ In what direction does the magnetic field point? Which is bigger, v or v’ ? electron

45. Exercise: answer B v v’v’ In what direction does the magnetic field point ? Into the page [F = -e v x B] Which is bigger, v or v’ ? v = v’ [B does no work on the electron, F  v] electron F

46. What is the orbital radius of a charged particle (charge q, mass m) having kinetic energy K, and moving at right angles to a magnetic field B, as shown below?. x x x x x x q m B K

47. What is the orbital radius of a charged particle (charge q, mass m) having kinetic energy K, and moving at right angles to a magnetic field B, as shown below?. x x x x x x q m B r K = (1/2) mv 2 q v B = m v 2 / r F = q v x B = m a and a = v 2 / r q B = m v / r  r q B = m v r = m v / (q B) r 2 = m 2 v 2 / (q B) 2 (1/2m) r 2 = K / (q B) 2  r = [2mK] 1/2 / (q B)

48. What is the relation between the intensities of the electric and magnetic fields for the particle to move in a straight line ? The magnetic field points into the picture. The direction of the electric field is not yet specified. x x x x x x q m B v E

49. What is the relation between the intensities of the electric and magnetic fields for the particle to move in a straight line ? x x x x x x q m B v E FEFE FBFB F E = q E and F B = q v B If F E = F B the particle will move following a straight line trajectory q E = q v B v = E / B

50. What is the relation between the intensities of the electric and magnetic fields for the particle to move in a straight line ?. x x x x x x q m B v E FEFE FBFB F E = q E and F B = q v B If F E = F B the particle will move following a straight line trajectory q E = q v B v = E / B So need E pointing to the right.

51. Trajectory of Charged Particles in a Magnetic Field What if the charged particle has a velocity component along B? unchanged Circular motion in xy plane. z x y