1. Magnetic Fields Chapter 29 (continued)

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

3. 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

4. ++++++++++++++++++++++++++++++++++++++++ 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.

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

6. 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

7. Trajectory of Charged Particles in a Magnetic Field What if the charged particle has a velocity component along B? z x y v

8. Trajectory of Charged Particles in a Magnetic Field What if the charged particle has a velocity component along B? F z =0 so v z =constant The force is in the xy plane. It acts exactly as described before, creating circular motion in the xy plane. z x y Result is a helix v

9. Trajectory of Charged Particles in a Magnetic Field What if the charged particle has a velocity component along B? F z =0 so v z =constant The force is in the xy plane. It acts exactly as described before, creating circular motion in the xy plane. z x y Result is a helix v

10. 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

11. 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 FBFB FEFE

12. The Electromagnetic Force + + + FBFB FEFE When B, E, and v are mutually perpendicular, as pictured here, F E and F B point in opposite directions. q The magnitudes do not have to be equal, of course. But by adjusting E or B you can set this up so the net force is zero. Set F E = qE equal to F B = qvB: qE = qvB Hence with the pictured orientation of fields and velocity, the particle will travel in a straight line if v = E / B.

13. The Hall Effect I vdvd x xxxxxx B Consider a conducting bar, carrying a current, with a perpendicular magnetic field into the picture.

14. The Hall Effect I vdvd x xxxxxx B Consider a conducting bar, carrying a current, with a perpendicular magnetic field into the picture. The electrons drifting to the right tend to move down because of the magnetic force.

15. The Hall Effect Consider a conducting bar, carrying a current, with a perpendicular magnetic field into the picture. The electrons drifting to the right tend to move down because of the magnetic force. Thus you get a charge separation: a net negative charge along the bottom edge, and positive along the upper. I vdvd - - - - - - + + + + + + x xxxxxx B

16. The Hall Effect Consider a conducting bar, carrying a current, with a perpendicular magnetic field into the picture. The electrons drifting to the right tend to move down because of the magnetic force. Thus you get a charge separation: a net negative charge along the bottom edge, and positive along the upper. This charge separation sets up an electric field, top to bottom, which pulls electrons up – opposing the magnetic force. I vdvd - - - - - - + + + + + + x xxxxxx B E

17. The Hall Effect Consider a conducting bar, carrying a current, with a perpendicular magnetic field into the picture. The electrons drifting to the right tend to move down because of the magnetic force. Thus you get a charge separation: a net negative charge along the bottom edge, and positive along the upper. This charge separation sets up an electric field, top to bottom, which pulls electrons up – opposing the magnetic force. The charge separation builds up until the two forces are equal: eE=ev d B I vdvd - - - - - - + + + + + + x xxxxxx B E

18. The Hall Effect The charge separation builds up until the two forces are equal: eE=ev d B I vdvd - - - - - - + + + + + + x xxxxxx B d

19. The Hall Effect The charge separation builds up until the two forces are equal: eE=ev d B This means an electric potential difference develops between the two edges:V H =Ed=v d Bd-the Hall voltage I vdvd - - - - - - + + + + + + x xxxxxx B d

20. The Hall Effect The charge separation builds up until the two forces are equal: eE=ev d B This means an electric potential difference develops between the two edges:V H =Ed=v d Bd-the Hall voltage This means that measuring the Hall voltage lets you work out the drift velocity. I vdvd - - - - - - + + + + + + x xxxxxx B d

21. The Hall Effect The charge separation builds up until the two forces are equal: eE=ev d B This means an electric potential difference develops between the two edges:V H =Ed=v d Bd-the Hall voltage This means that measuring the Hall voltage lets you work out the drift velocity. Moreover, using J=nev d and I=JA (with A the slab’s cross- sectional area) gives v d =I/(Ane) and so V H =IBd/Ane. Measuring the Hall voltage lets you find the density of conduction electrons. I vdvd - - - - - - + + + + + + x xxxxxx B d

22. The Hall Effect The Hall effect also lets you find the sign of the charge carriers that make up the current. Above is the picture for electrons. But if the charge carriers actually had a positive charge, the picture would look like this: I vdvd - - - - - - + + + + + + x xxxxxx B VHVH I vdvd - - - - - - + + + + + + x xxxxxx B VHVH The carriers would move to the bottom edge still, and the Hall voltage would point in the opposite direction.

23. Magnetic Force on a Current L A I = n q v d A Force on one charge F = q v d x B Forces on all charges in a length L of a conductor: F = n A L q v d x B Use I = n q v d A and define a vector L whose length is L, and has the same direction as the current I. Then I L F [F points out of the page] F = I L x B

24. Magnetic Force on a Current NS I L Example: A current, I=10 A, flows through a wire, of length L=20 cm, between the poles of a 1000 Gauss magnet. The wire is at  = 90 0 to the field as shown. What is the force on the wire?

25. Magnetic Force on a Current Example: A current, I=10 A, flows through a wire, of length L=20 cm, between the poles of a 1000 Gauss magnet. The wire is at  = 90 0 to the field as shown. What is the force on the wire? NS L (up) I

26. Magnetic Force on a Current Example: A current, I=10 A, flows through a wire, of length L=20 cm, between the poles of a 1000 Gauss magnet. The wire is at  = 90 0 to the field as shown. What is the force on the wire? NS L (up) I

27. Magnetic Force on a Current Loop I A current loop is placed in a uniform magnetic field as shown below. What will happen? N S L B

28. Magnetic Force on a Current Loop I N S F=BIL  L F’ B No net force – but a torque is imposed.

29. Magnetic Torque on a Current Loop Simplified view: F=BIL I d L 

30. Simplified view: F=BIL I d L  Magnetic Torque on a Current Loop

31. for a current loop F=BIL I d L  Magnetic Torque on a Current Loop A=vector with magnitude A=Ld and direction given by a RH rule.

32. By analogy with electric dipoles, for which: The expression, implies the a current loop acts as a magnetic dipole! Here is the magnetic dipole moment, and Magnetic Force on a Current Loop Torque & Magnetic Dipole (Torque on a current loop)

33. By further analogy with electric dipoles: So for a magnetic dipole (a current loop) Potential Energy of a Magnetic Dipole The potential energy is due to the fact that the magnetic field tends to align the current loop perpendicular to the field.

34. Nonuniform Fields and Curved Conductors So far, we have considered only uniform fields and straight current paths. If this is not the case, we must build up using calculus. Consider a small length, dL, of current path. The force on dL is: dF = dL I x B I a b dF

35. Nonuniform Fields and Curved Conductors I a b For a conductor of length L F = L I x B For a bit of length dL dF = dL I x B Then, for the total length of the curved conductor in a non-uniform magnetic field F =  dF =  dL I x B To find the force exerted by a non-uniform magnetic field on a curved current we divide the conductor in small sections dL and add (integrate) the forces exerted on every section dL.

36. Nonuniform Fields and Curved Conductors: Example What is the force on the current-carrying conductor shown?  LL R

37. Nonuniform Fields and Curved Conductors: Example  LL R 

38.  LL R  

39.  LL R  

40.  LL R  

41.  LL R  

42.  LL R  

43.  LL R  

44. Equal to the force we would find for a straight wire of length 2(R+L)  LL R  