Magnetic Fields Chapter 29 (continued)

1. Magnetic FieldsChapter 29(continued)

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

3. Trajectory of Charged Particlesin a Magnetic Field (B field pointsinto plane of paper.) v B B + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + v F F Magnetic Force is a centripetal force

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

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

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

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

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

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

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: The Lorentz force + + + + + + + + + + + + + + + q

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: The Lorentz force FB + + + + + + + + + + + + + + + FE q

12. The Electromagnetic Force When B, E, and v are mutually perpendicular, as pictured here, FE and FB point in opposite directions. + + + + + + + + + + + + + + + FB FE 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. qE = qvB Set FE = qE equal to FB = qvB: Hence with the pictured orientation of fields and velocity, the particle will travel in a straight line if v = E / B.

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

14. x x x x x B x x The Hall Effect vd I • 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. x x x x x B x x The Hall Effect + + + + + + + + + + + + vd I - - - - - - - - - - - - • 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.

16. x x x x x B x x The Hall Effect + + + + + + + + + + + + E vd I - - - - - - - - - - - - • 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.

17. x x x x x B x x The Hall Effect + + + + + + + + + + + + E vd I - - - - - - - - - - - - • 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=evdB

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

19. x x x x x B x x The Hall Effect + + + + + + + + + + + + d vd I - - - - - - - - - - - - • The charge separation builds up until the two forces are equal: • eE=evdB • This means an electric potential difference develops between the two edges: VH=Ed=vdBd -the Hall voltage

20. x x x x x B x x The Hall Effect + + + + + + + + + + + + d vd I - - - - - - - - - - - - • The charge separation builds up until the two forces are equal: • eE=evdB • This means an electric potential difference develops between the two edges: VH=Ed=vdBd -the Hall voltage • This means that measuring the Hall voltage lets you work out the drift velocity.

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

22. + + + + + + + + + + + + x x x x x x x x x x B B x x x x vd I - - - - - - - - - - - - The Hall Effect VH • 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: - - - - - - - - - - - - VH vd I + + + + + + + + + + + + • 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 A • Force on one charge • F = q vd x B • Forces on all charges in a length L of a conductor: • F = n A L q vd x B • Use I = n q vd A and define a vector L whose length is L, and has the same direction as the current I. Then L I = n q vd A F I • F = I L x B L [F points out of the page]

24. I 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 q = 900 to the field as shown. What is the force on the wire? L N S

25. I 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 q = 900 to the field as shown. What is the force on the wire? L N S (up)

26. I 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 q = 900 to the field as shown. What is the force on the wire? L N S (up) (up)

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

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

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

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

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

32. Magnetic Force on a Current Loop Torque & Magnetic Dipole 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 (Torque on a current loop)

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

34. a b I dF 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

35. Nonuniform Fields and Curved Conductors a b I 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? R I L L

37. Nonuniform Fields and Curved Conductors: Example R I q L L

38. Nonuniform Fields and Curved Conductors: Example q R I q L L

39. q R I q L L Nonuniform Fields and Curved Conductors: Example

40. q R I q L L Nonuniform Fields and Curved Conductors: Example

41. q R I q L L Nonuniform Fields and Curved Conductors: Example

42. q R I q L L Nonuniform Fields and Curved Conductors: Example

43. q R I q L L Nonuniform Fields and Curved Conductors: Example

44. q R I q L L Nonuniform Fields and Curved Conductors: Example Equal to the force we would find for a straight wire of length 2(R+L)