1. Chapter 26: Magnetism: Force and Field
Magnets

2. Magnetism
Magnetic forces

3. Magnetism
Magnetic field of Earth

4. Magnetism
Magnetic monopoles?
Perhaps there exist magnetic charges, just like electric charges.
Such an entity would be called a magnetic monopole (having + or magnetic charge).
How can you isolate this magnetic charge?
Try cutting a bar magnet in half: N S N S
Even an individual electron has a magnetic “dipole”!
Many searches for magnetic monopoles—the existence of which would explain (within framework of QM) the quantization of electric charge (argument of Dirac)
No monopoles have ever been found:

5. Magnetism
Source of magnetic field
What is the source of magnetic fields, if not magnetic charge?
Answer: electric charge in motion!
e.g., current in wire surrounding cylinder (solenoid) produces very similar field to that of bar magnet.
Therefore, understanding source of field generated by bar magnet lies in understanding currents at atomic level within bulk matter.
Orbits of electrons about nuclei
Intrinsic “spin” of electrons (more important effect)

6. Magnetism
Magnetic force: Observations

7. Magnetism
Magnetic force (Lorentz force)

8. Magnetism
Magnetic force (cont’d)
Components of the magnetic force

9. Magnetism B x x x x x x ® ® ® ® ® ­ ­ ­ ­ ­ ­ ­ ­ v ´ q F F = 0
Magnetic force (cont’d)
Magnetic force F
x x x x x x v B q
® ® ® ® ®
F = 0
­ ­ ­ ­ ­ ­ ­ ­

10. Magnetism
Magnetic force (cont’d)
Units of magnetic field

11. Magnetism
Magnetic force vs. electric force

12. Magnetic Field Lines and Flux

13. Magnetic Field Lines and Flux

14. Magnetic Field Lines and Flux
Magnetic field lines (cont’d)

15. Electric Field Lines of an Electric Dipole
Magnetic Field Lines and Flux
Magnetic field lines (cont’d)
Electric Field Lines of an Electric Dipole
Magnetic Field Lines of a bar magnet

16. Magnetic Field Lines and Flux
Magnetic field lines (cont’d)

17. Magnetic Field Lines and Flux
Magnetic field lines (cont’d)

18. Magnetic Field Lines and Flux
Magnetic flux
magnetic flux through a surface
Area A B  B B

19. Magnetic Field Lines and Flux
Magnetic flux (cont’d)
Units:
A=C/s, T=N/[C(m/s)]
-> Tm2=Nm/[C/s]=Nm/A
Gauss’s law for magnetism
No magnetic monopole has been observed!

20. Motion of Charged Particles in a Magnetic Field
Case 1: Velocity perpendicular to magnetic field
υ perpendicular to B
The particle moves at constant speed υ in a circle in the plane perpendicular to B.
F/m = a provides the acceleration to the center, so v R F B x

21. Motion of Charged Particles in a Magnetic Field
Case 1: Velocity perpendicular to magnetic field

22. Motion of Charged Particles in a Magnetic Field
Case 1: Velocity perpendicular to magnetic field
Velocity selector

23. Motion of Charged Particles in a Magnetic Field
Case 1: Velocity perpendicular to magnetic field
Mass spectrometer

24. Motion of Charged Particles in a Magnetic Field
Case 1: Velocity perpendicular to magnetic field
Mass spectrometer

25. Motion of Charged Particles in a Magnetic Field
Case 1: Velocity perpendicular to magnetic field
Mass spectrometer

26. Motion of Charged Particles in a Magnetic Field
Case 1: Velocity perpendicular to magnetic field
Mass spectrometer

27. Motion of Charged Particles in a Magnetic Field
Case 2: General case (v at any angle to B)

28. Motion of Charged Particles in a Magnetic Field
Case 2: General case (cont’d)
Since the magnetic field does not exert
force on a charge that travels in its direction,
the component of velocity in the magnetic
field direction does not change.

29. Magnetic Force on a Current-Carrying Conductor
Magnetic force on a current (straight wire)

30. Magnetic Force on a Current-Carrying Conductor
Magnetic force on a current (straight wire) (cont’d)

31. Magnetic Force on a Current-Carrying Conductor
Magnetic force on a current (curved wire)

32. Magnetic force on a current: Example1
Magnetic Force on a Current-Carrying Conductor
Magnetic force on a current: Example1

33. Magnetic force on a current: Example1 (cont’d)
Magnetic Force on a Current-Carrying Conductor
Magnetic force on a current: Example1 (cont’d)

34. Magnetic force on a current: Example1 (cont’d)
Magnetic Force on a Current-Carrying Conductor
Magnetic force on a current: Example1 (cont’d)

35. Magnetic force on a current: Example1 (cont’d)
Magnetic Force on a Current-Carrying Conductor
Magnetic force on a current: Example1 (cont’d)

36. Magnetic force on a current: Example1 (cont’d)
Magnetic Force on a Current-Carrying Conductor
Magnetic force on a current: Example1 (cont’d)

37. Magnetic force on a current: Example2
Magnetic Force on a Current-Carrying Conductor
Magnetic force on a current: Example2

38. Magnetic force on a current: Example2 (cont’d)
Magnetic Force on a Current-Carrying Conductor
Magnetic force on a current: Example2 (cont’d)

39. Magnetic force on a current: Example2 (cont’d)
Magnetic Force on a Current-Carrying Conductor
Magnetic force on a current: Example2 (cont’d)

40. Plane of loop is parallel to the magnetic field
Force and Torque on a Current Loop
Plane of loop is parallel to the magnetic field

41. Plane of loop : general case
Force and Torque on a Current Loop
Plane of loop : general case
if q=90o

42. Plane of loop and magnetic moment
Force and Torque on a Current Loop
Plane of loop and magnetic moment

43. Plane of loop : magnetic moment (cont’d)
Force and Torque on a Current Loop
Plane of loop : magnetic moment (cont’d)
The same magnetic dipole moment formulae work for any shape of planar loop.
Any such loop can be filled by a rectangular mesh as in the sketch. Each sub-loop is made to carry the current NI. You will now see that all the interior wires have zero current and are of no consequence. Nevertheless, each sub-loop contributes to μ in proportion to its area.

44. Plane of loop : magnetic moment (cont’d)
Force and Torque on a Current Loop
Plane of loop : magnetic moment (cont’d)

45. Potential energy of a magnetic dipole
Force and Torque on a Current Loop
Potential energy of a magnetic dipole
Work done by the torque when the magnetic
moment is rotated by df :
In analogy to the case of an electric dipole
in Chapter 22, we define a potential energy:
Potential energy of a magnetic
dipole at angle f to a magnetic
field

46. Applications
Galvanometer
We have seen that a magnet can exert a torque on a loop of current – aligns the loop’s “dipole moment” with the field.
In this picture the loop (and hence the needle) wants to rotate clockwise
The spring produces a torque in the opposite direction
The needle will sit at its equilibrium position
Current increased
μ = I • Area increases
Torque from B increases
Angle of needle increases
Current decreased
μ decreases
Torque from B decreases
Angle of needle decreases

47. Applications Slightly tip the loop Restoring force from the magnetic
Motor
Slightly tip the loop
Restoring force from the magnetic
torque
Oscillations
Now turn the current off, just as the loop’s μ is aligned with B
Loop “coasts” around until its μ is ~antialigned with B
Turn current back on
Magnetic torque gives another kick to the loop
Continuous rotation in steady state

48. Applications Even better
Motor (cont’d)
Even better
Have the current change directions every half rotation
Torque acts the entire time
Two ways to change current in loop:
Use a fixed voltage, but change the circuit (e.g., break connection every half cycle
 DC motors
2. Keep the current fixed, oscillate the source voltage
AC motors
VS I t

49. Applications
Motor (cont’d)
flip the current direction

50. Applications - - - + + + charges accumulate (in this case electrons)
Hall effect - - - + + +
Measuring
Hall voltage
(Hall emf)
In a steady state
qEH =qvdB
Charges move sideways until the Hall field EH grows to balance the force due to the magnetic field:
n can be measured

51. Applications
Electromagnetic rail gun

52. Exercises
Exercise 1
If a proton moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what is the speed of the proton and the frequency of motion? 1 v r x 2

53. Exercises
Exercise 2
Example of the force on a fast moving proton due to the earth’s magnetic field. (Already we know we can neglect gravity, but can we neglect magnetism?)
Let v = 107 m/s moving North.
What is the direction and magnitude of F?
Take B = 0.5x10-4 T and v B to get maximum effect.
(a very fast-moving proton) B N F v
vxB is into the paper (west). Check with globe

54. Magnetic Field of a Moving Charge
Magnetic field produced by a moving charge
Note the factor of μ0 /4π the constant of proportionality needed just as 1/(4πε0) is needed in electrostatics.

55. Magnetic Field of a Current Element
Magnetic field produced by a current element ds
For an element ds of a conductor carrying a current I there are n A ds charges with drift velocity υd (using priciple of superposition).
number of charge q ds ds

56. Magnetic Field of a Current Element
Biot-Savart law ds
Note: ds is dL in the textbook. ds
Note that ds is in the direction of I, but has a magnitude which is ds the length of wire considered.
Deduced by Biot and Savart c from experiments with coils

57. r Magnetic Field of a Current Element  dB P
Biot-Savart law (cont’d)  dB P r
The magnitude of the field dB is: I  ds
Total magnetic field at P is found by summing over all the current elements ds in the wire.

58. Magnetic Field of a Straight Current Carrying Conductor
A straight wire of length L
A thin straight wire of length L carries constant current I .
Calculate the total B field at P.  P x ds r I R y
 dB
So the magnitude of dB is given by:

59. Magnetic Field of a Straight Current Carrying Conductor
A straight wire of length L (cont’d)  P x ds r I R y
 dB

60. Magnetic Field of a Straight Current Carrying Conductor
A straight wire of length L (cont’d)  P x ds r I R y
 dB
In the limit (L/R) →∞
Magnetic field by a long straight wire

61. Magnetic Field of a Straight Current Carrying Conductor
A straight wire of length L (cont’d) B B B I B

62. Magnetic Field of a Straight Current Carrying Conductor
Example: A long straight wire
Iron filings

63. Note: B field at the centre of a loop, =2
Magnetic Field of a Current Element
Example
Calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two straight segments and a circular arc of radius R that subtends angle . A´ A ds C´
The magnetic field due to segments A´A and CC´ is zero because ds is parallel to along these paths. I  C R
Along path AC, ds and are perpendicular. O
Note: B field at the centre of a loop, =2

64. Force Between Parallel Conductors
Two parallel wires
At a distance a from the wire with current I1 the magnetic field due to the wire is given by

65. Force Between Parallel Conductors
Two parallel wires (cont’d)
Parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other.

66. Force Between Parallel Conductors
Definition of ampere
The chosen definition is that for a = L = 1m, The ampere is made to be such that F2 = 2×10−7 N when I1=I2=1 ampere 
This choice does two things (1) it makes the ampere (and also the volt) have very convenient magnitudes for every day life and (2) it fixes the size of μ0 = 4π×10−7. Note ε0 = 1/(μ0c2). All the other units follow almost automatically.

67. Magnetic Field of a Circular Current Loop
Magnetic field produced by a loop current
Use to find B field at the center of a loop of wire. R I
Loop of wire lying in a plane. It has radius R and total current I flowing in it.
First find
is a vector coming out of the paper at the same angle anywhere on the circle. The angle is constant.
Magnitude of B field at center of loop. Direction is out of paper. R i

68. Magnetic Field of a Circular Current Loop
Example 1:
Loop of wire of radius R = 5 cm and current I = 10 A. What is B at the center? Magnitude and direction I
Direction is out of the page.

69. Magnetic Field of a Circular Current Loop
Example 2:
What is the B field at the center of a segment or circular arc of wire? 0 R I
Total length of arc is S. P
where S is the arc length S =R0
0 is in radians (not degrees)
Why is the contribution to the B field at P equal to zero from
the straight section of wire?

70. Ampere’s Law Previously from the Biot-Savart’s law we had Ampere’s Law
Ampere’s law : A circular path
Consider any circular path of radius R centered on the wire carrying current I.
Evaluate the scalar product B·ds around this path.
Note that B and ds are parallel at all points along the path.
Also the magnitude of B is constant on this path. So the sum of all the B·ds terms around the circle is
Previously from the Biot-Savart’s law we had
Ampere’s Law
On substitution for B

71. Ampere’s Law For any path which encloses the wirez
Ampere’s Law ^ k
Ampere’s law : A general path ^ y r ^ x q q
Let us look at the integral along any shape of closed path in 3D. The most general ds is
Where unit vectors are used for the radial r and the tangential directions q and for z along the wire k. In this system we have ^ ^ ^
tangential component of ds
For any path which encloses the wire
For any path which does not enclose the wire

72. Ampere’s Law
Ampere’s law :
This law holds for an arbitrary closed path that is threaded by a
steady current.
I is the total current that passes through a surface bounded by the
closed path.

73. Ampere’s Law Electric Field General: Coulomb’s Law
Electric field vs. magnetic field
Electric Field
General: Coulomb’s Law
High symmetry: Gauss’s Law
Magnetic Field
General: Biot-Savart Law
High Symmetry: Ampère’s Law

74. Applications of Ampere’s Law
Magnetic field by a long cylindrical conductor
A long straight wire of radius R carries a steady current I that is uniformly distributed through the cross-section of the wire. Outside R.
In region where r < R choose a circle of radius r centered on the wire as a path of integration. Along this path, B is again constant in magnitude and is always parallel to the path.
Now Itot ≠ I.
However, current is uniform over the cross-section of the wire.
Fraction of the current I enclosed by the circle of radius r < R equals the ratio of the area of the circle of radius r and the cross section of the wire R2.

75. Applications of Ampere’s Law
Magnetic field by a long cylindrical conductor B r R

76. Applications of Ampere’s Law
Magnetic field by a circular current
Consider circular current carrying loop. Calculate B field at point P, a dist x from the centre of the loop on the axis of the loop.
Ids ds
Again in this case vector I ds is tangent to loopand perp to vector r from current element to point P. dB is in direction shown, perp to vectors r and I ds. Magnitude dB is: ds ds

77. Applications of Ampere’s Law
Magnetic field by a circular current (cont’d) ds ds
Ids
Integrate around loop, all components of dB perp to axis (e.g. dBy). integrate to zero.
Only dBx , the components parallel to axis contribute. ds
Field due to entire loop obtained by integrating:

78. B on the axis of a current loop
Applications of Ampere’s Law
Magnetic field by a circular current (cont’d)
But I, R and x are constant ds ds
Ids
B on the axis of a current loop

79. Applications of Ampere’s Law
Magnetic field by a circular current (cont’d)
Limits: x 0
x >>R
Compare case of electric field on axis of electric dipole far from dipole
vs.

80. Applications of Ampere’s Law
Magnetic field by a solenoid
When the coils of the solenoid are closely spaced, each turn can be regarded as a circular loop, and the net magnetic field is the vector sum of the magnetic field for each loop. This produces a magnetic field that is approximately constant inside the solenoid, and nearly zero outside the solenoid. I

81. Applications of Ampere’s Law
Magnetic field by a solenoid (cont’d)
The ideal solenoid is approached when the coils are very close together and the length of the solenoid is much greater than its radius. Then we can approximate the magnetic field as constant inside and zero outside the solenoid. I

82. Applications of Ampere’s Law
Magnetic field by a solenoid (cont’d)
Use Ampère’s Law to find B inside an ideal solenoid.

83. Applications of Ampere’s Law
Magnetic field by a toroid
A toroid can be considered as a solenoid “bent” into a circle as shown. We can apply Ampère’s law along the circular path inside the toroid.
N is the number of loops in the toroid, and I is the current in each loop

84. Exercises
Problem 1
The wire semicircles shown in Fig. have radii a and b. Calculate the net
magnetic field that the current in the wires produces at point P. I
Since point P is located at a symmetric position
with respect to the two straight sections where the
current I moves (anti)parallel to the radial direction.
So there is no contributions from these segments.
The contribution from the semicircle of radius a is
a half of that from a complete circle of the same radius:
Similarly the contribution from the semicircle of radius b is:
From principle of superposition, the net magnetic field at point P is: I P

85. Exercises Long, straight conductors with square cross sections and
Problem 2
Long, straight conductors with square cross sections and
each carrying current I are laid side-by-side to form an
infinite current sheet. The conductors lie in the xy-plane,
are parallel to the y-axis and carry current in the +y
direction. There are n conductors per unit length measured
along the x-axis. (a) What are the magnitude and direction
of the magnetic a distance a below the current sheets?
(b) What are the magnitude and direction of the magnetic
field a distance a above the current sheet? x z y

86. Exercises B Below sheet, all the magnetic
Problem 2 (cont’d) B
Below sheet, all the magnetic
field contributions from different
wires add up to produce a magnetic
field that points in the positive x-direction. Components in the z-direction
cancel. Using Ampere’s law, where we use the fact that the field is anti-
symmetric above and below the current sheets, and that the legs of the path
perpendicular provide nothing to the integral. So, at a distance a beneath the
sheet the magnetic field is:
b) The field has the same magnitude above the sheet, but points in the negative
x-direction. L B