1. Magnetic Fields Gauss’s Law vs. Ampere’s Law Ampere’s Law
Ampere’s Law for long straight wire
Examples
Magnetic properties of materials
Force between 2 parallel wires
Motors
Loudspeakers

2. Magnetism and Induction Flowchart
law
change
field
Current
force
direct
examples
Force Law 1
B →
qv →
F = qv x B
RHR 1
charge deflection picture tube
Force Law 2
il →
F = il x B
2 wires, motor, loudspeaker
Ampere’s Law
B = μoi/2πr
← i
RHR 2
electromagnet solenoid
Faraday Induction
d/dt →
Φ = B*A →
ε = dΦ/dt i = ε/R
Lentz
generator transformer
Current creates magnetic field

3. Gauss’s vs. Ampere’s Law
Gauss’s Law for closed surface
Flux lines penetrating closed surface equals charge enclosed.
Ampere’s Law for closed loop
Magnetic field integrated along closed loop equals current enclosed.

4. Ampere’s Law for long straight wire
𝐵∙𝑑𝑙= 𝜇 𝑜 𝐼
Symmetric loop around current I
𝐵2𝜋𝑟= 𝜇 𝑜 𝐼
𝐵= 𝜇 𝑜 𝐼 2𝜋𝑟
Units
μo = 4π x 10-7 T-m/A “permeability of free space”
Electric forms “spokes” magnetic forms “loops”
Electromagnetic field

5. Right hand rule II
Right hand rule for current carrying wire

6. Example 20-7 Ampere’s Law for long wire 𝐵= 𝜇 𝑜 𝐼 2𝜋𝑟
𝐵= 𝜇 𝑜 𝐼 2𝜋𝑟
= 4𝜋∙ 10 −7 𝑇∙𝑚 𝐴 25 𝐴 2𝜋 0.1 𝑚
=5∙ 10 −5 𝑇
Direction: down into the page at point P

7. Example 20-8 𝐵 1 = 4𝜋∙ 10 −7 𝑇∙𝑚 𝐴 5 𝐴 2𝜋 0.05 𝑚 =2∙ 10 −5 𝑇 𝑢𝑝
For B1
𝐵 1 = 4𝜋∙ 10 −7 𝑇∙𝑚 𝐴 5 𝐴 2𝜋 0.05 𝑚 =2∙ 10 −5 𝑇 𝑢𝑝
For B2
𝐵 2 = 4𝜋∙ 10 −7 𝑇∙𝑚 𝐴 7 𝐴 2𝜋 0.05 𝑚 =2.8∙ 10 −5 𝑇 𝑢𝑝
Total field (by inspection)
𝐵=2∙ 10 −5 𝑇+2.8∙ 10 −5 𝑇=4.8∙ 10 −5 𝑇

8. Example 20-9

9. Magnetism in materials
Spinning charge particle creates magnetic field
If all spinning particles align, you have a magnet

10. Force between parallel wires - 1
Wire 1 sets up field.
𝐵 1 = 𝜇 𝑜 𝐼 1 2𝜋𝑟
Wire 2 feels force of that field.
𝐹= 𝐼 2 𝑙 𝐵 1
= 𝜇 𝑜 𝐼 1 𝐼 2 2𝜋𝑟 𝑙
And vice versa (wire 2 on 1)
F21=-F12 by action-reaction

11. Force between parallel wires - 2
currents same direction
currents opposite direction
action-reaction pair

12. Magnetism and Induction Flowchart
law
change
field
Current
force
direct
examples
Force Law 1
B →
qv →
F = qv x B
RHR 1
charge deflection picture tube
Force Law 2
il →
F = il x B
2 wires, motor, loudspeaker
Ampere’s Law
B = μoi/2πr
← i
RHR 2
electromagnet solenoid
Faraday Induction
d/dt →
Φ = B*A →
ε = dΦ/dt i = ε/R
Lentz
generator transformer
Wire 1 creates magnetic field
Wire 2 feels force of that field

13. Example 20-10 Field of first wire
𝐵= 𝜇 𝑜 𝐼 2𝜋𝑟 = 4𝜋∙ 10 −7 𝑇∙𝑚 𝐴 8 𝐴 2𝜋 𝑚 =5.33∙ 10 −4 𝑇
Force felt on second wire
𝐹=𝐼𝐿𝐵= 8𝐴 2𝑚 5.33∙ 10 −4 𝑁 𝐴 𝑚 =8.5∙ 10 −3 𝑁

14. Example 20-11 Field of first wire
𝐵 1 = 𝜇 𝑜 𝐼 1 2𝜋𝑟 = 4𝜋∙ 10 −7 𝑇∙𝑚 𝐴 80 𝐴 2𝜋 0.2 𝑚 =8∙ 10 −5 𝑇
Force on second wire
𝐹 2 = 𝐼 2 𝐿 𝐵 1 =𝑚𝑔
𝐼 2 = 𝑚𝑔 𝐿 𝐵 1
= 0.12∙ 10 −3 𝑘𝑔 𝑚 𝑠 𝑚 8∙ 10 −5 𝑁 𝐴 𝑚 =14.7 𝐴

15. Problem 41

16. Problem 43

17. Problem 45

18. Rotating current loop Force on each side of loop (fig a/b)
𝑏 2 𝑠𝑖𝑛Θ Θ
Force on each side of loop (fig a/b)
Left 𝐹=𝐼𝑎𝐵 𝑢𝑝
Right 𝐹=𝐼𝑎𝐵 𝑑𝑜𝑤𝑛
Moment arm of each torque (fig c)
𝑟 𝑝𝑒𝑟𝑝 = 𝑏 2 𝑠𝑖𝑛𝜑= 𝑏 2 𝑠𝑖𝑛𝜃
Torque of both sides
𝜏=𝐼𝑎𝐵 𝑏 2 𝑠𝑖𝑛𝜃+𝐼𝑎𝐵 𝑏 2 𝑠𝑖𝑛𝜃
=𝐼𝑎𝑏𝐵𝑠𝑖𝑛𝜃=𝐼 𝐴𝑟𝑒𝑎 𝐵𝑠𝑖𝑛𝜃
For N loops of wire
𝜏=𝑁𝐼 𝐴𝑟𝑒𝑎 𝐵𝑠𝑖𝑛𝜃

19. DC motor
Current reverses every half turn
DC motor

20. DC Motor animation DC motor animation

21. Loudspeaker