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

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

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.

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

Right hand rule II Right hand rule for current carrying wire

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

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

Example 20-9

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

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

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

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

Example 20-10 Field of first wire 𝐵= 𝜇 𝑜 𝐼 2𝜋𝑟 = 4𝜋∙ 10 −7 𝑇∙𝑚 𝐴 8 𝐴 2𝜋 0.003 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

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 9.8 𝑚 𝑠 2 1𝑚 8∙ 10 −5 𝑁 𝐴 𝑚 =14.7 𝐴

Problem 41

Problem 43

Problem 45

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

DC motor Current reverses every half turn DC motor

DC Motor animation DC motor animation http://educypedia.karadimov.info/library/MccComplet.swf

Loudspeaker