For a current loop m=izR2=m专R2p先 ,Ao→i=m3 }7.8610 6.023×1026 ForR=10cm→i=9.3×102 56 1.05 x 105 Amperes Thus,an ordinary piece of iron can have the same magnetic moment as a current loop of radius 10 cm of 105 Amperes current. B.Magnetic Dipole Field H=Hom 2cos0ir+sin0io 4πr30 (multiply top bottom by Ho) Electric Dipole Field E= 2cos0 ir+sin0 io 4πc.r3 Analogy p→μom P=Np=M=Nm,N =of magnetic dipoles/volume 1 Polarization Magnetization II.Maxwell's Equations with Magnetization EQS MOS (e目)=pu-V.p 7.(可=-7（闪 Po=-V.P(Polarization or paired Pm=-V.(Ho M)(magnetic charge charge density) density） n[旧-E门=-n[F-p]+如 n[,-f】=-n[你-)】 6.641,Electromagnetic Fields,Forces,and Motion Lecture 8 Prof.Markus Zahn Page 3 of 136.641, Electromagnetic Fields, Forces, and Motion Lecture 8 Prof. Markus Zahn Page 3 of 13 For a current loop 2 3 0 0 B B 0 0 4 4 A A m=i R =m R i=m R 3 M 3M π πρ ⇒ ρ For R = 10 cm ( ) ⎛ ⎞ ( × ) ⇒× × ⎜ ⎟ ⎝ ⎠ 26 -24 3 4 6.023 10 i = 9.3 10 .1 7.86 10 3 56 = 1.05 x 105 Amperes Thus, an ordinary piece of iron can have the same magnetic moment as a current loop of radius 10 cm of 105 Amperes current. B. Magnetic Dipole Field _ _ 0 r 3 0 m H = 2 cos i sin i 4 r θ µ ⎡ ⎤ θ+ θ ⎢ ⎥ π µ ⎣ ⎦ (multiply top & bottom by µ0 ) Electric Dipole Field θ ⎡ ⎤ θ+ θ ⎢ ⎥ π ε ⎣ ⎦ _ _ r 3 0 p E = 2 cos i sin i 4 r Analogy p m → µ0 P = Np M = Nm ⇒ , N = # of magnetic dipoles / volume Polarization Magnetization II. Maxwell’s Equations with Magnetization EQS ∇ ρ− i i ( ) ε0 u E= P ∇ ρ −∇ i p = P (Polarization or paired charge density) ( ) ⎡ ⎤ ⎡ ⎤ − − −+ ⎢ ⎥ ⎣ ⎦ ⎢ ⎥ ⎣ ⎦ i i ε α b a b n E E =n P P 0 s σ u MQS ∇ − i i (µ0 0 H= M ) ∇ µ( ) ρ −∇ µ m = i( 0 M) (magnetic charge density) ( ) ( ) α b a b n H H =n M M 0 0 ⎡ ⎤⎡ ⎤ − −µ − ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦ i i µ