Lecture_14-Part_1

# Lecture_14-Part_1 - p dt = ω p = v R mv = F net ω –...

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The magnetic field of a current loop and the magnetic field of a bar magnet look the same. B atom = μ 0 4 π 2 z 3 , μ= R 2 I What is the direction? S N What is the average current I ? current=charge/second: I = e t T = 2 R v R ev I 2 = One loop: eRv R ev R 2 1 2 2 = = The Atomic Structure of Magnets

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eRv 2 1 = μ Magnetic dipole moment of 1 atom: Method 1: use quantized angular momentum Orbital angular momentum: Rmv L = L m e Rmv m e eRv 2 1 2 1 2 1 = = = Quantum mechanics: L is quantized: s J , × = = 34 10 05 . 1 n L If n =1: μ= 1 2 e m L = 0.9 × 10 23 A m 2 per atom Magnetic Dipole Moment
eRv 2 1 = μ Magnetic dipole moment of 1 atom: Method 2: estimate speed of electron Momentum principle: net F dt p d = Circular motion: d

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Unformatted text preview: p dt = ω p = v R mv = F net ω – angular speed 2 2 2 4 1 R e R mv πε = m/s 6 2 10 6 . 1 4 1 × ≈ = Rm e v ≈ 1.3 × 10 − 23 A ⋅ m 2 /atom Magnetic Dipole Moment p = p = const = v / R Magnetic dipole moment of 1 atom: /atom m A 2 ⋅ ≈ − 23 10 μ Mass of a magnet: m ~5g Assume magnet is made of iron: 1 mole – 56 g 6 . 10 23 atoms number of atoms = 5g /56g . 6 . 10 23 ~ 6 . 10 22 magnet ≈ 6 × 10 22 ⋅ 10 − 23 = . 6 A ⋅ m 2 Magnetic Dipole Moment...
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Lecture_14-Part_1 - p dt = ω p = v R mv = F net ω –...

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