Lecture_21_Part_1 - Magnetic Dipole Moment Potential Energy...

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A current carrying loop has a tendency to twist in magnetic field Compass needle: collection of atomic current loops B l I F m × Δ = Ihw IA = = μ Magnetic Torque on a Magnetic Dipole Moment
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B l I F m × Δ = IwB F m = θ sin IwB F = Torque ( τ ) = distance from the axle (lever arm) times perpendicular component of the force. θ θ τ sin sin 2 2 IwhB IwB h = = Ihw = μ θ μ τ sin B = B × = μ τ Works with loops of any shape! Magnetic Torque: Quantitative Analysis
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Calculate amount of work needed to rotate from angle θ I to θ f : dE = 2 F h 2 d θ = Δ = f i d h IwB U W m θ θ θ θ 2 sin 2 Δ U m = IwhB sin θ d θ θ i θ f = IwhB cos θ [ ] θ i θ f Δ U m = μ B cos θ f cos θ i θ μ cos B U m = Potential energy for a magnetic dipole moment B U m = μ Magnetic Dipole Moment: Potential Energy Ihw = μ
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Potential energy for a magnetic dipole moment B U m = μ U= min - µ B 0 max µ B 0 What is the energy difference between the highest and the lowest state? Picture of the U and µ in magnetic field – important in atomic
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Unformatted text preview: Magnetic Dipole Moment: Potential Energy B l Id F d × = dF IBdl dF = dF ⊥ θ sin IBdl dF = ⊥ ∫ = dl IB F net sin π sin 2 RIB F net = μ sin 2 R B F net = We don’t know θ Force on a Magnetic Dipole B U m ⋅ − = μ 2 , 1 , m m U U < _ > Δ − = Δ = Δ B U x F m us by dx dB x B x U F m us by − → Δ Δ − = Δ Δ = _ dx dU F x − = dx dB F x = <0 There is no force if field is uniform! Force on a Magnetic Dipole F x = − d ( − i B ) dx x dx dB F x μ = 3 2 4 x B bar x π ≈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≈ 3 2 4 x dx d F bar x 4 6 4 x F bar x μμ − ≈ Two magnets 4 2 1 6 4 x F x − ≈ Force on a Magnetic Dipole...
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