829_Physics ProblemsTechnical Physics

829_Physics ProblemsTechnical Physics - Chapter 29 P29.25...

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Unformatted text preview: Chapter 29 P29.25 Choose U = 0 when the dipole moment is at θ = 90.0° to the field. The field exerts torque of magnitude µB sin θ on the dipole, tending to turn the dipole moment in the direction of decreasing θ. According to Equations 8.16 and 10.22, the potential energy of the dipole-field system is given by U −0 = z θ a µB sin θ dθ = µB − cos θ f 90 .0 ° P29.26 169 (a) θ 90 .0 ° = − µB cos θ + 0 U = −µ ⋅ B . or The field exerts torque on the needle tending to align it with the field, so the minimum energy orientation of the needle is: pointing north at 48.0° below the horizontal e je j where its energy is U min = − µB cos 0° = − 9.70 × 10 −3 A ⋅ m 2 55.0 × 10 −6 T = −5.34 × 10 −7 J . It has maximum energy when pointing in the opposite direction, south at 48.0° above the horizontal e je j where its energy is U max = − µB cos 180° = + 9.70 × 10 −3 A ⋅ m 2 55.0 × 10 −6 T = +5.34 × 10 −7 J . e j (b) P29.27 U min + W = U max : W = U max − U min = +5.34 × 10 −7 J − −5.34 × 10 −7 J = 1.07 µJ (a) τ = µ × B, τ = µ × B = µB sin θ = NIAB sin θ so a fb τ max = NIAB sin 90.0° = 1 5.00 A π 0.050 0 m (b) g e3.00 × 10 Tj = 2 −3 118 µN ⋅ m U = − µ ⋅ B , so − µB ≤ U ≤ + µB af a fb Since µB = NIA B = 1 5.00 A π 0.050 0 m g e3.00 × 10 Tj = 118 µJ , 2 −3 the range of the potential energy is: −118 µJ ≤ U ≤ +118 µJ . *P29.28 (a) τ = µ × B = NIAB sin θ e ja fb g F 2π rad IJ FG 1 min IJ = N ⋅ mb3 600 rev mingG H 1 rev K H 60 s K τ max = 80 10 −2 A 0.025 m ⋅ 0.04 m 0.8 N A ⋅ m sin 90° = 6.40 × 10 −4 N ⋅ m (b) Pmax = τ maxω = 6.40 × 10 −4 (c) In one half revolution the work is b g = 2 NIAB = 2e6.40 × 10 N ⋅ mj = 1.28 × 10 J In one full revolution, W = 2e1.28 × 10 Jj = 2.56 × 10 W = U max − U min = − µB cos 180°− − µB cos 0° = 2 µB −4 −3 −3 (d) Pavg = W 2.56 × 10 −3 J = = 0.154 W ∆t 1 60 s bg The peak power in (b) is greater by the factor π . 2 −3 J. 0.241 W ...
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