861_PartUniversity Physics Solution

861_PartUniversity Physics Solution - Electromagnetic...

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Electromagnetic Induction 29-11 (b) 3 C 6.00 10 A. dq i dt == × (c) CC D0 C 0 , dE i i j Kj dt K A A = = PP P so 3 DC 6.00 10 A. ii × EVALUATE: , = so Kirchhoff&s junction rule is satisfied where the wire connects to each capacitor plate. 29.38. IDENTIFY and SET UP: Use C / iq t = to calculate the charge q that the current has carried to the plates in time t . The two equations preceeding Eq.(24.2) relate q to the electric field E and the potential difference between the plates. The displacement current density is defined by Eq.(29.16). EXECUTE: (a) 3 C 1.80 10 A i 0 at 0 qt The amount of charge brought to the plates by the charging current in time t is 361 0 C (1.80 10 A)(0.500 10 s) 9.00 10 C qi t −−− × × = × 10 5 12 2 2 4 2 00 9.00 10 C 2.03 10 V/m (8.854 10 C / N m )(5.00 10 m ) q E A σ −− × == = ×⋅ × 53 (2.03 10 V/m)(2.00 10 m) 406 V VE d × × = (b) 0 / E qA = P 3 11 C 12 2 2 4 2 /1 . 8 0 1 0 A 4.07 10 V/m s (8.854 10 dE dq dt i dt A A × = = × × Since C i is constant dE / dt does not vary in time. (c) dE j dt = P (Eq.(29.16)), with P replaced by 0 P since there is vacuum between the plates.) 12 2 2 11 2 D (8.854 10 C / N m )(4.07 10 V/m s) 3.60 A/m j × = 24 2 3 DD (3.60 A/m )(5.00 10 m ) 1.80 10 A; ij A i i × = EVALUATE: CD . = The constant conduction current means the charge q on the plates and the electric field between them both increase linearly with time and D i is constant. 29.39. IDENTIFY: Ohm&s law relates the current in the wire to the electric field in the wire. D . dE j dt = P Use Eq.(29.15) to calculate the magnetic fields. SET UP: Ohm&s law says E J ρ = . Apply Ohm&s law to a circular path of radius r . EXECUTE: (a) 8 62 (2.0 10 m)(16 A) 0.15 V/m. 2.1 10 m I EJ A ×Ω === = × (b) 8 2.0 10 m (4000 A/s) 38 V/m s. dE d ρ I ρ dI dt dt A A dt ⎛⎞ = = ⎜⎟ × ⎝⎠ (c) 10 2 0 (38 V/m s) 3.4 10 A/m . dE j dt = × (d) 10 2 6 2 16 (3.4 10 A/m )(2.1 10 m ) 7.14 10 A. A ==× × Eq.(29.15) applied to a circular path of radius r gives 16 21 0D 0 D (7.14 10 A) 2.38 10 T, 2 2 (0.060 m) I B r μμ ππ × = × and this is a negligible contribution. 5 0C 0 C (16 A) 5.33 10 T. 2 2 (0.060 m) I B r = × EVALUATE: In this situation the displacement current is much less than the conduction current. 29.40. IDENTIFY: Apply Ampere’s law to a circular path of radius , rR < where R is the radius of the wire. SET UP: The path is shown in Figure 29.40. 0C 0 E d dI dt μ Φ ⋅+ Bl = ! ! o P Figure 29.40
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29-12 Chapter 29 EXECUTE: There is no displacement current, so 0C dI μ Bl = ! ! o The magnetic field inside the superconducting material is zero, so 0. d = ! ! o But then Ampere&s law says that C 0; I = there can be no conduction current through the path. This same argument applies to any circular path with r < R , so all the current must be at the surface of the wire. EVALUATE: If the current were uniformly spread over the wire&s cross section, the magnetic field would be like that calculated in Example 28.9. 29.41. IDENTIFY: A superconducting region has zero resistance. SET UP: If the superconducting and normal regions each lie along the length of the cylinder, they provide parallel conducting paths.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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861_PartUniversity Physics Solution - Electromagnetic...

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