836_PartUniversity Physics Solution

# 836_PartUniversity Physics Solution - 28-14 Chapter 28...

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28-14 Chapter 28 28.43. IDENTIFY and SET UP: Use the appropriate expression for the magnetic field produced by each current configuration. EXECUTE: (a) 0 2 I B r μ π = so 2 6 7 0 22 ( 2 . 0 0 1 0 m ) ( 3 7 . 2 T ) 3.72 10 A 41 0 T m / A B I ππ μπ × == = × ×⋅ . (b) 0 2 N I B R = so 5 7 0 2 2(0.210 m)(37.2 T) 1.24 10 A (100)(4 10 T m/A) RB I N = × . (c) 0 N B I L = so 7 0 (37.2 T)(0.320 m) 237 A (4 10 T m/A)(40,000) BL I N = . EVALUATE: Much less current is needed for the solenoid, because of its large number of turns per unit length. 28.44. IDENTIFY: Example 28.10 shows that outside a toroidal solenoid there is no magnetic field and inside it the magnetic field is given by 0 . 2 μ NI B π r = SET UP: The torus extends from 1 15.0 cm r = to 2 18.0 cm. r = EXECUTE: (a) r = 0.12 m, which is outside the torus, so B = 0. (b) r = 0.16 m, so 3 00 (250)(8.50 A) 2.66 10 T. 2 2 (0.160 m) NI B r μμ = × (c) r = 0.20 m, which is outside the torus, so B = 0. EVALUATE: The magnetic field inside the torus is proportional to 1/ r , so it varies somewhat over the cross- section of the torus. 28.45. IDENTIFY: Example 28.10 shows that inside a toroidal solenoid, 0 2 NI B r = . SET UP: 0.070 m r = EXECUTE: 3 (600)(0.650 A) 1.11 10 T. 2 2 (0.070 m) NI B r = × EVALUATE: If the radial thickness of the torus is small compared to its mean diameter, B is approximately uniform inside its windings. 28.46. IDENTIFY: Use Eq.(28.24), with 0 replaced by m0 K = , with m 80. K = SET UP: The contribution from atomic currents is the difference between B calculated with and B calculated with 0 . EXECUTE: (a) 0 (80)(400)(0.25 A) 0.0267 T. 2 ( 0 . 0 6 0 m ) NI K NI B rr = = (b) The amount due to atomic currents is 79 79 (0.0267 T) 0.0263 T. 80 80 BB = EVALUATE: The presence of the core greatly enhances the magnetic field produced by the solenoid. 28.47. IDENTIFY and SET UP: 2 K NI B r = (Eq.28.24, with 0 replaced by K ) EXECUTE: (a) m 1400 K = 2 7 0m 2 (2.90 10 m)(0.350 T) 0.0725 A (2 10 T m/A)(1400)(500) rB I KN × = (b) m 5200 K = 2 7 2 0.0195 A (2 10 T m/A)(5200)(500) rB I × = EVALUATE: If the solenoid were air-filled instead, a much larger current would be required to produce the same magnetic field. 28.48. IDENTIFY: Apply 2 K μ NI B π r = . SET UP: m K is the relative permeability and mm 1 K χ = is the magnetic susceptibility. EXECUTE: (a) m 2 2 (0.2500 m)(1.940 T) 2021. (500)(2.400 A) π rB π K μ NI μ = (b) 1 2020. K =− = EVALUATE: Without the magnetic material the magnetic field inside the windings would be 4 /2021 9.6 10 T. B The presence of the magnetic material greatly enhances the magnetic field inside the windings.

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Sources of Magnetic Field 28-15 28.49. IDENTIFY: The magnetic field from the solenoid alone is 00 . B μ nI = The total magnetic field is m0 . B KB = M is given by Eq.(28.29). SET UP: 6000 turns/m n = EXECUTE: (a) (i) 13 0 (6000 m ) (0.15 A) 1.13 10 T. B μ nI μ −− == = × (ii) 36 m 0 15 1 9 9 (1.13 10 T) 4.68 10 A m. K MB μμ × = × (iii) 3 (5200)(1.13 10 T) 5.88 T. BKB × = (b) The directions of B ! , 0 B ! and M ! are shown in Figure 28.49. Silicon steel is paramagnetic and 0 B !
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836_PartUniversity Physics Solution - 28-14 Chapter 28...

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