ssm_ch30 - Chapter 30 5 The magnitude of the magnetic field...

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Chapter 30 5 The magnitude of the magnetic field inside the solenoid is B = ¹ 0 ni s ,whe re n is the number of turns per unit length and i s is the current. The field is parallel to the solenoid axis, so the flux through a cross section of the solenoid is © B = A s B = ¹ 0 ¼r 2 s ni s ,whe re A s (= ¼r 2 s ) is the cross-sectional area of the solenoid. Since the magnetic field is zero outside the solenoid, this is also the flux through the coil. The emf in the coil has magnitude E = Nd © B dt = ¹ 0 ¼r 2 s Nn di s dt and the current in the coil is i c = E R = ¹ 0 ¼r 2 s Nn R di s dt ; where N is the number of turns in the coil and R is the resistance of the coil. The current changes linearly by 3 : 0A in 50ms, so di s =dt =(3 : 0A) = (50 £ 10 ¡ 3 s) = 60 A = s. Thus i c = (4 ¼ £ 10 ¡ 7 T ¢ m = A) ¼ (0 : 016 m) 2 (120)(220 £ 10 2 m ¡ 1 ) 5 : 3 ­ (60 A = s) = 3 : 0 £ 10 ¡ 2 A : 21 (a) In the region of the smaller loop, the magnetic field produced by the larger loop may be taken to be uniform and equal to its value at the center of the smaller loop, on the axis. Eq. 29–26, with z = x andmuchg rea te rthan R ,g ives B = ¹ 0 iR 2 2 x 3 for the magnitude. The field is upward in the diagram. The magnetic flux through the smaller loop is the product of this field and the area ( ¼r 2 )o fthesma l le rloop : © B = ¼¹ 0 ir 2 R 2 2 x 3 : (b) The emf is given by Faraday’s law: E = ¡ d © B dt = ¡ μ ¼¹ 0 ir 2 R 2 2 d dt μ 1 x 3 = ¡ μ ¼¹ 0 ir 2 R 2 2 ¶μ ¡ 3 x 4 dx dt = 3 ¼¹ 0 ir 2 R
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the field of the larger loop. It is counterclockwise as viewed from above, in the same direction as the current in the larger loop. 29 Thermal energy is generated at the rate E 2 =R ,wh e r e E is the emf in the wire and R is the resistance of the wire. The resistance is given by R = ½L=A ,where ½ is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area of the wire. The resistivity can be found in Table 26–1. Thus R = ½L A = (1 : 69 £ 10 ¡ 8 ­ ¢ m)(0 : 500 m) ¼ (0 : 500 £ 10 ¡ 3 m) 2 =1 : 076 £ 10 ¡ 2 ­ : Faraday’s law is used to find the emf. If B is the magnitude of the magnetic field through the loop, then E = AdB=dt ,whe re A is the area of the loop. The radius r of the loop is r = L= 2 ¼ and its area is ¼r 2 = ¼L 2 = 4 ¼ 2 = L 2 = 4 ¼ .Thu s E = L 2 4 ¼ dB dt = (0 : 500 m) 2 4 ¼ (10 : 0 £ 10 ¡ 3 T = s) = 1 : 989 £ 10 ¡ 4 V : The rate of thermal energy generation is P = E 2 R = (1 : 989 £ 10 ¡ 4 V) 2 1 : 076 £ 10 ¡ 2 ­ =3 : 68 £ 10 ¡ 6 W : 37 (a) The field point is inside the solenoid, so Eq. 30–25 applies. The magnitude of the induced electric field is
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This note was uploaded on 01/19/2009 for the course PHYS 171.102 taught by Professor Heckman during the Fall '98 term at Johns Hopkins.

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ssm_ch30 - Chapter 30 5 The magnitude of the magnetic field...

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