846_PartUniversity Physics Solution

846_PartUniversity Physics Solution - 28-24 Chapter 28...

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28-24 Chapter 28 28.77. IDENTIFY: Use the current density J to find dI through a concentric ring and integrate over the appropriate cross section to find the current through that cross section. Then use Ampere’s law to find B ! at the specified distance from the center of the wire. (a) SET UP: Divide the cross section of the cylinder into thin concentric rings of radius r and width dr , as shown in Figure 28.77a. The current through each ring is 2 . dI J dA J rdr π = = Figure 28.77a EXECUTE: () 22 00 24 1/ 2 . II dI r a r dr r a r dr aa ⎡⎤ =− = ⎣⎦ The total current I is obtained by integrating dI over the cross section 2 42 0 0 44 1 1 / , a Id I r a r d r rr a I ⎛⎞ == = = ⎜⎟ ⎢⎥ ⎝⎠ ∫∫ as was to be shown. (b) SET UP: Apply Ampere’s law to a path that is a circle of radius r > a , as shown in Figure 28.77b. (2 ) dBr Bl = ! ! o encl 0 = (the path encloses the entire cylinder) Figure 28.77b EXECUTE: 0e n c l dI μ = ! ! o says ) B rI πμ = and . 2 I B r = (c) SET UP: Divide the cross section of the cylinder into concentric rings of radius r and width , dr as was done in part (a). See Figure 28.77c. The current dI through each ring is 2 0 2 4 1 Ir dI r dr Figure 28.77c EXECUTE: The current I is obtained by integrating dI from 0 to : r = = 2 2 11 0 0 / r r I I r d r r r a a ′′ = 2 2 4 (/ 2 / 4) 2 r r r a a = (d) SET UP: Apply Ampere’s law to a path that is a circle of radius r < a , as shown in Figure 28.77d. ) = ! ! o 0 encl 2 r I (from part (c)) Figure 28.77d EXECUTE: 2 0 n c l 0 2 says (2 ) (2 / ) B r r a a μπ ⋅= = ! ! o and 2 / ) 2 B ra a EVALUATE: Result in part (b) evaluated at : . 2 I raB a Result in part (d) evaluated at 2 : (2 / ) . Ia I μμ = The two results, one for r > a and the other for r < a , agree at r = a . 28.78. IDENTIFY: Apply Ampere&s law to a circle of radius r . SET UP: The current within a radius r is = J A ! ! , where the integration is over a disk of radius r .
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Sources of Magnetic Field 28-25 EXECUTE: (a) () / / / / 0 0 0 22 2 1 . a a ra a b Id e r d r d b e d r π b δ eb δ e r δδ δ θπ π −− ⎛⎞ =⋅= = = = ⎜⎟ ⎝⎠ ∫∫ JA ! ! (0.050/ 0.025) 0 2 (600 A/m) (0.025 m) (1 ) 81.5 A. I π e =− = (b) For 0e n c l 00 , 2 d B π rI I μμ ≥⋅ = == Bl ! ! o and . 2 I B r μ = (c) For , / / / 0 0 2 2 . r r b I r d e rdrd be d r π b δ e r ′′ = = ! ! / / / / () 2 ( ) 2 ( 1 ) a a r Ir π b δ ee π b δ = and / 0 / (1 ) . ) r a e Ir I e = (d) For , / n c l / ) ()2 ) r a e dB r π I e ⋅= = = ! ! o and / / ) . 2( 1 ) r a Ie B re = (e) At 0.025 m, r 4 0 / 0.050/ 0.025 ( 1) (81.5 A) ( 1.75 10 T 2 ( 2 (0.025 m) ( a μ e B πδ e π e = × . At 0.050 m, / 4 0 / )( 8 1 . 5 A ) 3.26 10 T. 1 )2 ( 0 . 0 5 0 m ) a a B ae ππ = × At 2 0.100 m, 4 0 (81.5 A) 1.63 10 T. 2 2 (0.100 m) I B r = × EVALUATE: At points outside the cylinder, the magnetic field is the same as that due to a long wire running along the axis of the cylinder. 28.79. IDENTIFY: Evaluate the integral as specified in the problem. SET UP: Eq.(28.15) says 2 0 3 / 2 2( ) x Ia B xa = + .
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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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846_PartUniversity Physics Solution - 28-24 Chapter 28...

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