This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 32 1. (a) The flux through the top is +(0.30 T) π r 2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the sides must be negative and exactly cancel the total of the previously mentioned fluxes. Thus (in magnitude) the flux though the sides is 1.1 mWb. (b) The fact that it is negative means it is inward. 2. We use n Bn = ∑ = 1 6 Φ to obtain Φ Φ B n Bn 6 1 5 1 2 3 4 5 3 = =  ++= + = ∑ Wb Wb Wb Wb Wb Wb . b g 3. (a) We use Gauss’ law for magnetism: z ⋅ = B dA . Now, z ⋅ = + + B dA C Φ Φ Φ 1 2 , where Φ 1 is the magnetic flux through the first end mentioned, Φ 2 is the magnetic flux through the second end mentioned, and Φ C is the magnetic flux through the curved surface. Over the first end the magnetic field is inward, so the flux is Φ 1 = –25.0 μ Wb. Over the second end the magnetic field is uniform, normal to the surface, and outward, so the flux is Φ 2 = AB = π r 2 B , where A is the area of the end and r is the radius of the cylinder. It value is Φ 2 2 3 5 0120 160 10 7 24 10 72 4 = × = + × = +π . . . . . m T Wb Wb b gc h Since the three fluxes must sum to zero, Φ Φ Φ C = == 1 2 250 72 4 47 4 . . . . Wb Wb Wb Thus, the magnitude is   47.4 Wb. C Φ = (b) The minus sign in c Φ indicates that the flux is inward through the curved surface. 4. From Gauss’ law for magnetism, the flux through S 1 is equal to that through S 2 , the portion of the xz plane that lies within the cylinder. Here the normal direction of S 2 is + y . Therefore, 1283 CHAPTER 32 ( 29 ( 29 ( 29 ( 29 1 2 left 1 2 2 2 2 ln3 . r r r B B r r r i S S B x L dx B x L dx L dx r x iL μΦ = Φ = = = π= π ∫ ∫ ∫ 5. We use the result of part (b) in Sample Problem 321: ( 29 2 0 0 for 2 R dE B r R r dt μ ε = ≥ to solve for dE / dt : ( 29 ( 29 ( 29 ( 29 ( 29 2 2 7 3 13 2 2 12 3 C 0 0 N m 2 2.0 10 T 6.0 10 m 2 V 2.4 10 . m s 4 T m A 8.85 10 3.0 10 m dE Br dt R7⋅ × × = = = × ⋅ π×10 ⋅ × × 6. From Sample Problem 321 we know that B ∝ r for r ≤ R and B ∝ r –1 for r ≥ R . So the maximum value of B occurs at r = R , and there are two possible values of r at which the magnetic field is 75% of B max . We denote these two values as r 1 and r 2 , where r 1 < R and r 2 > R . (a) Inside the capacitor, 0.75 B max / B max = r 1 / R , or r 1 = 0.75 R = 0.75 (40 mm)=30 mm. (b) Outside the capacitor, 0.75 B max / B max = ( r 2 / R ) –1 , or r 2 = R /0.75 = 4 R /3 = (4/3)(40 mm) = 53 mm. . (c) From Eqs. 3215 and 3217, B i R i R d max . . . = = = × ⋅ = ×7 5 2 2 4 10 60 2 0 040 30 10 π π π π T m A A m T....
View
Full
Document
This note was uploaded on 12/17/2009 for the course PHY 108 taught by Professor Iashvili during the Spring '08 term at SUNY Buffalo.
 Spring '08
 IASHVILI

Click to edit the document details