This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1243 Chapter 32 1. We use 6 1 Bn n to obtain 5 6 1 1Wb 2 Wb 3Wb 4 Wb 5Wb 3Wb . B Bn n 2. (a) The flux through the top is +(0.50 T) 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 1.3 mWb. (b) The fact that it is negative means it is inward. 3. (a) We use Gauss law for magnetism: r B d r A 0. Now, r B d r A 1 2 C , where 1 is the magnetic flux through the first end mentioned, 2 is the magnetic flux through the second end mentioned, and C is the magnetic flux through the curved surface. Over the first end the magnetic field is inward, so the flux is 1 = 25.0 Wb. Over the second end the magnetic field is uniform, normal to the surface, and outward, so the flux is 2 = AB = r 2 B , where A is the area of the end and r is the radius of the cylinder. Its value is 2 0.120m 2 1.60 10 3 T 7.24 10 5 Wb 72.4 Wb . Since the three fluxes must sum to zero, C 1 2 250 72 4 47 4 . . . . Wb Wb Wb Thus, the magnitude is   47.4 Wb. C (b) The minus sign in C indicates that the flux is inward through the curved surface. CHAPTER 32 1244 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, 1 2 left 1 ( ) ( ) ( ) 2 ( ) 2 ln 3 . 2 2 r r r B B r r r i iL S S B x L dx B x L dx L dx r x 5. We use the result of part (b) in Sample Problem Magnetic field induced by changing electric field, 2 , 2 R dE B r R r dt to solve for dE / dt : 7 3 13 2 2 12 2 2 3 2 2.0 10 T 6.0 10 m 2 V 2.4 10 . m s 4 T m A 8.85 10 C /N m 3.0 10 m dE Br dt R 6. The integral of the field along the indicated path is, by Eq. 3218 and Eq. 3219, equal to i d enclosed area total area (0.75 A) (4.0 cm)(2.0 cm) 100 cm 2 75 nT m . 7. (a) Inside we have (by Eq. 3216) 2 1 / 2 d B i r R , where 1 0.0200 m, r 0.0300 m, R and the displacement current is given by Eq. 3238 (in SI units): 12 2 2 3 14 (8.85 10 C /N m )(3.00 10 V/m s) 2.66 10 A E d d i dt ....
View
Full
Document
This document was uploaded on 04/22/2011.
 Spring '11

Click to edit the document details