21_InstSolManual_PDF_Part1

21_InstSolManual_PDF_Part1 - ELECTROMAGNETIC INDUCTION 21...

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Unformatted text preview: ELECTROMAGNETIC INDUCTION 21 10. D 11. C 12. A 13. D 14. C 15. A Answers to Multiple-Choice Problems 1. A 2. B 3. C 4. B 5. B 6. A 7. C 8. A 9. C Solutions to Problems 21.1. Set Up: FB 5 BA cos f. The normal to the loop is in the z direction. The loop has area Solve: (a) f 5 0°. FB 5 BA 5 1 0.230 T 2 1 1.33 3 1022 m2 2 5 3.06 3 1023 Wb. (b) f 5 53.1°. FB 5 BA cos 53.1° 5 1 3.06 3 103 Wb 2 cos 53.1° 5 1.84 3 1023 Wb. (c) f 5 90°. FB 5 0. Reflect: The flux is a maximum when the magnetic field is perpendicular to the plane of the loop and the flux is zero when the magnetic field is parallel to the plane of the loop. A 5 pr 2 5 p 1 6.50 3 1022 m 2 2 5 1.33 3 1022 m2. 21.2. Set Up: FB 5 BA cos f, where f is the angle between B and the normal to the surface. Solve: (a) The normal to the surface is parallel to the x axis. f 5 90° and FB 5 0. (b) The normal to the surface is parallel to the z axis and f 5 0°. FB 5 1 0.128 T 2 1 0.300 m 2 2 5 0.0115 Wb. The flux is into the enclosed volume. S (c) The normal to the surface and the direction of B are shown in Figure 21.2. 40.0 cm tan f 5 and f 5 53.1°. 30.0 cm FB 5 BA cos f 5 1 0.128 T 2 1 0.300 m 2 1 0.500 m 2 cos 53.1° 5 0.0115 Wb. The flux is out of the shaded volume. (d) The net flux is zero, since the magnitude of the flux into the volume equals the magnitude of the flux out of the volume. 30.0 cm f x S 40.0 cm f normal B y Figure 21.2 21-1 ...
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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