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Unformatted text preview: brown (twb493) – Ch22-h1 – chiu – (56565) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In a cylindrical region of height h = 8 cm and diameter d = 8 cm, vector B is found to point downwards and measured to have: B TOP = 1 . 1 T, B BOTTOM = 3 . 3 T, and B CY L = 2 . 2 T. 8 cm 8 cm 1 . 1 T 2 . 2 T 3 . 3 T What can you conclude from these mea- surements? List all that apply, separated by commas. If none apply, enter “none”. A. Gauss’ law for magnetism says net mag- netic flux Φ B through a closed surface is zero, so the measurements must be in- correct. B. This pattern of magnetic field indicates the existence of a magnetic monopole, something that has never been found to exist. C. There is a current through the region that is not uniform. D. The region encloses only half of a mag- netic dipole. Correct answer: A, B. Explanation: Statement A is correct, and is simply a statement of Gauss’ law for magnetism. Statement B is also correct, and is a valida- tion of Gauss’ law for magnetism. Statement C is an incorrect application of Ampere’s law. Statement D is incorrect. Even if the region did enclose half of a magnetic dipole, contintegraldisplay vector B · d vector A = 0 still holds true, and the field shown is not a dipole field pattern. 002 10.0 points The magnetic field has been measured to be horizontal everywhere along a rectangu- lar path l = 25 cm long and h = 1 cm high, as in the following figure. l h . 00013 T 9 × 10- 5 T 7 × 10- 5 T Along the bottom, the average magnetic field B 1 = 0 . 00013 T, along the sides the average magnetic field B 2 = 9 × 10- 5 T, and along the top the average magnetic field B 3 = 7 × 10- 5 T. Defining out of the page (toward you) as positive, determine the mag- nitude and direction of the current through this region. Remember that μ = 4 π × 10- 7 T · m / A . Correct answer: 11 . 9366 A. Explanation: Apply Ampere’s Law to the path. Start in the top right corner and go around the path counterclockwise. Note that integraldisplay B · d vector ℓ on the left and right sides of the path is zero since vector B is perpendicular to the path. contintegraldisplay vector B · d vector ℓ = μ I inside path − B 3 L + B 1 L = μ I inside path ( B 1 − B 3 ) L = μ I inside path Since B 1 > B 3 , then I inside path is positive. We can conclude (using the right-hand rule around the path), that current flows through the surface enclosed by this path in the + z brown (twb493) – Ch22-h1 – chiu – (56565) 2 direction (out of the page). We can also calculate the current....
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This note was uploaded on 12/01/2011 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
- Fall '08