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22-1 - brown(twb493 Ch22-h1 chiu(56565 This print-out...

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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 10.0points 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 CYL = 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 10.0points 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 0 . 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 0 = 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 on the left and right sides of the path is zero since vector B is perpendicular to the path. contintegraldisplay vector B · d vector = μ 0 I inside path B 3 L + B 1 L = μ 0 I inside path ( B 1 B 3 ) L = μ 0 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
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brown (twb493) – Ch22-h1 – chiu – (56565) 2 direction (out of the page). We can also calculate the current. I inside path = ( B 1 B 3 ) L μ 0 = 11 . 9366 A . 003(part1of2)10.0points The electric field is measured all over a cubical surface, and the pattern of field detected is shown in the figure below. On the right side of the cube and the bottom of the cube, the electric field has the value ( 390 , 390 , 0 ) N / C .
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