Ch22-h1-solutions

Ch22-h1-solutions - pan(zp695 Ch22-h1 chiu(56565 This...

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pan (zp695) – Ch22-h1 – chiu – (56565) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points In a cylindrical region oF height h =9cm and diameter d , ± B is Found to point downwards and measured to have: B TOP = 1T, B BOTTOM =3T,and B CYL =2T . 9cm 1T 2T 3T 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 ±ux Φ B through a closed surFace is zero, so the measurements must be in- correct. B. This pattern oF magnetic feld 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 CisanincorrectapplicationoFAmpere’slaw. Statement D is incorrect. Even iF the region did enclose halF oF a magnetic dipole, ± ± B · d ± A =0 still holds true, and the feld shown is not a dipole feld pattern. 002 10.0 points The magnetic feld has been measured to be horizontal everywhere along a rectangu- lar path l =25cmlongand h =1cmh igh ,as in the Following fgure. l h 0 . 00013 T 0 . 0001 T 5 × 10 - 5 T Along the bottom, the average magnetic feld B 1 . 00013 T, along the sides the average magnetic feld B 2 . 0001 T, and along the top the average magnetic feld B 3 =5 × 10 - 5 T. Defning 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: 15 . 9155 A. Explanation: Apply Ampere’s Law to the path. Start in the top right corner and go around the path counterclockwise. Note that ² B · d ± ² on the leFt and right sides oF the path is zero since ± B is perpendicular to the path. ± ± B · d ± ² = 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 ±ows through the surFace enclosed by this path in the + z
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pan (zp695) – 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 = 15 . 9155 A . 003 (part 1 of 2) 10.0 points The electric Feld is measured all over a cubical surface, and the pattern of Feld detected is shown in the Fgure below. On the right side of the cube and the bottom of the cube, the electric Feld has the value ± 375 , - 375 , 0 ² N / C .
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Ch22-h1-solutions - pan(zp695 Ch22-h1 chiu(56565 This...

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