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Solutions Set 1

# Solutions Set 1 - Solutions Problem Set 1 Ch 33 Conceptual...

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Unformatted text preview: Solutions Problem Set # 1 Ch. 33 Conceptual Questions 33.]. (”Thespheteisnoteﬂeendbymem Glueisnotnupetic. (”Mimmmkheemhghummmmeimmfmmmmw “models. Minnow” ammumamomewumcmum 33.2. lieu-mt Hut-aunnmabumdem Minimal-bia- mmnuwuumumwmuumm 33A. mummdummmmm.umhuu www.mmmmormmumwwnummg.mm materiallouoﬂhem 33.3. Yes, if both cylinders are magnets they will repel if one of them is turned 180°. If only one cylinder is a magnet, then they will still attract after one of them is turned around. Ch. 33 Problems 335. Md: mmﬂtﬂisdﬂoﬂm‘uywm Vbuh: Hm Match": 903.6. Solve TheB’m-Sevmln'u help-u [I0 TmMXlw-Io cum-Io wu)ul”‘ ‘1 " (2.0-Io n) [Lo-lo n)‘ I 2.83- lo “ T mm-mmuumwmmmmmudumm bus-no ‘lt. 31.". Ml: mmmamwdummauuwwnmu \W Mahatma)“; I‘M: newMdhmuhmwlammhmmjcammahmm mdhm.kmmmmmmu¢hmuThem-tantalum.» ldmmuddnmiqbwmmm-hdmktdhulhtkmtdwma Mahalﬂimkmuwgﬂmaﬁcﬂhumoﬂkpumw«he “thﬁmﬁﬂmmolmﬂmtﬁchhMWMd MMMMMﬁm)clhu-Mkfuhoﬂhcmadmcmmmm “newMMnummMmh6.AkW.M¢cﬂcﬂluun ‘- uh 33.14. Model: Assume the wires are inﬁnitely long. Visualize: Please refer to Figure EX33.14. Solve: The magnetic field strength at point a is “o M d , out of page) + ( 21rd , into page) top bottom 1 \ _ —7 (—_— ) J'(2"10 Tm/A)(10A)l2.0x10'2m 6.0x10'2mJ _ (4.0 + 2.0 cm => Em = (6.7 x10'5 T, out ofpage) At points b and c, 33.43. Model: Assume that the wires are infinitely long and that the magnetic field is due to currents in both the wires. Visualize: Point 1 is a distance dl away from the two wires and point 2 is a distant d2 away from the two wires. A right triangle with a 75° degree angle is formed by a straight line from point 1 to the intersection and a line from point 1 that is perpendicular to the wire. Likewise, point 2 makes a 15° right triangle. d. = (4.0 cm) sin75" Solve: First we determine the distances dl and d2 of the points from the two wires: 1 = (4.0 cm)sin75° = 3.86 cm = 0.0386 2 = (4.0 cm)sin15° = 1.04 cm = 0.0104 At point 1, the fields from both the wires point up and hence add. The total field is ) (4 x10'7 T m/A)(5.0 A) = — = 5.2 x10—5 0.0386 In In vector form, '= (5.2 x 10'5 T, out of page). Using the right-hand rule at point 2, the fields are in opposite directions but equal in magnitude. So, -I'. 33.53. Model: The magnetic field is that of a current in the wire. Visualize: Please refer to Figure P3353. Solve: As given in Equation 33.6 for a current carrying small segment the Biot-Savart law is For the straight sections, “because both and point along the same line. That is not the case with the curved section over which ‘ where we used '_ = RA is R d!‘ for the small arc length As. Integrating to obtain the total magnetic field at the center of the semicircle, ...
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Solutions Set 1 - Solutions Problem Set 1 Ch 33 Conceptual...

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