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## PHYSICS II

by: Donato Hoeger Jr.

57

0

11

# PHYSICS II PHGN 200

Donato Hoeger Jr.
GPA 3.94

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
11
WORDS
KARMA
25 ?

## Popular in Engineering Physics

This 11 page Class Notes was uploaded by Donato Hoeger Jr. on Monday October 5, 2015. The Class Notes belongs to PHGN 200 at Colorado School of Mines taught by Staff in Fall. Since its upload, it has received 57 views. For similar materials see /class/219613/phgn-200-colorado-school-of-mines in Engineering Physics at Colorado School of Mines.

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Date Created: 10/05/15
PHGNQOO All Sections Recitation 6 March 277 2007 1 Those damn in nite and semi in nite half in nite wires a Consider a semi in nite wire carrying current I7 see Fig 1 Find the magnetic eld7 g in the rst quadrant7 ie7 z 2 0y gt 02 0 You may nd 4d 1 4 1 o Ciw2n2l32 w Nmcfnz useful Figure 1 A semi in nite wire from 0 to 00 carrying current I is shown in blue Solution The magnetic eld is given by no Idl7gtltF g 3 7 2 E r where F Ff 7775 and dishould really be dFs Against all of my wishes7 we will NOT use d7S because your book and lecture notes call it di and its important to have uniform notation Recall that 71 is the source variable and Ff is the eld variable Computing Ff yields Ff 963 ya 3 since we want to know the magnetic eld anywhere in the rst quadrant7 ie7 z 2 0y gt 02 0 Computing 71 yields F5 5 67 since the wire that produces the eld lies on the s axis why is there an s subscript on x in 47 Taking the difference between 3 and 4 yields F 7 sayj 5 PHGNQOO All Sections Recitation 6 March 277 2007 dZis computed in the usual way 7 ie7 Z 57 compare this to d2 7 i dxS 7 d2 dzsz 6 Substituting 5 and 6 into 2 yields E i 01 dzsk 7 y 4 0 96 7 969 112 Finally7 evaluating the integral in 7 Via 1 yields a 01 1 A 3 7 k s 4 y y x2y2 b Consider an in nite wire carrying current I7 see Fig 2 Find the magnetic eld7 E in the rst quadrant7 ie7 z 2 0y gt 02 0 You may nd A w 2 A C 7 w 772 d n 9 useful Figure 2 An in nite wire from foo to 00 carrying current I is shown in blue Page 2 PHGNQOO March 277 2007 All Sections Recitation 6 Solution The magnetic eld due to an in nite wire is given by 7 but with different limits of 10 dxsic integration why7 Thus7 a M01 y 4 we ltz 7 m2 W2 Finally7 evaluating the integral in 10 via 9 yields a 0 A B L k 11 27Ty Where does the magnetic eld due to a semi in nite wire equal one half the magnetic eld of an in nite WiFe7 187 Bsemiiin nite in nite2 Solution The magnetic eld due to a semi in nite wire is equal to the magnetic eld of an in nite wire divided by two when a Bi nie Bsemiiin nite 0 1 A 01 A M 7 k M k 47f y y z y 47w x 0 yxW x 07 ie7 on the y axis What is so special about the y axis Page 3 PHGNQOO All Sections Recitation 6 March 277 2007 2 A blue wire carrying current I OP3 is wound evenly on a torus of rectangular cross section There are N turns of the blue wire in all A red wire is thrown over the torus and is connected to a resistor7 R7 see Fig 3 Find the magnitude and direction clockwise or counterclockwise of the current in the red wire7 Iredwire Figure 3 A blue wire carrying current I OP3 is wound evenly on a torus of rectangular cross section7 with inner radius 71 and outer radius Q There are N turns of the blue wire in all A red wire is thrown over the torus and is connected to a resistor7 R Solution The magnetic eld produced by the blue wire can be found Via Arnpere7s Law7 fg 39 Molenclm where 5le is the current enclosed by the Arnperian loop We choose the Arnperian loop to be a circle of radius r centered at the origin why7 see Fig 4 Page 4 PHGNQOO All Sections Recitation 6 March 27 2007 f 7 1 7 2 Figure 4 Arnperian loop of radius r laying in the torus77 is shown By symmetry g is parallel to dig thus 12 yields f Bde Mound 13 By symmetry magnetic eld B is constant on the Arnperian loop thus 13 yields B f MoIencl B27177 MoIencl B 0131ch 27139 where and NI Thus the magnetic eld produced by the blue wire is given by 0N B L 14 27139 The magnetic ux through the area enclosed by the red Wire is given by ltIgtB EM 72 B wdr 15 7 1 Page 5 PHGNQOO All Sections Recitation 6 March 277 2007 Why are the limits of integration from T1 to r2 if we are calculating the magnetic ux through the area enclosed by the red Wire Substituting 14 into 15 and integrating yields ptwa r2 lt1gt 71 7 16 B 27139 n n The magnitude of the induced emf is given by 17 Substituting 16 into 17 yields 2 T1 on ln If 18 27139 71 Substituting 18 into Ohm7s law yields ON 10 red Wire M w In tzv 27TR 71 where red wire ows in the clockwise direction why Page 6 PHGNQOO All Sections Recitation 6 March 277 2007 3 Consider a conducting rod sitting on the top of an incline The top of the incline is made from pair of frictionless conducting rails There is a resistor7 R7 that connects the two rails7 and a constant magnetic eld directed vertically upwards with a magnitude B07 see Fig 5 The separation distance between the two frictionless conducting rails is L If at time t 07 the rod is released from rest7 nd the velocity of the rod as a function of time Top View Side View 6 C Figure 5 Top and side Views of the conducting incline are shown Notice that the pair of frictionless conducting rails7 the conducting rod and the resistor form a complete circuit Page 7 PHGN200 All Sections Recitation 6 March 277 2007 Solution First7 we write the magnetic eld in terms of the given coordinate systern7 see Fig 57 B0 7 sin6icos6j 19 The magnetic ux through the area enclosed by the resistor7 rails and the rod is given by lt1gtB 33M de j BoLcos0 dx mo BoLcos0z 7 0 20 The magnitude of the induced ernf is given by dltlgtB 151 dt 21 Substituting 20 into 21 and identifying as velocity 1 yields 8 BoLcost9v 22 Substituting 22 into Ohrn7s law yields I B LS6v 23 where I is the current in the rod owing in the positive 1 direction why The magnetic force on the rod is given by g aw L A a Idzk gtlt B0 isintcost9j7 used 19 for B 0 7B01Lcos0 sin6j 24 where I is given by 23 Writing the sum of all forces in the 3 direction yields BELZ cos2 6 6 7 d1 25 R 1 my sin 7 m dt We must solve 25 for 1 so we rewrite 25 in the more manageable forrn d1 7 b 7 26 cm df Page 8 PHGN200 All Sections Recitation 6 March 277 2007 2 2 2 where a W m comments d1 7 dt a1 7 b 7t0 lncw7b a 016 av 7b 91 1 7 6quot 1 Finally7 substituting a and b into 27 yields ng sint9 mt BZL2 cos26 B2L2c0529 17 6 O Rm t It is interesting to plot 28 for some parameter values7 see Fig 6 Figure 6 The velocity ofthe rod is shown for the following parameter values 9 and b gsin t9 Now7 it is a trivial matter to nd 1 so we proceed without any Notice that the graph is at for roughly t gt 5 Can you explain this at region physically 27 28 132 29 i 1andgs1n6 2 Page 9 PHGNQOO All Sections Recitation 2 I 139 Figure 1 For the above electric dipole p 2amp1 February 6 2007 1 Dipole in a 2 D world because the 3 D world is too damn hard a Find the electric potential anywhere in the xy plane see Fig 1 7 i 1 7 1 Ans V 7 4m W W b Find the sicornponent of the electric eld from the expression you have obtained for the electric potential in part a ls your result consistent with what you obtained during the rst recitation 7 i m 7 m Ans Em 4m lbw70213 z2ltyzgt2132l D Given a curve y fx where 1 3 x 3 2 in the xyiplane nd the electric potential at some eld point ab Assume that the curve has a linear charge density given by A m 2 Ans V i fm WV 1 f zl dx whereA g A sphere with radius R and volume charge density p porn where p0 gt 0 and n 2 0 is centered at the origin of a coordinate system 00 a Find the electric eld inside the sphere ie r lt R OTn1 Ans Em rltR b Find the electric eld outside the sphere ie r gt R i poRn3 Ans Eim rgtR c Find the electric potential difference between the center and surface of the sphere ie Vcenter 7 7 7 AnS Vicenter surface 7 n2n3 o Vrsurface 4 A cylinder with radius R length Z and volume charge density p porn where p0 gt 0 and n 2 0 is shown in Fig PHGN200 All Sections Recitation 2 February 6 2007 l l w f Figure 2 A cylinder with radius R and length Z where ltlt 1 The gure is not to scale a Find the electric eld inside the cylinder ie r lt R Ans E r lt R b Find the electric eld outside the cylinder ie r gt R i 0Rn2 Ans E i m r gt R c Find the electric potential difference between the center and the surface of the cylinder ie Vcenm 7 lsulface poRn2 Ans Vcenter 7 V u ace m Page 2

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