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# PHYSICS II PHGN 200

Colorado School of Mines

GPA 3.94

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This 13 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 38 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 5 March 6 2007 1 Find the torque around the ziaxis F that the blue wire carrying a current I see Fig 1 experiences in the presence of a uniform magnetic eld B B0 j I 9 gt Figure 1 A wire carrying current I is shown in blue L is the length of the diagonal77 part of the wire Solution The differential form of the torque equation is given by review Physics I d F gtlt dF 1 where F is the lever arm77 and dF is the differential force that the wire experiences dF is the magnetic force it7s given by d Idix E 2 The horizontal parts of the wire do not contribute to the torque because the lever arm77 is zero The vertical part of the wire does not contribute to the torque because it experiences no force dz is anti parallel to E hence IdZgtlt g 0 Thus we only have to worry about the diagonal77 part of the wire ie y tan0 where tant9 is the slope Computing dZyields Zx yj Z tan6j 1 12tan0j dz dz tan0 dxj 3 Substituting 3 into 2 yields dF Badge 1 4 PHGNQOO All Sections Recitation 5 March 6 2007 Before we substitute 4 into 1 we need to know F the lever arrn77 1n The lever arrn77 is the vertical distance between the xiaxis and the wire it is given by Finally substituting 5 and 4 into 1 yields d Botan6xdx i Lcos9 F Botan0 z dzi 0 2 i 7 IBOL s1r120 cost9 239 2 Find the net force on a wire bent into a serni circle shape with radius R and carrying a current I see Fig 2 in the presence of a non uniform magnetic eld B arctant9z k Make sure you evaluate any integrals that may arise 6 Figure 2 A serni circular wire with radius R and current I lies in the z 0 plane Solution The magnetic force on the wire is given by d Idix E 6 Page 2 PHGNQOO All Sections Recitation 5 March 67 2007 where dZis computed in the usual way 7 ie7 Z Rcos0 i Rsin0j 0 d i A A E i 7Rs1nt9 z Rcost9 j dZ iRsin0 d0 Rcost9 d j 7 Substituting 7 into 6 yields d I i Rsin0 d i Rcos6 dej gtlt B V F OWIiRsin6d6 Rcos6d6j gtlt E 8 Now7 we need to substitute E given by g arctant9zl c7 into However7 8 implicitly states that g must be evaluated on the wire semi circle Evaluating B on the wire semi circle yields Finally7 putting 9 into 8 yields 3 Set up an expression for the net force on the wire curve y fx7 where 1 3 x 3 x2 lies in the z 0 plane and carries a current I see Fig 3 Assume that there is a non uniform magnetic eld B Byz in the region of space where the wire lies Figure 3 An arbitrary wire curve7 y fx7 carrying a current I as indicated on the diagram is shown Page 3 PHGNQOO All Sections Recitation 5 March 6 2007 Solution The magnetic force on the wire is given by 1131de B 10 where dZis computed in the usual way ie 7 A A 95 12 f 96 dZd392fx dsj 11 where f denotes a derivative of fx wrt x Evaluating g on the wire curve yields mag72 7yf96720 12 Finally substituting 11 and 12 into 10 yields E E dF dz fs dsj gtlt xy fz 0 11 F m21dx392f dj gtlt xy fz 0 1 50pm light we will learn this later in the course Write an expression for the magnetic eld B that the particle generates in terms of the electric eld E and speed of light 0 4 A particle with charge q is traveling with constant velocity 17 where 1 ltlt 0 and c is the speed of Solution The magnetic eld due to a slow moving charge is approximately here approximately means we are ignoring retardation effects dont worry if you dont know what this means given by a Mo 117 X F B 7 13 M TB 7 and the electric eld is given by a g F E 7 14 47TEO r3 Solving 14 for 7 and substituting the result into 13 yields a 17 gtlt E7 B CZ 15 Page 4 PHGNQOO All Sections Recitation 5 March 67 2007 Equation 15 is a cute equation relating the electric and magnetic elds Later on in the course7 we will learn that electric and magnetic elds are fundamentally related quantities this is largely due to the work of James Clerk Maxwell7 httpenwikipediaorgwikiJames lerkj laxwell 5 Helmholtz coil This problem is NOT on the quiz a Without using any symmetry arguments7 nd the magnetic eld anywhere on the ziaxis I y z 20 plane 6 Z Figure 4 A circular wire with radius R and current I lies in the z 20 plane Solution The magnetic eld is given by no Idl7gtlt F 3 7 16 537 47139 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 z 1 17 since we want to know the magnetic eld anywhere on the ziaxis Computing 71 yields 71 Rcost97Rsint9j2017 18 since the current loop that produces the eld lies in the zoiplane Taking the difference between 17 and 18 yields F iRcos6 7Rsin6j z 720 k 19 Page 5 PHGNQOO All Sections Recitation 5 March 6 2007 dZis computed in the usual way ie l7 Rcost9 i Rsin0j 20 I compare this to 18 d2 1 A E i 7Rs1nt9 z Rcost9 j dZ iRsin0 d0 Rcos0 d0j 20 Substituting 19 and 20 into 16 and integrating the z and y components of the eld recall that fOZW sin0 d0 fOZW cost9 d0 0 yields a IR2 A B k 21 2R2 z 7 202 You could have skipped the calculation of the z and y components of the eld because both of them are zero as can be shown by a symmetry argument However symmetry arguments are dif cult to explain and possibly even more dif cult to understand hence the algebra intensive exercise b Find the magnetic eld due to two circular coils each having a radius R and carrying current I in the same direction Assume the center of the rst circular coil is at 000 and the center of the second circular coil is at 0 0R Furthermore assume that N1 is the number of wire loops in the rst coil and N2 is the number of wire loops in the second coil Solution Using 21 with 20 0 yields the magnetic eld produced by the rst coil NMOIRZ A B1 2R22232 k 22 Using 21 with 20 R yields the magnetic eld produced by the second coil 2 A 32 MW k 23 2R2 z 7 By The total magnetic eld is given by g i N1M01R2 I NZMOIRZ I 7 32 32 2 R2 22 2 R2 z 7 By This arrangement of coils is known as the Helmholtz coil httpenwikipediaorgwikiHelmholtzmoil and produces a surprisingly uniform magnetic eld on the ziaxis see Fig 5 Page 6 PHGNQOO All Sections Recitation 5 March 67 2007 09 7 r r r r r w r r r r r r 77 r r r r r r is r r r r r w r r r r r r r W 08 7 r r r r r w r r r r r r a r r r r r r r r r r r r 77 r r r r r r r if r r r r w r r r r r r w r r r r o5 r r r r r r VJ r r r r r r VJ r r r r r r J r r r r r VJ r r r r r r r v 777777 777777 777777 77777 7777777 v 777777 777777 777777 77777 7777777 v 777777 777777 777777 77777 7777777 v Figure 5 NMOI NW0 1 and R 1 Page 7 PHGNQOO All Sections Recitation 1 January 23 2007 Figure 1 For the above electric dipole p 2amp1 1 Dipole in a 2 D world because the 3 D world is too damn hard a Find the electric eld anywhere in the xy plane see Fig 1 Solution The electric eld due to N point particles is given by N a a 1 r E7 4 whereFF7F 1 47TEO gin37 f 52 Ff is called the eld point and F9 is the z source point We will call the positive charge the rst charge and the negative charge the second charge Thus Ff 965 M F91 if 7792 51 Computing all quantities needed for 1 yields 3741 1274 womb5 ziy 2 T1llFf lll 72ll77f 7792ll wyiaz wyaz a Finally substituting 2 and 3 into 1 yields 1 q MHz75 7 z y 4 F l 7WWzl WW2 D11 PHGNQOO All Sections Recitation 1 January 237 2007 b Evaluate the electric eld found in part a on a circle with radius R see Fig 1 Solution To nd the electric eld on the circle7 we set the eld point on the circle7 ie7 z Rcos0 and y Rsin0 in 47 which yields Rcos6iRsin67 j Rcos06Rsin6Zj 5 q R2 2mm 9 2 E 4m R2 7 2mm 9 2 c Find an approximate expression for the x component of the electric eld on the circle if R lt 1 see 7 if lel lt 1 You can drop Fig 1 Hint Expand the denominator in Taylor series7 1 6 1n6 any terms containing square or higher powers of because if is srnall7 then f is super srnall Solution Frorn 57 we see that the x cornponent of the electric eld on the circle is given by 1 1 J t 6 chos6 E 32 R2 2ZRsin 6 2 m 4m 122 i magma 2 We rewrite the denorninators as follows 1 R2 up magma 2 R2 up magma 2 732 7 R21 2esm6 e 2 If R R 732 2mm e 2 11T E 2esm0 e 2 193 1 32 h 7 l e 7 w ere E R R 732 1353 R B 17E 3mm 3a 11777 i 73 R R 2 R BZsinQ R 3 117 l R l Finally7 substituting the above result into 6 and simplifying yields 3 E1 FfRS sin0 cost97 where p 2amp1 Page 2 PHGNQOO All Sections Recitation 1 January 237 2007 d Find the total charge enclosed by the circle Find the unit norrnal to the circle Solution The total charge enclosed by the circle is given by genclosed 11 i 12 q q 0 We will nd the radially outward unit norrnal for the upper half of the circle you should do bothl The equation for the circle upper half with radius R is given by R xz2y2 The non unit normal is given by review Calculus 7 1 7 63A 7 3s 6y 1 A M 117 7 2 yZ Substituting z Rcos0 and y Rsint into 7 yields 73 cos 6 i sin 6 j 8 It is trivial to con rm that 7 given in 8 is actually 7 we got luckyl 2 Hard integrals made easy a Set up an integral expression for the electric eld at a eld point7 0 07 due to a ring of charge with linear charge density A A0 sin 9 where A0 is a positive constant see Fig 2 y o 9 2 0 R Figure 2 The circle has a linear charge density given by A A0 sin 0 where A0 is a positive constant Page 3 PHGN200 All Sections Recitation 1 January 23 2007 Solution The electric eld is given by where F Ff 7 775 From Fig 2 we have Thus dq is given by where d7quotS is the differential element of eld points hence we use r5 NOT rf also be a scalar One way to nd d7quotS is d d 7Rsint9 2 Rcost9j d7quotS i d7S d0 7 d0 drs R2 cos2 6 sin2 0 d0 dr 5 R d0 drs Rd6 13 Finally substituting 10 11 and 13 into 9 yields a R 2 A 0 i R 9 2 AR 9 E W Cmdgi 32de 14 47T o 0 R2 x3 7 2x0 cos 9 0 R2 x3 7 2Rcos 0 where A A0 sin 0 1 47TEO E7 F dq7 Ff 02 03 F5 Rcost9392 Rsin0j F 0 7 Rcos 97 7 Rsin j r xR2 x3 7 2Rcost9 dq Adrs 12 the source variable we are summing up the source points NOT the Also note that dq is a scalar thus the right hand side of 12 must as follows 71 Rcos6 Rsin0j b Using only a symmetry argument nd the direction of the electric eld Page 4 PHGNQOO All Sections Recitation 1 January 23 2007 Solution Green Blue Red Figure 3 We know that the electric eld points radially outward from a positive charge shown in red and radially inward from a negative charge shown in blue The linear charge density is given by A A0 sin 0 thus the upper half of the circle is positively charged and the lower half of the circle is negatively charged By drawing a few representative charges on the circle we readily see that the electric eld has ONLY a yicornponent Moreover we see that the electric eld points down ie in the negative y direction It is interesting to note that we have effectively calculated the sicornponent of the integral given by 14 just by drawing a picture 3 Given a curve y f where 1 3 x 3 2 in the xyiplane nd the electric eld at some eld point ab Assume that the curve has a linear charge density given by A Page 5 PHGNQOO All Sections Recitation 1 January 237 2007 Solution The electric eld is given by a 1 77 Effgm 15gt 47TEO where F Ff 7 775 First we nd Ff and 71 as follows Ff 1fo 1 779 xi 1 z392 fzj recall that y f Thus7 F1ali1bif1f 16 7 1a121bf9612 17 dq is given by dq xdrS and can be found in the usual rnanner7 j 1 may we ds 7 ds 11f 12 d7quotS 11 f 2d 18 Note that 18 is just an arc length formula from Calculus Finally7 substituting 167 17 and 18 into 15 yields a 1 2 a i 95 2 E 32 1 1 d c Z 4713960m1 a i z2 b fx21 1 2 M 1 UM d A 4776011 a 7 z2 b f952132 j 19 where A Page 6

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