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Physics Assignment

Physics Assignment - Phys 122 Assignment 3 Due March 03...

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Unformatted text preview: Phys 122 Assignment 3 Due - March 03 Name: Print full name Show all work. Include diagrams whenever applicable. Place your solutions in the space below each ' problem. Highlight or box the final answers. It makes it much easier to mark. All assignment pages have to be stapled together WITH A STAPLER. Include units with your final answers. 1. An electron, having a speed of 2.5 x 107 m/s is projected into a uniform magnetic field B = 0.01 T ( in x- direction), with its velocity making an angle of 80° with E . (see the diagram below) I — a 1 . of 475mm" -. memo "4/5 t)” = Um) 90° : 0.;an :07 “‘/5 pi X ciaelcusa After entering the magnetic field the electron will move along the helical trajectory. (remember the demonstration in the phys. lab) 21) Sketch this trajectory in the diagram above and indicate clearly whether the electron would move clockwise or counter—clockwise as seen when looking along the +x direction [2 marks] b) Calculate the radius of the circular Component of the trajectory. [3 marks] mlflzi B R , - R, mu . may wax/0’ .. 0.0mm- ipb (mono-W) [0.0mm : “MM c) Calculate the period, T, of one revolution. [2 marks] film = 5.5594045 - ’—-—'.”_/ 1".1. d) Calculate the pitch (the distance between the “coils” of the trajectory. [2 marks] 5 " 7C7 sac/55", x= (Prick r {E'Ti‘01361/O )[asguc u , = I, 65 am 1 2. Two long wires shown below carry currents of 5.0 A in opposite directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic field at: , a. 7 ya (2 1‘ 3,2 I r _ B, ? ga'EESLOOAzP .7 g = 43.: .2 i l 01 _ io—m.0cm——ff-10.dcmféfi-10.0cm-fi‘g % 2 I t, - Z a) Midpoint between the ires. [2 marks] Wk 78, M {mm} {wa 2 _ l 52; — ~5 é? : {57' + is: I g x fl—Ll : “fix '0 :F (5.00) : Q.mx/0 T— ” XTTQ 1T ,‘ (ammo-L) m 4L4: b) At point P1 (10.0 cm to the right of the wire on the right) [3 marks] r686 . l5 t :5 -7391: _4:L : 927x10? Map—15:00 («9‘ P‘ 51 (9 ’27 ’22 Q: .271“ am aw i «=7 -6... V abfoox/o / more Me = 39, t W c) At point P2 (20.0 cm to the left of the wire on the left) [3 marks] is“ = {23, _ T52 : t 97116;; #300) Q]; Q Q; 02” RI Ra m \Orza) 0.300, —6 E C X T (Quit: xf +18%» at WE ‘ A conductor is suspended by two flexible wires (as shown below) has a mass per unit length of 0.040 kg/m. What current must exist in the conductor for the tension in the supporting wires to be zero when the magnetic field, B is 3.60 T into the page? Draw a RED. and show the required direction of the current in the diagram. [4 marks] l I I: 32’ (36 (/. 0 C) 0. /07%}_ H II 4. Two long, parallel conductors carry currents 1] =3 .00 A and 1: =3.00 A, both directed into the page as in the picture below. Determine the resultant magnetic field, E , ( in components) at point P. Sketch a diagram showing El and 132 . Clearly show your coordinate system (x,y). _, [6 marks] ’1 ’ \\ 5.00 cm \ I [I ‘JIP 15.0 cm 't / l ,’ i ,’ 12.0 cm . 3 / First, observe that (5.00 cm)2 + (12 .0 cm)2 = (13.0 emf. Thus, the G ‘1 ' triangle shown in dashed lines is a right triangle giving , _, 12.0 cm) = = .4° , = . °— = . ° a sm[13.ocm 67 and}? 900 a 226 At point P, the field due to wire 1 is _!_1L11__(41rx10'7 T-m/A)(3.oo A) _ Bl ‘ 2m “ 2::(5.00x10‘z m) _ 12'0 'dr and it is directed from P toward wire 2, or to the left and at 67.4° below the horizontal. The field due to wire 2 has magnitude ~&_(4zx10‘7 T-m/A)(3.oo A) _ B’ ‘ 2m, ' 2::(12.00><10-2 m) ‘5'00 yr and at P is directed away from wire 1 or to the right and at 22.6° below the horizontal. ,1 Thus, 3,, = 431 cos67.4° = —4.62 ,tr 3], = —3, sin 67.4° = —11.1 ,uT BB = B,cos22.6° = +4.62 pr B2y = — 2 sir122.6° = —1.92;zr and B, =le+Bh =0,while 3, =31, +132, =—13.o;tr. The resultant field at P is E: (—9.61,"'~'0)/IT ’73:; (“1.62 I—I.q2)/,r .9 EM: (0, ‘l5.0)/47— in the picture below, the current in the long wire is 11 = 5.00 A and the wire lies in the plane of the rectangular loop, which carries current I =10.0 A. The dimensions are: c = 0.100 m. a = 0.150 m, and 1 =0.45 111. Find the magnitude and the direction of the net force on the loop exerted by the magnetic field due .to the straight wire. [6 marks] 7 .2 £11,,- _,—) 47 J) E :_//0I,[2, L yo 1711 C 5 mp“) JZITC W( Max) 7\ F : /Z/ojvli.2 _’__ , J— t H) V W (9fo 0+0" 0 X F : (Wuo‘ilzgooymo (a 950) L __I'__ M; 2“, ‘6225‘0 0.!00 > ...
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