Question

A wire formed in the shape of a right triangle with base Lab = 38 cm and height Lbc = 63 cm carries current I = 245 mA as shown in Position 1. The wire is located in a region containing a constant magnetic field B = 0.8 T aligned with the positive z-axis. OOO OO OO BIT oooooo Position 1 C Position 2 1) What is Fac,x, the x-component of the force on the segment of the wire that connects points a and c in Position 1? N Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 2) What is Facy, the y-component of the force on the segment of the wire that connects points a and c in Position 1? N Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 3) The wire is now rotated 180° about the y-axis to Position 2, as shown. What is AU12, the change in potential energy of the wire? Note that AU12 is a signed number. AU 12 is positive if the potential energy in Position 2 is higher than the potential energy in Position 1. J Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 4) The wire is now rotated back 90° about the y-axis towards position 1. If the wire is released from this position, how would it move? It would rotate towards Position 1 It would rotate towards Position 2 It would remain stationary Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 3) The wire is now returned to Position 1 and then rotated 180° about the x-axis to Position 3, as shown. What is AU13, the change in potential energy of the wire? If the potential energy increases in going from Position 1 to Position 3, the change in potential energy is positive. olo o a O a Olo O O OVO Position 3 J Submit You currently have 0 submissions for this question. Only 10 submission are allowed.

Answer 1

- Lab - 38cm = 0.38 m, Lbc 2 63 cm = 0.63 m. I = 245 mA = 0.245 A , B=0.8T (a Fac = I Lac B = IB / Lab + Lloc Facx = -Fac sine = - IB Last Loc / Loc (hab tubo Fac,x= - IBL bc u Fagna -(0.245) (0.8) (0.63) Fac, x = -0.12 35 N This is negative, (6) Fac, y = fac cose = IBLab Fagya (0-245)(0.8)(0.38) Facy = 0.07 448 NI (C) UE= - M. B = M B coso - MB cos180" - upe MB AU = UF - Ui= MB- (-MB) = 2 MB AU = 2 (tLab Lbc) B .? AU = I Lab Lbc B AU = (0:245)(0.38) (0.63) (0-8) AU = 0.047J 1

(d) It would rotuve towards position ! of As position I is at lower energy than 2. 's when released loop will move towards lower potential energy. - (4) 43-ll,) B AU = Uz-u, a 40=-(-2 MB) AU= 2 MB = 2 I AU = I Lab Lb GB AU= (0,245 (0.38) (0.63) 20.8] Lau= 0.047 J/