1) 2) 3) 4) 5) A wire formed in the shape of a right triangle with base Lab = 39
ID: 2142363 • Letter: 1
Question
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A wire formed in the shape of a right triangle with base Lab = 39 cm and height Lbc = 64 cm carries current I = 559 mA as shown in Position 1. The wire is located in a region containing a constant magnetic field B = 1.34 T aligned with the positive z-axis. 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? What is Fac,y, the y-component of the force on the segment of the wire that connects points a and c in Position 1? The wire is now rotated 180o about the y-axis to Position 2, as shown. What is ?U12, the change in potential energy of the wire? Note that ?U12 is a signed number. ?U12 is positive if the potential energy in Position 2 is higher than the potential energy in Position 1. The wire is now rotated back 90o 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 The wire is now returned to Position 1 and then rotated 180o about the x-axis to Position 3, as shown. What is ?U13, 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.Explanation / Answer
magnetic field causes zero net force on current carrying loop.
Fab+Fbc+Fca=0
Fca=-(Fab+Fbc)
Fab= i(L X B)
Fab=Fx=0.559(0.39X1.39)-j =-0.3033 N j
Fbc=Fy=0.559(0.64X1.39) i = 0.4972 Ni
c. pot energyU =-u.B = -(iA)B
Uinitial =iABcos 0 =iAB
Ufinal=iABcos(180) = -iAB
pot energy = -iAB-iAB =-2iAB
=-2*0.559 *0.5* 0.39*0.64*1.34
=-0.187 J
d. U = U1=U2 = -0.187 J
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