A wire formed in the shape of a right triangle with base L_ab = 12 cm and height

Question

A wire formed in the shape of a right triangle with base L_ab = 12 cm and height L_bc = 68 cm carries current I = 644 mA as shown in Position 1. The wire is located in a region containing a constant magnetic field B = 0.58 T aligned with the positive z-axis. What is F_ac_x, the x-component of the force on the segment of the wire that connects points a and c in Position 1? N What is F_ac_y, the y component of the force on the segment of the wire that connects points a and c in Position 1? N The wire is now rotated 180 degree about the y axis to Position 2, as shown. What is Delta U_12, the change in potential energy of the wire Note that Delta U_12 is a signed number. Delta U_12 is positive if the potential energy in Position 2 is higher than the potential energy in Position 1. J The wire is now rotated back 90 degree 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 180 degree about the x-axis to Position 3, as shown. What is Delta U_13, 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. J

Explanation / Answer

a )

Lab = 12 cm = 0.12 m

Lbc = 68 cm = 0.68 m

B = 0.58 T

i = 644 mA = 0.644 A

Fab = i X Lab X B

= 0.644 X 0.12 X 0.58

= 0.04482 (- j )

Fbc = i X Lbc X B

= 0.644 X 0.68 X 0.58

= 0.2539 ( - i )

b )

Fab= 0.04482 (- j )

c )

U = - i A B

= i A B cos 0

and

i A B cos180

dU1 = i A B ( cos0 - cos180 )

= - 2 i A B

dU1 = -2 X 0.644 X 0.5 X 0.12 X 0.68 X 0.83

dU1 = -0.0436 J

d )

U = - i A B

= i A B cos 0

and

i A B cos 90

dU2 = i A B ( cos0 - cos90 )

= - i A B

dU2 = - 0.644 X 0.5 X 0.12 X 0.68 X 0.83

dU2 = -0.0218 J

e )

dU = dU2 - dU1

= -0.0218 + 0.0436

= 0.0218 J

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