Right Triangular Current Loop 1 2 3 1 5 6 A wire formed in the shape of a right
ID: 1789200 • Letter: R
Question
Right Triangular Current Loop
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2
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A wire formed in the shape of a right triangle with base Lab = 24 cm and height Lbc = 65 cm carries current I = 429 mA as shown in Position 1. The wire is located in a region containing a constant magnetic field B = 0.65 T aligned with the positive z-axis.
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
Your submissions:
-0.18
Computed value:
-0.18
Submitted:
Wednesday, October 25 at 1:14 PM
Feedback:
Your answer has been judged correct; the exact answer is: -0.1812525.
2)
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?
N
Your submissions:
0.067
Computed value:
0.067
Submitted:
Wednesday, October 25 at 1:14 PM
Feedback:
Your answer has been judged correct; the exact answer is: 0.066924.
3)
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.
J
4)
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
5)
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.
Position 1 Position 2Explanation / Answer
a)
Fac,x = -Fac*sin(theta) = -IBLbc
= -(0.429)(0.65)(0.65)
= -0.181 N
b)
Fac,y = Fac*cos(theta) = IBLab
= (0.429)(0.65)(0.24)
= 0.067 N
c)
dU = I*Lab*Lbc*B
= (0.429)(0.65)(0.24)(0.65)
= 0.0435 J
d)
Rotate towards Position 1.
e)
dU = I*Lab*Lbc*B
= (0.429)(0.65)(0.24)(0.65)
= 0.0435 J
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