Rotating Triangle in a Magnetic Field A conducting wire formed in the shape of a
ID: 2060264 • Letter: R
Question
Rotating Triangle in a Magnetic FieldA conducting wire formed in the shape of a right triangle with base b = 25 cm and height h = 75 cm and having resistance R = 2.4 O, rotates uniformly around the y-axis in the direction indicated by the arrow (clockwise as viewed from above (loooking down in the negative y-direction)). The triangle makes one complete rotaion in time t = T = 1.5 seconds. A constant magnetic field B = 1.1 T pointing in the positive z-direction (out of the screen) exists in the region where the wire is rotating.
1)
What is ?, the angular frequency of rotation?
radians/second
Your submissions:
4.18879
Computed value:4.18879SubmittedTuesday, March 13 at 5:17 PM
Feedback: Correct
2)
What is Imax, the magnitude of the maximum induced current in the loop?
A
3)
At time t = 0, the wire is positioned as shown. What is the magnitude of the magnetic flux F1 at time t = t1 = 0.5625 s?
T-m2
4)
What is I1, the induced current in the loop at time t = 0.5625 s? I1 is defined to be positive if it flows in the negative y-direction in the segment of length h.
A
5)
Which of the folowing statements about Fo, the magnitude of the flux through the loop at time t = to = 0.375 s, and Io, the magnitude of the current through the loop at time t = to = 0.375 s, is true? Fmax and Imax are defined to be the maximum values these quantites achieve during the complete rotation.
Fo = 0 and Io = 0
Fo = 0 and Io = Imax
Fo = Fmax and Io = 0
Fo = Fmax and Io = Imax
6)
Suppose the frequency of rotation is now doubled. How do Fmax, the maximum value of the flux through the loop, and Imax, the maximum value of the induced current in the loop change?
Fmax and Imax both double
Fmax doubles and Imax remains the same
Fmax remains the same and Imax doubles
Both Fmax and Imax remain the same
7)Below is some space to write notes on this problem
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Explanation / Answer
Given base of the triangle b= 38 cm heigth of the triangle h = 62 cm area of the triangle A = (1/2) ( 38 *62 10-4m2) = 0.1178 m2 resistance of the triangle R = 1.3 ? angular velocity of triangel = 2? / 1.6 = 3.9269 rad / sec magnitue of magnetic field B = 1.9 T maximum emf induced in the loop ? = NBA ? = (1) ( 1.9 ) ( 0.1178 ) ( 3.9269 ) = 0.8789 V maximum current in the triangle I = ? / R = 0.8789 / 1.3 = 0.6760 A maximu flux at time t = 0.6 s ?B = BA cos ?t = ( 1.9)(0.1178 ) cos (3.9269*0.6 ) = 0.2236 Wb ______________________________________ emf in duced in the loop at t = 0.6 s ?' =N BA ? sin ?t = 0.8789 sin ( 3.9269*0.6 ) = 0.0361 V induced current at t = 0.6 s I ' = ? ' / R = 0.0361 / 1.3 = 0.02779 A I'm sorry this is someone elses work but It's good to share the help aroundRelated Questions
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