1. What is ?, the angular frequency of rotation? 2. What is I max , the magnitud
ID: 2295693 • Letter: 1
Question
1. What is ?, the angular frequency of rotation?
2. What is Imax, the magnitude of the maximum induced current in the loop?
3)
At time t = 0, the wire is positioned as shown. What is the magnitude of the magnetic flux ?1 at time t = t1 = 0.4125 s?
4)
What is I1, the induced current in the loop at time t = 0.4125 s? I1 is defined to be positive if it flows in the negative y-direction in the segment of length h.
5)
Which of the folowing statements about ?o, the magnitude of the flux through the loop at time t = to = 0.275 s, and Io, the magnitude of the current through the loop at time t = to = 0.275 s, is true? ?max and Imax are defined to be the maximum values these quantites achieve during the complete rotation.
6)
Suppose the frequency of rotation is now doubled. How do ?max, the maximum value of the flux through the loop, and Imax, the maximum value of the induced current in the loop change?
Explanation / Answer
Similar Problem but with different value ...Hope this will help!!!base of the triangle b= 38 cm heigth of the triangle h = 62 cm area of the triangle A = (1/2) ( 38 *62 10-4 m 2 ) = 0.1178 m 2 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 Similar Problem but with different value ...Hope this will help!!!
base of the triangle b= 38 cm heigth of the triangle h = 62 cm area of the triangle A = (1/2) ( 38 *62 10-4 m 2 ) = 0.1178 m 2 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
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