course FxP̘ٜhOassignment #010
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18:22:33 Query Principles and General Physics 21.04. A circular loop of diameter 9.6 cm in a 1.10 T field perpendicular to the plane of the loop; loop is removed in .15 s. What is the induced EMF?
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RESPONSE --> Start with magnetic flux=(B)(A), where B=strength of magnetic field and A=Area of the circular loop. Diameter of 9.6 cm=.096m, .960/2=.048m=radius. Area of the loop=pi(.048m^2)=.0072^2m Thus, (.0072)(1.10)=.0079 Volts. The induced EMF is the calculated voltage above/time in seconds: .0079V/.15 sec=5.3E-2.
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18:24:28 The average induced emf is the average rate of change of the magnetic flux with respect to clock time. The initial magnetic flux through this loop is {}{}flux = magnetic field * area = 1.10 T * (pi * .048 m)^2 = .00796 T m^2.{}{}The flux is reduced to 0 when the loop is removed from the field, so the change in flux has magnitude .0080 T m^2. The rate of change of
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RESPONSE --> ok
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18:24:37 flux is therefore .0080 T m^2 / (.15 sec) = .053 T m^2 / sec = .053 volts.
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RESPONSE --> ok
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18:48:49 query gen problem 21.23 320-loop square coil 21 cm on a side, .65 T mag field. How fast to produce peak 120-v output? How many cycles per second are required to produce a 120-volt output, and how did you get your result?
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RESPONSE --> I am a little lost on this one, I must admit. But I think since we are dealing with coils we would start with the formula: E=NBomegaA., where n=number of coils in the loop, B=strength of the magnetic field, omega=radians per second, A=Area of the loop. So, looking for cycles per second, so solve for omega=E/NBA=120V/320(.65)(.210)=2.74 radians per second.
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18:56:15 The average magnitude of the output is peak output/sqrt(2) . We find the average output as ave rate of flux change. The area of a single coil is (21 cm)^2 = (.21 m)^2 and the magnetic field is .65 Tesla; there are 320 coils. When the plane of the coil is perpendicular to the field we get the maximum flux of fluxMax = .65 T * (.21 m)^2 * 320 = 19.2 T m^2. The flux will decrease to zero in 1/4 cycle. Letting t_cycle stand for the time of a complete cycle we have ave magnitude of field = magnitude of change in flux / change in t = 9.17T m^2 / (1/4 t_cycle) = 36.7 T m^2 / t_cycle. If peak output is 120 volts the ave voltage is 120 V / sqrt(2) so we have 36.7 T m^2 / t_cycle = 120 V / sqrt(2). We easily solve for t_cycle to obtain t_cycle = 36.7 T m^2 / (120 V / sqrt(2) ) = .432 second.+ A purely symbolic solution uses maximum flux = n * B * A average voltage = V_peak / sqrt(2), where V_peak is the peak voltage giving us ave rate of change of flux = average voltage so that n B * A / (1/4 t_cycle) = V_peak / sqrt(2), which we solve for t_cycle to get t_cycle = 4 n B A * sqrt(2) / V_peak = 4 * 320 * .65 T * (.21 m)^2 * sqrt(2) / (120 V) = .432 second.
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RESPONSE --> i don't remember seeing where average voltage=V_peak/sqrt2. Also, where did we get t=9.17T m^2?
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rPdǦ}ضx assignment #010 010. `Query 31 Physics II 06-26-2007 ӔxmjFc assignment #010 010. `Query 31 Physics II 06-26-2007"