course phy122 This was a short asst. Tried to skip problem 3 (said Univ. physics) and it seemed to end.Exam 1 next. Got your message about better self-critique. You can safely assume the better the self-critique the better the understanding I have of the concepts being tuaght (and vice verca.) A poor self critique is not me being lazy. I am working daily with my tutor as I just need a quicker turn around time this week to have a hope of passing the exam. Self critiques will be better for parts 2,3, and 4 provided I am able to get that far. assignment #010
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15:30:06 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 --> d=9.6 cm = 9.6*10^-2m B=1.1T t=0.15s k'=10^-7 Tesla/(amp meter) B=(2k' Pi I) / a I=(Ba)/(2k'pi) V=4 * A * E * f EMF V=4 pi r^2 B * (1/t) r=9.6 * 10^-6=0.096/2 V=(4 pi * (0.048)^2 * 1.1 * 1) / 0.15 V=(4 pi (0.048)^2 * 1.1)/ 0.15 = 0.21V
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15:30:36 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 --> ?
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15:36:58 flux is therefore .0080 T m^2 / (.15 sec) = .053 T m^2 / sec = .053 volts.
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RESPONSE --> 4 was for sphere. since it is a loop (circufrence) I should of multiplied by 1.10 instead of 4. (1.1 * pi * (0.048^2) * 1.1)/0.15=.058V (pi as 3.14)
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15:44:24 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 --> 320 loop square coil 21cm Area=0.21m V=A*B*f f=V/(AB) = 120/((0.21)^2 * 65) f=120/(320*(0.21)^2*65 =0.13/sec
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17:46:16 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 --> We chose F as 1/t and it should be 1/4t.
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