#$&* course PHY 202 031. `Query 31 was submitted 25 Jul 2011 around 5:00 PM. 031. `Query 31*********************************************
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Given Solution: 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. Flux is designated by the Greek letter phi. 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 flux is therefore `d phi / `dt = .0080 T m^2 / (.15 sec) = .053 T m^2 / sec = .053 volts. STUDENT COMMENT OK so its in Volts. I understand INSTRUCTOR RESPONSE You had the right number. You should also carry the units throughout the calculation. A Tesla is a N / (amp m) so the unit T m^2 / sec becomes N m / (amp sec) = J / (C/s * s) = J / C, or volts. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 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: fluxMax = .65 T * (.21 m)^2 * 320 fluxMax = 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 ave magnitude of field = 9.17T m^2 / (1/4 t_cycle) ave magnitude of field = 36.7 T m^2 / t_cycle If peak output is 120 V 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 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 t_cy cle = 4 * 320 * .65 T * (.21 m)^2 * sqrt(2) / (120 V) t_cycle = 0.432 sec confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique Rating: 3