#$&* course Phy 122 9:15 pm July 22 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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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: ok ********************************************* Question: A 320-loop square coil 21 cm on a side rotates about an axis perpendicular to a .65 T mag field. What frequency of oscillation will produce a peak 120-v output? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: maximum flux = .65 T * ( .21 m ) ^2/ loop * 320 loops = 9.173 T m ^2 = 9.173 T m ^2 * sin (2 pi f t) = 9.173 T m ^2 * 2 pi f cos (2 pi f t) 9.173 T m ^2 * 2 pi f = 120 V so.. = 120 V / (9.173 T m ^2 * 2 pi) = 2.08 V / (T m ^2) = 2.08 T m^2 / (s) / (T m ^2) = 2.08 s ^-1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: I wouldn't advocate using a formula from the book to solve this problem. Common sense, starting from the premise that the voltage function is the derivative of the flux function, is much more efficient (way fewer formulas to remember and less chance of using the wrong one). The maximum flux is .65 T * (.21 m)^2/ loop * 320 loops = 9.173 T m^2. So the flux as a function of clock time could be modeled by phi(t) = 9.173 T m^2 * sin(2 pi f t). The voltage induced by changing flux is the rate of change of flux with respect to clock time. So the voltage function is the t derivative of the flux: V(t) = phi ' (t) = 9.173 T m^2 * 2 pi f cos(2 pi f t). Maximum voltage occurs when cos(2 pi f t) = 1. At this instant the voltage is max voltage = 9.173 T m^2 * 2 pi f. Setting this equal to the peak voltage we get 9.173 T m^2 * 2 pi f = 120 V so that f = 120 V / (9.173 T m^2 * 2 pi) = 2.08 V / / (T m^2) = 2.08 T m^2 / s / (T m^2) = 2.08 s^-1. We can generalize this symbolically by replacing 9.173 T m^2 by phi_max, which represents the maximum flux. So a generator with maximum flux phi_max, rotating a frequency f has flux function phi(t) = phi_max cos(2 pi f t) with t derivative V(t) = phi ' (t) = phi_max cos(2 pi f t). Everything follows easily from this formulation, with no need to memorize the formulas that result. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!