#$&* course PHY 202 031. `Query 31
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Given Solution: The average induced emf is the average rate of change of the magnetic flux with respect to clock time. The initial magnetic flux through this loop is flux = magnetic field * area = 1.10 T * (pi * .048 m)^2 = .00796 T m^2. Flux is designated by the Greek letter phi. The flux is reduced to 0 when the loop is removed from the field, so the change in flux has magnitude .0080 T m^2. The rate of change of flux is therefore `d phi / `dt = .0080 T m^2 / (.15 sec) = .053 T m^2 / sec = .053 volts. STUDENT COMMENT OK so its in Volts. I understand INSTRUCTOR RESPONSE You had the right number. You should also carry the units throughout the calculation. A Tesla is a N / (amp m) so the unit T m^2 / sec becomes N m / (amp sec) = J / (C/s * s) = J / C, or volts. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery gen problem 21.23 320-loop square coil 21 cm on a side, .65 T mag field. How fast to produce peak 120-v output? How many cycles per second are required to produce a 120-volt output, and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Not sure, the equation in the book is e = NB'omegaA sin 'omega t. N = 320 Loops, B = 0.65 T, A = (0.21 m^2) = 0.0441 m^2, e = 120V / sqrt(2) = 84.9 v 'omega = e / NBA = 84.9 V / (320 * 0.65 T * 0.0441 m^2) = 9.25 rad / s This is my best guess at the moment. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Now I see how this was approached. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `quniv query 29.54 (30.36 10th edition) univ upward current I in wire, increasing at rate di/dt. Loop of height L, vert sides at dist a and b from wire. When the current is I what is the magnitude of B at distance r from the wire and what is the magnetic flux through a strip at this position having width `dr? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: General College Physics Student confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The magnetic field due to the wire at distance r is 2 k ' I / r. The field is radial around the wire and so by the right-hand rule (thumb in direction of current, fingers point in direction of field) is downward into the page. The area of the strip is L * `dr. The magnetic flux thru the strip is therefore 2 k ' I / r * (L `dr). The total magnetic field over a series of such strips partitioning the area is thus sum(2 k ' I / r * L `dr, r from a to b). Taking the limit as `dr -> 0 we get integral (2 k ' I / r * L with respect to r, r from a to b). Our antiderivative is 2 k ' I ln | r | * L; the definite integral therefore comes out to flux = 2 k ' L ln | b / a | * I. If I is changing then we have rate of change of flux = 2 k ' L ln | b / a | * dI/dt. This is the induced emf through a single turn. You can easily substitute a = 12.0 cm = .12 m, b = 36.0 cm = .36 m, L = 24.0 cm = .24 m and di/dt = 9.60 A / s, and multiply by the number of turns. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: A 320-loop square coil 21 cm on a side rotates about an axis perpendicular to a .65 T mag field. What frequency of oscillation will produce a peak 120-v output? Your Solution: General College Physics Student confidence rating #$&*:32;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): ------------------------------------------------ Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!