course Phys 202 UiHzxFassignment #031
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18:39:59 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 --> The question went straight to the answer before I could answer so here is what I did beforereading the answer. The area of one coil is (21cm)^2=441cm^2 and because there is 320 loops, that makes it: 320*441cm^2=141120cm Flux=mag field*area= .65T*1411.2m^2=9.17 T This is where I was stuck. I got 9.17 not 19.2 for the answer of the flux max. It did not match the answer given. How do we know that the flux will end in 1/4 a cycle?
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18:40:07 univ 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?
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18:40:09 ** 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. **
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