course Phy 202 In completing this Query a box came up that said Run Time Error and the query ended. I wasn't for sure if I completed the whole query or not due to this error message.
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11:31:15 query zhen Phy 21.23 720-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?......!!!!!!!!...................................
RESPONSE --> fluxMax = .65 T * (.21 m)^2 * 720 ave magnitude of field =.65 T * .(21) m^2 * 720 / (1/4 T). .65 T * (.21 m)^2 * 720 / (1/4 T) = 120 V / sqrt(2). T = .65 T * (.21 m)^2 * 720 * sqrt(2) / (120 V) = .17
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11:41:28 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 720 coils. When the plane of the coil is perpendicular to the field we get the maximum flux of fluxMax = .65 T * (.21 m)^2 * 720 . The flux will decrease to zero in 1/4 cycle. Letting T stand for the time of a complete cycle we have ave magnitude of field = magnitude of change in flux / change in t = .65 T * .(21) m^2 * 720 / (1/4 T). If peak output is 120 volts the ave voltage is 120 V / sqrt(2) so we have .65 T * (.21 m)^2 * 720 / (1/4 T) = 120 V / sqrt(2). We easily solve for T to obtain T = .65 T * (.21 m)^2 * 720 * sqrt(2) / (120 V) =......!!!!!!!!...................................
RESPONSE --> ok i understand
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11:41:34 *&*& verify numbers using DERIVE and maybe change to symbolic approach *&*&
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RESPONSE --> ok
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11:41:38 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?......!!!!!!!!...................................
RESPONSE --> dna
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11:41:42 ** 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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