course Phy 202 ??x???w???+??P|??assignment #010010. `Query 31
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09:59:55 Query Principles and General Physics 21.04. A circular loop of diameter 9.6 cm in a 1.10 T field perpendicular to the plane of the loop; loop is removed in .15 s. What is the induced EMF?
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RESPONSE --> The emf is the magnetic flux change in terms of time. So, first we have to calculate the basic magnetic flux: flux =magnetic field*area = 1.10 T * (pi * .048 m)^2 = .00796 T m^2. The change in flux is then .00769 and the time is given as .15 s, so .00769Tm^2/.15 s=.0512 volts
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10:00:14 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.{}{}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
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RESPONSE --> I just rounded differently, that is the reason for my different responses. Otherwise, my answer is generally the same.
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10:00:22 flux is therefore .0080 T m^2 / (.15 sec) = .053 T m^2 / sec = .053 volts.
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RESPONSE --> I understand.
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22:20:08 query 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?
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RESPONSE --> I know we have to find the average rate of output for which I have to know the average flux. Since the coil is square, then I know we can find the area by squaring one of the sides, which we are given. Of course centimeters don't work, so I have to convert the 21 cm to meters, which is .21 m. So, .21m^2=.0441 m. Now, since there are 320 loops, the total area of the coil is .0441m * 320=about 14 m^2 I can now use this to find the maximum flux since I am given the mag. field. Max flux=.65 T*14 m^2=9.17 T m^2 Since the magnetic field is only able to work when perpendicular, when the magnetic field becomes zero on the square the flux, in turn, will be zero (flux=m. field strength*area). So, I know the change is 9.17 T * m^2; but other than that, I am not sure what to do.
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19:53:06 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 --> I see now that I should have finished out the problem by using time for the cycle as a variable and have solved for it by dividing the average magnetic field by the time of the cycle.
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19:53:07 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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RESPONSE -->
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19:53:09 ** 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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RESPONSE -->
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