course PHY 202 쓢ƂGgׇvkʪassignment #008
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19:22:09 Query introductory problem set 54 #'s 8-13 Explain how to determine the magnetic flux of a uniform magnetic field through a plane loop of wire, and explain how the direction of the field and the direction of a line perpendicular to the plane of the region affect the result.
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RESPONSE --> Flux is a measure of how much of a magnetic field goes through a certain area. To determine the magnetic flux, one must use the formula: flux=(B)(A) Where B=strength of the magnetic field A=Area of the plane loop of wire If the the plane loop of wire is perpendicular to the magnetic field, then the equation above will calculate the flux. If the plane is oriented at an angle to the magnetic field then, one must use the formula: flux=(B)(A)cos(theta). Where theta is the angle that the magnetic field makes with the line perpendicular to the loop.
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19:23:29 To do this we need to simply find the area of the plane loop of wire. If we are given the radius we can find the area using Pi * r ^2 Then we multiply the area of the loop (In square meters ) by the strength of the field (in tesla). This will give us the strength of the flux if the plane of the loop is perpendicular to the field. If the perpendicular to the loop is at some nonzero angle with the field, then we multiply the previous result by the cosine of the angle.
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RESPONSE --> If given the radius of the loop vs it's length in meters use the formula Pi*r^2 to get the area of the loop.
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19:35:34 Explain how to determine the average rate of change of magnetic flux due to a uniform magnetic field through a plane loop of wire, as the loop is rotated in a given time interval from an orientation perpendicular to the magnetic field to an orientation parallel to the magnetic field.
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RESPONSE --> The average rate of change of a magnetic flux can found by using the following steps: 1) First you must obtain the time in seconds needed to complete one time interval from an orientation perpendicular to a parallel orientation. For example of the time to complete the cycle is 6 hz, then that means 6 cycles per second. 1/6=.1666 seconds. Then .1666 cycles/4=.04165 seconds. 2)Next, you must multiply the squared length of the plane loop of wire(meters)by the strength of the magnetic field(in Tesla). This calculation will give the voltage produced. 3) WHen the loop is oriented parallel to the magnetic field, then the flux at that point is zero. Thus the averate rate of flux change will be the voltage calculated in step 2. 4) Take the voltage calculated in step 2* the time interval in seconds. This will give you the overall average rate of change of the magnetic flux.
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19:42:02 ** EXPLANATION BY STUDENT: The first thing that we need to do is again use Pi * r ^ 2 to find the area of the loop. Then we multiply the area of the loop (m^2) by the strength of the field (testla) to find the flux when the loop is perpendicular to the field. Then we do the same thing for when the loop is parallel to the field, and since the cos of zero degrees is zero, the flux when the loop is parallel to the field is zero. This makes sense because at this orientation the loop will pick up none of the magnetic field. So now we have Flux 1 and Flux 2 being when the loop is perpendicular and parallel, respectively. So if we subtract Flux 2 from flux 1 and divide this value by the given time in seconds, we will have the average rate of change of magnetic flux. If we use MKS units this value will be in Tesla m^2 / sec = volts. **
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RESPONSE --> -Calculate Flux 1 and Flux 2. Flux one is when the loop is perpendicular to the field. Flux 2 is when the lopp is parallel to the field. -Then subtract Flux 2 from Flux 1 and divide this value by the given time in seconds. More clear steps then the ones I gave.
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19:47:21 Explain how alternating current is produced by rotating a coil of wire with respect to a uniform magnetic field.
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RESPONSE --> The charges in the rotating coil of wire are moving at a certain velocity. The charges will experience a force from the magnetic field according to the formula: F=qvB. Where q is the electric charge, v is the velocity and B is the magnetic field. THe forces make the charges move which means voltage is generated, so an electric field will be acting along the bar.
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19:49:17 ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: Y ou rotate a coil of wire end over end inside a uniform magnetic field. When the coil is parallel to the magnetic field, then there is no magnetic flux, and the current will be zero. But then when the coil is perpendicular to the field or at 90 degrees to the field then the flux will be strongest and the current will be moving in one direction. Then when the coil is parallel again at 180 degrees then the flux and the current will be zero. Then when the coil is perpendicular again at 270 degrees, then the flux will be at its strongest again but it will be in the opposite direction as when the coil was at 90 degrees. So therefore at 90 degrees the current will be moving in one direction and at 270 degrees the current will be moving with the same magnitude but in the opposite direction. COMMENT: Good. The changing magnetic flux produces voltage, which in turn produces current. **
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RESPONSE --> The changing magnetic flux produces voltage which produces current. ok
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19:51:46 Query Principles and General College Physics 18.04. 120V toaster with 4.2 amp current. What is the resistance?
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RESPONSE --> R=V/I, where R-Resistance, V=Voltage and I=Current. Thus: 120V/4.2=28.57 Ohms
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20:00:59 Query Principles and General College Physics 18.28. Max instantaneous voltage to a 2.7 kOhm resistor rated at 1/4 watt.
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RESPONSE --> GIven the formula: P=v^2/R, where P=Power in Watts, V=Voltage, and R=Resistance in Ohms Solving for Voltage=sqrt(R)(P). Thus: sqrt(2700)(.25)=26 Volts
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20:02:36 Voltage is energy per unit of charge, measured in Joules / Coulomb. {}Current is charge / unit of time, measured in amps or Coulombs / second.{}Power is energy / unit of time measured in Joules / second.{}{}The three are related in a way that is obvious from the meanings of the terms. If we multiply Joules / Coulomb by Coulombs / second we get Joules / second, so voltage * current = power. In symbols this is power = V * I.{}{}Ohm's Law tells us that current = voltage / resistance.In symbols this is I = V / R. So our power relationship power = V * I can be written {}{}power = V * V / R = V^2 / R. {}{}Using this relationship we find that {}{}V = sqrt(power * R), so in this case the maximum voltage (which will produce the 1/4 watt maximum power) will be{}{}V = sqrt(1/4 watt * 2.7 * 10^3 ohms) = 26 volts.
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RESPONSE --> ok
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20:10:52 Query general college physics problem 18.39; compare power loss if 520 kW delivered at 50kV as opposed to 12 kV thru 3 ohm resistance.** The current will not be the same at both voltages. It is important to understand that power (J / s) is the product of current (C / s) and voltage (J / C). So the current at 50 kV kW will be less than 1/4 the current at 12 kV. To deliver 520 kW = 520,000 J / s at 50 kV = 50,000 J / C requires current I = 520,000 J/s / (50,000 J/C) = 10.4 amps. This demonstrates the meaning of the formula P = I V. To deliver 520 kW = 520,000 J / s at 12 kV = 12,000 J / C requires current I = 520,000 J/s / (12,000 J/C) = 43.3 amps. The voltage drops through the 3 Ohm resistance will be calculated as the product of the current and the resistance, V = I * R: The 10.4 amp current will result in a voltage drop of 10.4 amp * 3 ohms = 31.2 volts. The 43.3 amp current will result in a voltage drop of 40.3 amp * 3 ohms = 130 volts. The power loss through the transmission wire is the product of the voltage ( J / C ) and the current (J / S) so we obtain power losses as follows: At 520 kV the power loss is 31.2 J / C * 10.4 C / s = 325 watts, approx. At 12 kV the power loss is 130 J / C * 43.3 C / s = 6500 watts, approx. Note that the power loss in the transmission wire is not equal to the power delivered by the circuit, which is lost through a number of parallel connections to individual homes, businesses, etc.. The entire analysis can be done by simple formulas but without completely understanding the meaning of voltage, current, resistance, power and their relationships it is very easy to get the wrong quantities in the wrong places, and especially to confuse the power delivered with the power loss. The analysis boils down to this: I = P / V, where P is the power delivered. Ploss = I^2 R, where R is the resistance of the circuit and Ploss is the power loss of the circuit. So Ploss = I^2 * R = (P/V)^2 * R = P^2 * R / V^2. This shows that power loss across a fixed resistance is inversely proportional to square of the voltage. So that the final voltage, which is less than 1/4 the original voltage, implies more than 16 times the power loss. A quicker solution through proportionalities: For any given resistance power loss is proportional to the square of the current. For given power delivery current is inversely proportional to voltage. So power loss is proportional to the inverse square of the voltage. In this case the voltage ratio is 50 kV / (12 kV) = 4.17 approx., so the ratio of power losses is about 1 / 4.17^2 = 1 / 16.5 = .06. Note that this is the same approximate ratio you would get if you divided your 324.5 watts by 5624.7 watts. **
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RESPONSE --> wow...a little complicated. Had to read a few times for comprehension.
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уؒ̐ŋ assignment #008 008. `Query 29 Physics II 06-25-2007"