#$&* course Phy 122 10:10 pm July 15 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qExplain 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Again, you would find the area when the radius is available. You do this by taking pi and multiplying it by the radius squared. After you have the area, you multiply the area of the loop by the strength of the field. The strength of the field is measured in teslas and the area of the loop in square meters. You do this same thing when the loop is parallel to the field. Here you have Flux 1 and Flux 2 when the loop is parallel and perpendicular. So you subtract Flux 2 from Flux 1. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qExplain how alternating current is produced by rotating a coil of wire with respect to a uniform magnetic field. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Changing of magnetic flux produces both voltage and current. coil parallel to the magnetic field = 0 current (no magnetic flux) coil perpendicular to the field (90 degrees) = strongest flux, current moving in one direction coil parallel again ( this time at 180 degrees) = 0 current, 0 flux coil perpendicular again (this time at 270 degrees) = flux in the opposite direction than it was at 90 degrees, but again at its strongest 90 degrees = current moves in one direction 270 degrees = current moves in the opposite direction but with the same magnitude confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: Y ou rotate a coil of wire end over end inside a uniform magnetic field. When the coil is parallel to the magnetic field, then there is no magnetic flux, and the current will be zero. But then when the coil is perpendicular to the field or at 90 degrees to the field then the flux will be strongest and the current will be moving in one direction. Then when the coil is parallel again at 180 degrees then the flux and the current will be zero. Then when the coil is perpendicular again at 270 degrees, then the flux will be at its strongest again but it will be in the opposite direction as when the coil was at 90 degrees. So therefore at 90 degrees the current will be moving in one direction and at 270 degrees the current will be moving with the same magnitude but in the opposite direction. COMMENT: Good. The changing magnetic flux produces voltage, which in turn produces current. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qQuery Principles and General College Physics 18.04. 120V toaster with 4.2 amp current. What is the resistance? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: For this problem, you use Ohm's Law which says that current = voltage divided by resistance current is directly proportional to voltage current is also inversely proportional to resistance I = v/ r Here we are solving for resistance, so we can rearrange the formula to R = v/ i R = 120 volts / 4.2 amps = 28.57 ohms confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: current = voltage / resistance (Ohm's Law). The common sense of this is that for a given voltage, less resistance implies greater current while for given resistance, greater voltage implies greater current. More specifically, current is directly proportional to voltage and inversely proportional to resistance. In symbols this relationship is expressed as I = V / R. In this case we know the current and the voltage and wish to find the resistance. Simple algebra gives us R = V / I. Substituting our known current and voltage we obtain R = 120 volts / 4.2 amps = 29 ohms, approximately. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qQuery Principles and General College Physics 18.28. Max instantaneous voltage to a 2.7 kOhm resistor rated at 1/4 watt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I = V / R power = V^2 / R. V = sqrt (1/4 watt * 2.7 * 10 ^3) = 26 v confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!