#$&* course Phy 242 7/21 11 029. `Query 29
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Given Solution: To do this we need to simply find the area of the plane loop of wire. If we are given the radius we can find the area using Pi * r ^2 Then we multiply the area of the loop (In square meters ) by the strength of the field (in tesla). This will give us the strength of the flux if the plane of the loop is perpendicular to the field. If the perpendicular to the loop is at some nonzero angle with the field, then we multiply the previous result by the cosine of the angle. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qExplain how to determine the average rate of change of magnetic flux due to a uniform magnetic field through a plane loop of wire, as the loop is rotated in a given time interval from an orientation perpendicular to the magnetic field to an orientation parallel to the magnetic field. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Flux of the magnetic field is found as usual by multiplying the magnitude of the magnetic field by the area of the loop (B*A). This is the flux when the loop is perpendicular to the field. If the field is rotated so tha tis it parallel to the field, cos(0) = 0 so there is 0 flux To find the average rate of change of magnetic flux, we subtract the parallel from the perpendicular flux and divide by time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: This simply expands on the concept mentioned in the previous question. When the coil is perpendicular to the field, the flux is the strongest. When it is parallel to the field, the flux is zero. As a coil rotates, every 90 degrees of the full rotation is either a maximum or minimum magnetic flux which switches direction, therefore creating an alternating current. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: Y ou rotate a coil of wire end over end inside a uniform magnetic field. When the coil is parallel to the magnetic field, then there is no magnetic flux, and the current will be zero. But then when the coil is perpendicular to the field or at 90 degrees to the field then the flux will be strongest and the current will be moving in one direction. Then when the coil is parallel again at 180 degrees then the flux and the current will be zero. Then when the coil is perpendicular again at 270 degrees, then the flux will be at its strongest again but it will be in the opposite direction as when the coil was at 90 degrees. So therefore at 90 degrees the current will be moving in one direction and at 270 degrees the current will be moving with the same magnitude but in the opposite direction. COMMENT: Good. The changing magnetic flux produces voltage, which in turn produces current. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qQuery Principles and General College Physics 18.04. 120V toaster with 4.2 amp current. What is the resistance? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I = V/R R = V/I R = 120 volts / 4.2 amps R = 29 ohms confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: voltage * current = power I = V/R Therefore Power = V * V/R Power= v^2 / R Solving for V: V = sqrt(power * R) V = sqrt( 1/4 watt * 2.7 e^3 ohms) V = 26 volts confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Voltage is energy per unit of charge, measured in Joules / Coulomb. Current is charge / unit of time, measured in amps or Coulombs / second. Power is energy / unit of time measured in Joules / second. The three are related in a way that is obvious from the meanings of the terms. If we multiply Joules / Coulomb by Coulombs / second we get Joules / second, so voltage * current = power. In symbols this is power = V * I. Ohm's Law tells us that current = voltage / resistance.In symbols this is I = V / R. So our power relationship power = V * I can be written power = V * V / R = V^2 / R. Using this relationship we find that V = sqrt(power * R), so in this case the maximum voltage (which will produce the 1/4 watt maximum power) will be V = sqrt(1/4 watt * 2.7 * 10^3 ohms) = 26 volts. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qQuery general college physics problem 18.39; compare power loss if 520 kW delivered at 50kV as opposed to 12 kV thru 3 ohm resistance.** The current will not be the same at both voltages. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: N/A confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. ** STUDENT COMMENT This is confusing. I will have to keep reading over the problem until I understand the solution….I’m afraid that I will have problems with this test…we will see……. INSTRUCTOR RESPONSE The concept of thinking these ideas through in terms of the units is developed in the introductory problem sets, and may be helpful. A key idea here is that power (which you should understand is energy / time interval, measured in Joules/second or watts) is the product of current (in C / sec or amps) and voltage (in J / C or volts). If you multiply the number of Coulombs per second, by the number of Joules per Coulomb, you get the number of Joules per second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): N/A ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery univ students with 12th and later editions should attempt this problem before reading the solution; 11th edition 25.62 (26.50 10th edition) A rectangular block of a homogeneous material (i.e., with constant resistivity throughout) has dimensions d x 2d x 3d. We have a source of potential difference V. To which faces of the solid should the voltage be applied to attain maximum current density and what is the density? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: R = rho * L/A Face d * 2d: R = rho * 3d / 2d^2 = 3rho / 2d I = V/R I = v / (3 rho / 2d) I = 2dV / 3 rho Current density = I/A = (2 dV / 3 rho) / 2d^2 = 1/3 V / rho d Face d * 3d: R = rho * 2d / 3d^2 = 2 rho / 3d I = V / 2 rho / 3d I = 3dV / 2 rho Current density = I/A = (3dV / 2rho) / 3d^2 = 1/2 V / rho d Face 2d * 3d: R = rho * d / 6d^2 = rho / 6d I = V / (rho / 6d) I = 6dV / rho Current density = I/A = (6dV/rho) / 6d^2 = V / rho d Max current density is V / rho d therefore the voltage should be applied to the 2d * 3d face, which is the largest face. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** First note that the current I is different for diferent faces. The resistance of the block is proportional to the distance between faces and inverely proportional to the area, so current is proportional to the area and inversely proportional to the distance between faces. Current density is proportional to current and inversely proportional to the area of the face, so current density is proportional to area and inversely proportional to the distance between faces and to area, leaving current inversely proportional to distance between faces. For the faces measuring d x 2d we have resistance R = rho * L / A = rho * (3d) / (2 d^2) = 3 / 2 rho / d so current is I = V / R = V / (3/2 rho / d) = 2d V / (3 rho). Current density is I / A = (2 d V / (3 rho) ) / (2 d^2) = V / (3 rho d) = 1/3 V / (rho d). For the faces measuring d x 3d we have resistance R = rho * L / A = rho * (2d) / (3 d^2) = 2 / 3 rho / d so current is I = V / R = V / (2/3 rho / d) = 3 d V / (2 rho). Current density is I / A = (3 d V / (2 rho) ) / (3 d^2) = V / (2 rho d) = 1/2 V / (rho d). For the faces measuring 3d x 2d we have resistance R = rho * L / A = rho * (d) / (6 d^2) = 1 / 6 rho / d so current is I = V / R = V / (1/6 rho / d) = 6 d V / (rho). Current density is I / A = (6 d V / (rho) ) / (6 d^2) = V / (rho d) = V / (rho d). Max current density therefore occurs when the voltage is applied to the largest face. ** STUDENT QUESTION I don’t really understand why current isn’t constant throughout the object. I can follow the solution, but I don’t understand that one point. My solution was based on the constant current which I see is not correct. INSTRUCTOR RESPONSE If the voltage is applied to different faces then you will get different currents. The potential gradient is greater if the faces are closer together. This effect would tend to result in greater current for faces that are closer together. If the faces are closer together the areas of the faces will be greater, so the cross-sectional area will be greater. This effect would result in greater current. So the faces closest to together will experience the greatest potential gradient, and result in the greatest cross-sectional area, resulting in the greatest current. However the question asks about current density, not current. Intuitively, we know that the potential gradient is the electric field responsible for accelerating the charges, so that the charge density will just be proportional to the potential gradient, so that if the voltage is applied to the faces with the least separation the current density will be the greatest. In fact, since the separations are d, 2d and 3d we can see that the potential gradients will be V / d, 1/2 V / d and 1/3 V / d so the other two current densities will be 1/2 and 1/3 as great as the maximum. This can be worked out symbolically for the general case: R = rho * L / A, and current is I = V / R = V * A / (rho L) so current density is I / A = V / (rho L) The greatest current density occurs for the least value of L, which for the given situation is d, the separation between the largest faces. Working out the details more fully is be unnecessary but might be instructive: The faces closest together are separated by distance d, and have cross- sectional area 2 d * 3 d = 6 d^2. So the resistance is R = rho * L / A = rho * d / (6 d^2) = rho / (6 d) = 1/6 * rho / d The faces furthest apart are separated by distance 3d, and have cross- sectional area d * 2 d = 2 d^2. So the resistance is R = rho * L / A = rho * d / (2 d^2) = rho / (2 d) = 1/2 * rho / d Similar analysis shows that the resistance between faces separated by 2 d is rho / (3/2 d) = 2/3 * rho / d. Thus the currents would be I = V / R = 6 V * d / rho for the closest faces I = V / R = 2 V d / rho for the most widely separate faces and I = V / R = 2 /3 V d / rho for the in-between separation. The respective current densities, current / area, would be (6 V d / rho) / (6 d^2) = V / (rho d) (2 V d / rho) / (3 d^2) = 2/3 V / (rho d) (2/3 V d / rho) / (2d^2) = 1/3 V / (rho d). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!