#$&* course Phy 242 029. `Query 29 *********************************************
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Given Solution: To do this we need to simply find the area of the plane loop of wire. If we are given the radius we can find the area using Pi * r ^2 Then we multiply the area of the loop (In square meters ) by the strength of the field (in tesla). This will give us the strength of the flux if the plane of the loop is perpendicular to the field. If the perpendicular to the loop is at some nonzero angle with the field, then we multiply the previous result by the cosine of the angle. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):good ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qExplain how to determine the average rate of change of magnetic flux due to a uniform magnetic field through a plane loop of wire, as the loop is rotated in a given time interval from an orientation perpendicular to the magnetic field to an orientation parallel to the magnetic field. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The average change of flux can be found in a quarter turn of the coil. You take the flux while the point is perpendicular and then while it is parallel. Since we know that at parallel cos(90)=0, so we just need to take the maximum flux and subtract zero(still leaving the maximum flux) then divide that by the time it took to make the quarter turn. This will be the average rate for not just the quarter but the entire thing since it will just continue to repeat this pattern as it turns each quarter turn. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):good, forgot units ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qExplain how alternating current is produced by rotating a coil of wire with respect to a uniform magnetic field. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The variations of magnetic flux produce voltage, where there is voltage there must be current. As previously discussed, each quarter turn of a coil dramatically changes the flux felt by a charge, the movement of that creates currents in the other object which is shown by the presence of voltage. 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): good ------------------------------------------------ Self-critique Rating:3 ********************************************* 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: 3R=rho*L/A=rho*2d/3d^2=2/3rho/d I=V/R=V/1/6rho/d=6dV/rho 1 R=rho*L/A= rho*3d/2d^2=3/2rho/d I=V/R=V/3/2rho/d=2dV/3rho. 2R=rho*L/A=rho*3d/2d^2=3/2rho/d I=V/R=V/3/2rho/d=2dV/3rho I/A=2dV/3 rho/2 d^2=V/(3rho d)=1/3V/rho d From this you can see the maximum current density is established when using the biggest surface. 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 inversely 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 fae. ** 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):good ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: