#$&* course phy 242 july 27 Question: `quniv query 29.54 (30.36 10th edition) univ upward current I in wire, increasing at rate di/dt. Loop of height L, vert sides at dist a and b from wire.
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Given Solution: ** The magnetic field due to the wire at distance r is 2 k ' I / r. The field is radial around the wire and so by the right-hand rule (thumb in direction of current, fingers point in direction of field) is downward into the page. The area of the strip is L * `dr. The magnetic flux thru the strip is therefore 2 k ' I / r * (L `dr). The total magnetic field over a series of such strips partitioning the area is thus sum(2 k ' I / r * L `dr, r from a to b). Taking the limit as `dr -> 0 we get integral (2 k ' I / r * L with respect to r, r from a to b). Our antiderivative is 2 k ' I ln | r | * L; the definite integral therefore comes out to flux = 2 k ' L ln | b / a | * I. If I is changing then we have rate of change of flux = 2 k ' L ln | b / a | * dI/dt. This is the induced emf through a single turn. You can easily substitute a = 12.0 cm = .12 m, b = 36.0 cm = .36 m, L = 24.0 cm = .24 m and di/dt = 9.60 A / s, and multiply by the number of turns. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: A 320-loop square coil 21 cm on a side rotates about an axis perpendicular to a .65 T mag field. What frequency of oscillation will produce a peak 120-v output? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: max flux = .65 *.21^2*320 = 9.17 T m^2 phi = 9.17*sin(2*pi*f*t) V = phi` = 9.17*2pi*f*cos(2pi*f*t) Max V = 9.17*2pi*f 120 = 9.17*2pi*f F = 2.1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: I wouldn't advocate using a formula from the book to solve this problem. Common sense, starting from the premise that the voltage function is the derivative of the flux function, is much more efficient (way fewer formulas to remember and less chance of using the wrong one). The maximum flux is .65 T * (.21 m)^2/ loop * 320 loops = 9.173 T m^2. So the flux as a function of clock time could be modeled by • phi(t) = 9.173 T m^2 * sin(2 pi f t). The voltage induced by changing flux is the rate of change of flux with respect to clock time. So the voltage function is the t derivative of the flux: • V(t) = phi ' (t) = 9.173 T m^2 * 2 pi f cos(2 pi f t). Maximum voltage occurs when cos(2 pi f t) = 1. At this instant the voltage is • max voltage = 9.173 T m^2 * 2 pi f. Setting this equal to the peak voltage we get • 9.173 T m^2 * 2 pi f = 120 V so that • f = 120 V / (9.173 T m^2 * 2 pi) = 2.08 V / / (T m^2) = 2.08 T m^2 / s / (T m^2) = 2.08 s^-1. We can generalize this symbolically by replacing 9.173 T m^2 by phi_max, which represents the maximum flux. So a generator with maximum flux phi_max, rotating a frequency f has flux function • phi(t) = phi_max cos(2 pi f t) with t derivative • V(t) = phi ' (t) = phi_max cos(2 pi f t). Everything follows easily from this formulation, with no need to memorize the formulas that result. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: