#$&* course Phy 242 031. `Query 31 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: 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): good Self-critique rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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: flux at cos(0)= .65*.21^2*320=9.173 V(t)=9.173*sin(2*pi*freq*t) V(max)=9.173*2*pi*freq=120 Solving for frequency we get 2.08/s 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): good Self-critique rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!