Chapter 22, Problem 19 Your answer is partially correct. Try again. A rectangula
ID: 1417134 • Letter: C
Question
Chapter 22, Problem 19 Your answer is partially correct. Try again. A rectangular loop of wire with sides 0.220 and 0.407 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawing). The magnetic field has a magnitude of 0.673 T and is directed parallel to the normal of the loop's surface. In a time of 0.252 s, one-half of the loop is then folded back onto the other half, as indicated in part b of the drawing. Determine the magnitude of the average emf induced in the loop. xXx tn (al Units v Number 0.11956Explanation / Answer
1)
I assume that neither you nor your teacher knows anything about AC circuit theory. If you did you would recognise that this question cannot be answered since we are told nothing about the power factor of the load. The point is that in order to estimate the power lost in the resistance of the power lines we must calculate the current in them. In an AC circuit this cannot be calculated directly from the power and the voltage. The best we can do is to assume that the power factor is intended to be 1, which should be stated in the question; it is rarely 1 in practice.
With this assumption we can calculate the current as (1.8*10^6)/1600 = 1125A
You have not given us the units of the resistance per unit length - I shall assume that the units are ohm/km.
Total resistance of power lines is R = 2*5.4*4.9*10^-2 = 0.5292 ohms.
Power list in wires = R*I^2 = 0.5292 x 1125^2 = 669.77 kw
b)
I suppose we are intended to assume that the transformer is ideal, and has no losses. With this assumption, the current will be reduced by a factor of 1/100, so the power lost in the wires will be reduced by a factor of 10^-4, giving the new value of 66.977 W
2)
The way it's drawn suggests that the area of the loop becomes zero. so,
emf = delta(flux)/deltat
emf = B delta area / delta t = 0.673 x 0.220 x 0.407 / 0.252
emf = 0.239 V
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