Electricity Demystified, 2nd edition
Stan Gibilisco
Explanations for Quiz Answers in Chapter 11
1. When we examine the right-hand portion of Fig. 11-6, we can see that EM energy with a wavelength of 600 nm occupies the visible-light spectrum. The correct choice is C.
2. In an oscilloscope, the vertical deflection coils cause the electron beam to move up and down on the screen. The horizontal deflection coils cause the beam to sweep across the screen from left to right. If we want to analyze a signal waveform, we'll usually apply the signal to the vertical deflection coils, so we get a graphical rendition of the signal's amplitude (going up and down) as a function of time (going from left to right). The correct choice is A.
3. In an electromagnetic (EM) field, the electric lines of flux run perpendicular to the magnetic lines of flux at every point in space. The correct choice is B. In some special situations, the electric and magnetic lines of flux both rotate in the same sense as the EM field propagates, so choice D technically holds true in those instances. However, the flux lines still maintain their right-angle orientation at every point. Most EM fields don't have rotating lines of flux, so choice B is the only one of the four that's always true.
4. If we reverse the direction of the DC in a galvanometer coil, the needle will deflect the other way. We're told that the needle turns toward the east in the first place, so the needle must turn toward the west in the second case. Now we know that the correct choice must be either A or B. When we reverse the DC, we also reduce it from 40 mA to 10 mA. Therefore, the compass needle will rotate away from geomagnetic north to a lesser extent in the second case than it did originally. We'll see less than the 32º of needle deflection that we observed in the first case. The correct choice is B.
5. To find the free-space wavelength (in meters) for an EM wave at a particular frequency (in hertz), we must divide 3.00 x 108 by the number of hertz. That's the equivalent of dividing 300 by the number of megahertz. In this situation, the frequency equals 1.50 MHz, so we can calculate the wavelength as 300 / 1.50 = 200 m. The correct choice is B.
6. If we shield all the interconnecting cables in a home-entertainment system, we can reduce the likelihood that it will suffer from electromagnetic interference (EMI). We can't completely eliminate the risk, but of the four choices given here, B is the right one. Choices A and C are absolutely wrong; if we do either of those things, we'll increase the risk of EMI to our system.
7. In order for an EM field to arise, electrical charge carriers must accelerate. That is to say, they must move at a speed that changes with time. None of the three choices A, B, or C indicate such a state of affairs, so we have to take D, "None of the above."
8. In a cathode-ray tube (CRT), the anodes focus the electron beam and increase its speed (accelerate it). The answer is A.
9. Remember than a nanometer (1 nm) equals one thousand-millionth of a meter (10-9 m). The Ångström represents 10-10 m, which works out to 0.1 nm. The correct choice is B.
10. To figure out the frequency (in hertz) for an EM field having a certain wavelength (in meters), we must divide 3.00 x 108 by the number of meters. If we divide 300 by the number of meters, we get megahertz. In this case, the wavelength equals 12 m, so we can calculate the frequency as 300 / 12 = 25 MHz. The answer is D.