Consider the following configuration of field lines. This could be a... A: E-field. B: B-field. C: Either E or B

Transcription

1 (EMWaves-1) Consider the following configuration of field lines. This could be a... A: E-field B: B-field C: Either E or B Answer: either E or B. Both B-field lines and E-field lines generated by changing B-field (a al Faraday) close on themselves. See concept test This could represent the field lines of a A: E-field B: B-field C: Either E or B Answer: Only E-field lines from a charge could produce field lines like this. There is no such thing as "magnetic charge"; B-field lines always close on themselves.

2 (EMWaves-2) For a charging capacitor, the dφ displacement current ε E o dt between the plates equals the conduction current I in the wires leading to the capacitor. Assume the wires are long and straight, and the capacitor gap is so small that "fringing effects" are negligible. I I A L At point A, between the capacitor plates, there is A: a changing E-field and a constant B-field B: a changing E-field and a changing B-field C: a changing B-field and a constant E-field D: a changing E-field and zero B-field E: some other answer/don't know Answer: The plate is steadily charging up (because there's a constant I=dq/dt feeding it) so the E field is also steadily increasing. but the B field arises from d(phi)/dt, and since phi is proportional to E which is increasingly steadily (i.e. is proportional to time), then d(phi)/dt will just be a *constant*. As you move along line L, the B-field is... A: constant B: not constant Answer: Its a constant, it's just exactly the same as if the capacitor wasn't there at all.

3 (EMWaves-3) A radio wave of wavelength 2 meters passes by a person with a radio receiver. The person watches the electric and magnetic fields go up and down as the wave travels past. After 1 second, the number of waves that moved past the person is: A: 1 wave B: 3x10 8 waves C: 1.5x10 8 waves D: 6x10 8 waves E: Not enough information Answer: lambda*frequency = c, so here frequency = c/lambda = 3E8 m/s / 2 m = 1.5 E8 /s. So I expect to see 1.5E8 cycles/second.

4 (EMWaves-4) A radio wave of wavelength 2 meters passes by a person with a radio receiver. The person watches the E and B fields go up and down as the wave travels by. Later, a new radio wave passes the person. They observe that the E and B fields go up and down 10 times faster than the original wave. What is the best conclusion? A: Second wave has a wavelength of 0.2 m B: Second wave has a wavelength of 20 m C: Second wave travels 10 times faster. D: Second wave has 1/10 the frequency. E: None of these is correct. Answer: lambda*freq = c. If it goes up and down 10 times faster, it means freq is 10 times higher, so lambda is 10 times SMALLER. The new wavelength is 0.2 m.

5 (EMWaves-5) We say that "concert A" corresponds to a frequency of 440 Hz. Does that mean that a flute playing concert A is emitting low frequency electromagnetic radiation which we "hear"? A: Yes, sound is an electromagnetic wave. B: No, sound is not an electromagnetic wave. Answer: Sound is a pressure wave in air. It has nothing to do with EM waves - they are different things "waving".

6 (EMWaves-6) Under cover of night, a Girl scout signals her friends on a distant hill by alternately dimming and brightening her flashlight. This signal is most accurately described as.. A: Frequency modulation. B: Amplitude modulation. Answer: frequency modulation in the first case, amplitude modulation in the second case. The brightness of the light is a measure of the amplitude of the electric field in the E/M waves from the light. If the Girlscout signaled by changing the color of the light (keeping the brightness fixed), this would be frequency modulation, because different frequencies of visible light correspond to different colors. dim bright less bright AM FM

7 (EMWaves-7) The electric fields of two EM waves are described by E 1 = E 10 cos(k 1 x ω 1 t)ˆ j. E 2 = E 20 cos(k 2 x ω 2 t)ˆ j The "wave number" k of wave 1 is larger than that of wave 2, k 1 > k 2. Which wave has the larger frequency f? A: Wave 1 B: Wave 2 C: impossible to tell/don't know. Answer: k = 2 pi/lambda. So larger k means SMALLER lambda. But since lambda*frequency=c, then smaller lambda means BIGGER frequency. So larger k means larger frequency. (In fact, omega = c*k. Can you work this out for yourself, starting from the relation of omega to frequency f, and k to lambda?) For either wave, at x=0 and t=0, what is the direction of the B-field? A: + x ˆ B: y ˆ C: x ˆ E: none of these/don't know. D: + ˆ z Answer: E points in the j direction, but it TRAVELS in the x direction (I see that from the functional form, as t increases, x increases if you follow a "spot" on the wave. I know that E x B tells me the direction an EM wave travels, so B must point in the + z direction.

8 (EMWaves-8) A radio transmitter has a vertical antenna. I) Which is best for the receiving antenna? A: It should also be vertical B: It should be horizontal C: Makes absolutely no difference Answer: A is correct: the electric field will be strongest in the direction of oscillation of the electrons. A vertical antenna tends to "polarize" the emitted radiation, and so you'll pick it up best if you orient the antennae parallel each other. II) Same question, but for the case where you are using a "loop" antenna? Note: Loop antennas use the Faraday effect: a changing B field through the loop induces an EMF (which makes currents flow) Answer: Same logic - you want the antennas parallel.

9 (EMWaves-9) A point souce of radiation emits power P o isotropically (in all directions uniformly). A detector of area a d is located a distance R away from the source. What is the power p received by the detector? A: P o 4πR 2 a d R detector a d B: P o a d 2 R 2 P o C: P o a dr E: Don t know D: P o πr 2 a d Po Answer: R a d. At a distance R from the source, the power P 2 o is now spread out uniformly 4π over a sphere of radius R, area 4πR 2. So at distance R, the energy flux or intensity, which is power/area is P o /4πR 2. The power received by the detector is power area Po area of detector = R a 4 2 π d

10 (EMWaves-10) Two radio dishes receive signals from a radio station sending out radio waves in all directions with power P. Dish 2 is twice as far away as Dish 1, but has twice the diameter. Dish 1 Dish 2 Which dish receives more power? A: Dish 1 B: Dish 2 C: Both receive the same power Answer: both receive the same power. Dish 2 has twice the diameter, but 4 times the area (area of disk = πd 2 /4). The energy flux = power/area at any distance R is P o /4πR 2. So at Disk 2, the flux is 1/4 the flux at Disk 1. Power received = (power/area) (detector area) (see concept test above) The factor 4 increase in disk area is just cancelled by the factor of 1/4 decreased in flux.

11 (EMWaves-11) Light is sent through polarizer #1. It leaves with the electric field polarized in the plane of the board. Polarizer #2 is set with its axis 180 degrees from the axis of polarizer #1. What is the ratio of the light intensity leaving the second polarizer to the light intensity entering the second polarizer? First Polarizer Second Polarizer Unpolarized Light incident on Second Polarizer Light exits. A: zero B: 1/2 C: 1/4 D: 1 Answer: 1. Polarization refers to an axis, the "up" or "down" arrow doesn't really mean anything, just the line it represents matters!

12 (EMWaves-12) An unpolarized beam of light passes through 2 polaroid filters oriented at 45 o with respect to each other. The intensity of the original beam is I o. What is the intensity of the light coming through both filters? I o c A: (1/2)Io B: (1/4)Io C: (1/8)Io D: (1/16)Io E: None/don't know Hints: cos45 o = 1/ 2. The average of cos 2 θ, (over all values of θ) is cos 2 θ =1 / 2.) Answer: (1/4)I o After the first filter, the intensity is I 1 = I o (1/2). Reason: If the original beam were polarized, then after passing through one filter the intensity would be I 1 = I o cos 2 θ, but here the original beam is unpolarized so all angles θ are in the beam. The intensity after one filter is then I 1 = I o (cos 2 θ) average = I o (1/2). After the second filter the intensity is I 2 = I 1 cos 2 45 = I 1 (1/2) = I o (1/2)(1/2) = I o (1/4).

13 (EMWaves-13) A polarized beam of light passes through three ideal polaroid filters. The filters, in order 1 st, 2 nd, 3 rd, are tilted at 0 o, 45 o, and 90 o with respect to the incoming beam's axis. Does any light get through the all the filters and come out the other side? E c A: Some light gets through. B: No light gets through. Answer: some light gets through. If the 2 nd filter were not present, then no light would get thru. After the first filter, the E-field is unchanged, E 1 =Eo. After the 2 nd filter the E-field is E 2 = E of cos(45 o ). After the third filter the E-field is E 3 = E 2 cos(45o)= E of cos(45 o ) cos(45 o ) = E o /2. The intensity of the light (intensity I = energy flux S ) is proportional to the E 2, so the final intensity is I o /4.

14 (EMWaves-14) A polarized beam of light passes through a series of ideal polaroid filters, whose axes are vary continuously from along the E-field of the original beam to perpendicular to the E-field direction, as shown. Does any light get through the all the filters and come out the other side? E c A: Some light gets through. B: No light gets through. Answer: See the last question - some light makes it through! In fact, it's a fun math puzzle. The more filters you put in like this, the more will make it through. This is a "light rotator"!

15 (EMWaves 15) A ray of light passes through 3 regions labeled I, II, and III, as shown. How do the indices of refraction of regions I and III compare? I II III 40 o 25o 25o 35o A: n I > n III B n I < n III C: n I = n III D: Impossible to tell. Answer: If niii was equal to ni, then the outgoing angle would be the same as the original, 40 degrees. Since it is smaller, the outgoing light is not bending up as much, so niii must be bigger than Ni (it must be closer to the Nii value, which is clearly bigger than Ni, because light from I to II bends towards the normal)

16 (EMWaves-16) A group of sprinters gather at point P on a parking lot bordering a beach. They must run across the parking lot to a point Q on the beach as quickly as possible. I) Which path from P to Q takes the least time? (Consider the relative speeds of the sprinters on the hard surface of the parking lot and on loose sand) P Parking lot A B C E D Beach (loose sand) Q.All paths take the same time. Answer: Path D. There's some compromise here: Path C is shorter, but you spend more distance (and thus WAY more time) in the slow sandy part. E might look tempting, but now the parking lot path is so much longer that you've wasted extra time there. There is always a "path of least time" which involves some compromise between shortest overall length and shortest length in the "slow" medium. If you could run equally fast in sand, then C would be correct. If you went SUPER slow in sand, then E would be correct. But in between, you'd pick path D. By the way - my dog Sasha knew this! If I threw the ball into the reservoir, she would run a path much like "D". Smart dog, eh? II) How about if you run from Q towards P? Answer: Same answer, makes not difference.

17 (EMWaves-17) A ray of light passes thru a sheet of glass that is thick at the bottom and thin at the top. Which way is the ray traveling after it has passed through the glass? glass, n=1.5 A B air, n=1 C A: bent toward the thick end B: straight through, no change in direction. C: bent toward the thin end Answer: Draw the normal at each boundary. Bend TOWARDS the normal at the first interface (which bends you up) and then on the far side of the glass, bend AWAY from the normal (which is AGAIN up) giving you path A.

18 (EMWaves-18) A light ray inside glass is totally internally reflected from an air-glass interface as shown. air glass The air surrounding the glass is replaced with water. With the same light ray in the glass, the total internal reflection will now... A: definitely not occur. B: definitely occur. C: not enough information to know/ don't know. Answer: TIR occurs when you go from larger n to smaller n. The bigger the difference, the more likely you are to have TIR. So if you replace air with water, TIR COULD still occur (if the angle is "grazing" enough, but isn't *guaranteed* to happen any more. In the limit that you use n=1.5 on the outside, then no TIR at all would occur. HINT: n(air)=1 n(water)=1.33 n(glass) = 1.5 (Actually, this depends on the glass type but use this value for the problem)