Electromagnetism..BR

[BeatMCAT]

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So, 8.8b in my book:

To increase the oscillation frequency of an electron (isolated in vacuum) that oscillates cuz it is stimulated by an electromagnetic wave, one must:

1) decrease the wavelength of the wave (OBVIOUSLY)
2)increase the electric field amplitude of the wave (huh? not right)
3)increase the magnetic filed amplitude of the wave (damn..its fancy but not right)
4) increase the speed of the wave (WHY IS THIS WRONG?)

from above, i know 1 is right but why is 4 wrong. While i'm typing, i think i realized the answer but just want to make sure.
is it cuz its in a vacuum and speed is 3.0 x 10^8 m/s in vacuum always. Is that the reason why 4 is wrong? Thanks!

 

[indianjatt]

Exactly. In order to increase the frequency you can't increase the speed of the wave because the speed is only dependent on the medium (i.e. you can't change it without changing mediums and in this case the medium didn't change so that is not an option).

 

[Majik]

1.) Frequency of the wave is determined by the source. Changing the wavelength will change the velocity of the wave, but the frequency will remain constant. wave velocity=(frequency)(wavelength)
2.) The amplitude of the wave is independent of both the frequency and the period of the wave. Changing the amplitude will not affect the frequency or period. What we can say if a wave is displaced further from equilibrium (higher amplitude) is that it has more energy, however this alone will not cause the wave to oscillate any differently. It will propagate at the same pace.
3.) For the same reason 2 is wrong. An EM wave consists of an E-field and B-field perpendicular to each other. For an EM wave, the amplitude is the E-field and B-field (both perpendicular to each other and the direction of motion). In other words, an EM wave is a transverse wave.
4.) Increasing the speed of the wave will increase the wavelength of the wave, but will not change the frequency.

The reason the answer is D is because none of these changes alter the frequency of the source being emitted. The only way to alter the source frequency is if the source emitting it changed it. For example, "red" light and "green" light both have very different frequencies. Green light is higher in energy than red light. This in turn would oscillate the electron more frequently, the above changes will not. So if one of the choices mentioned that the energy of the light source changed, (E=hf), changing an energy must mean that the frequency changed as well (h is a constant).

Hope this helps.

 

[BerkReviewTeach]

For example, "red" light and "green" light both have very different frequencies.

They also have different wavelengths, which invalidates your explanation for eliminating #1 above. The wavelength is also set by the source, while the speed depends on the medium.

To the OP
Question 8.8b deals with the right-hand rule. One short cut to help is that if you want the particle to be deflected out of the plane of the page, then both the particle velocity and the B-field have to be in directions perpendicular to the normal to the page. In other words, they have to be in directions within the plane of the page. This should help to get rid of choices B and C right away without having to actually do the right-hand rule.

 

[Majik]

I understand there's an inverse relationship between frequency and wavelength when velocity is fixed (as is the case in a vacuum), however, I'm having trouble understanding how you can decrease the wavelength of an EM wave within a given medium. For a transverse wave, like a rope tied to a tree, you could oscillate the wave more (increasing frequency decreases wavelength, while v stays constant since the medium remains the same). How would you go about doing this for an EM wave? Maybe I'm misinterpreting the answer choice but is choice A essentially saying, "By using a EM wave of different wavelength." If that's the case, I feel choice A is a bit misleading. Otherwise, I'm a bit confused.

 

[BerkReviewTeach]

I understand there's an inverse relationship between frequency and wavelength when velocity is fixed (as is the case in a vacuum), however, I'm having trouble understanding how you can decrease the wavelength of an EM wave within a given medium. Maybe I'm misinterpreting the answer choice but is choice A essentially saying, "By using a EM wave of different wavelength." If that's the case, I feel choice A is a bit misleading. Otherwise, I'm a bit confused.

Think about bouncing a garden hose up and down and the wave that eminates from it. With the garden hose, if you bounce it up and down slowly (at a low frequency), then the wave you send out can travel pretty far before you start the next pulse. In this scenario, the wave speed through the hose is constant, but the freqeuncy and distance between pulses (wavelength) is up to you. As you bounce it up and down at a greater cadence, the frequency is increasing causing the wavelength (distance it travels before the next pulse) to decrease. This visually explains that speed is constant with frequency and wavelength being inverses that are up to the source.

So now let's take your green vs. red light example. Let's say both lights strike that electron in the question. The red light is generated by a lower energy process than green light, which results in a lower frequency and longer wavelength for red light than green light. When red light strikes the electron, it will transfer its energy and thus its oscillation frequency. Using green light rather than red light, it will transfer more energy and thus give the electron a greater oscillation frequency. This makes choice 1 a valid statement.

For this question, choice 1 is basically saying giving the electron a higher frequency can be done by bombarding it with a photon of shorter wavelength (which equates to higher frequency and higher energy).

As an aside, the answers for the b-questions are at the end of your book. Depending on how old your book is, the answers are in a different place and of a different level of detail (ranging from letters only to explanations). In the new physics book (from late 2010 and 2011), the b-answers are all printed with the question.

 

[BeatMCAT]

Thanks for the explainations!!! I have the old book, that only has the answer choices and no explainations, so this def helped!

 

[Majik]

Think about bouncing a garden hose up and down and the wave that eminates from it. With the garden hose, if you bounce it up and down slowly (at a low frequency), then the wave you send out can travel pretty far before you start the next pulse. In this scenario, the wave speed through the hose is constant, but the freqeuncy and distance between pulses (wavelength) is up to you. As you bounce it up and down at a greater cadence, the frequency is increasing causing the wavelength (distance it travels before the next pulse) to decrease. This visually explains that speed is constant with frequency and wavelength being inverses that are up to the source.

So now let's take your green vs. red light example. Let's say both lights strike that electron in the question. The red light is generated by a lower energy process than green light, which results in a lower frequency and longer wavelength for red light than green light. When red light strikes the electron, it will transfer its energy and thus its oscillation frequency. Using green light rather than red light, it will transfer more energy and thus give the electron a greater oscillation frequency. This makes choice 1 a valid statement.

For this question, choice 1 is basically saying giving the electron a higher frequency can be done by bombarding it with a photon of shorter wavelength (which equates to higher frequency and higher energy).

As an aside, the answers for the b-questions are at the end of your book. Depending on how old your book is, the answers are in a different place and of a different level of detail (ranging from letters only to explanations). In the new physics book (from late 2010 and 2011), the b-answers are all printed with the question.

Thank you for your explanation. My interpretation was that (for example), red light in a vacuum has a fixed frequency and wavelength. Within this same medium, the experimenter cannot change the frequency by decreasing the wavelength, but rather the wavelength decreases as a result of increasing the frequency of oscillation. This is essentially what happens when you use a different (more energetic) frequency of light.

If on the other hand, the experimenter passed light through some other medium (say from a vacuum to glass for example), the speed of light would decrease and so would the wavelength. The frequency however, would remain fixed because the oscillation of the wave is determined by source which emitted it. This would have no effect on the oscillation of the electron.

Even if light was being passed through different mediums, the electron is located in a vacuum. Therefore, light passing from some other medium to a vacuum (where speed of light is at its maximum) would result in an increase, not decrease in wavelength. This is why I ruled out choice A.