Suggested Solutions (Including Photonics as Detailed Study)

[Chavi]

Question 10 in light was bloody hard.  . . needless say, I figured out the answer in the last 2 minutes!!! phew, , ,

[m@tty]

I agree with everythign but light and matter Q2. I think its option 2, because I saw in a trial paper solution I did once that "intensity proportional to photoelectric current" is an explanation that is supported by both wave and particle model.

The question clearly stated that the wave model could account for 2 of the observations but not a third. Clearly then they only have one correct answer in mind.
I agree that it was 2.

[DuBistKomisch]

I agree with everythign but light and matter Q2. I think its option 2, because I saw in a trial paper solution I did once that "intensity proportional to photoelectric current" is an explanation that is supported by both wave and particle model.

For light and matter question 2, I thought that only 2 was acceptable - as the wave theory does stipulate that photo current depends on the intensity of incident light. . .
but it CANNOT explain why Ek is solely dependent on frequency. . .

I guess the "only" in the second observation makes a difference. The explanation you would use is the same for each though so I guess I'll lose a mark or two.

[Chavi]

for 10 I pretty sure voltage was -3V (negative)

Hmm, good point. I'm not sure if the sign should really matter though since we're not told the polarity of the voltmeter. In fact I don't remember being told whether it is measured at the voltmeter at all...

Voltage is a vector - because Flux is a vector (and Field is a vector)

[taiga]

I think it may be Lenz's law rather than faradays. Faraday's would not imply any direction, Lenz's law takes into account the fact that voltage is above and below the axis.

[DuBistKomisch]

I think it may be Lenz's law rather than faradays. Faraday's would not imply any direction, Lenz's law takes into account the fact that voltage is above and below the axis.

Wrong. Faraday's Law includes the negative sign which gives the sign of the induced emf. Lenz's law is for determining the direction of the current.

[Ovalteenies]

I think it may be Lenz's law rather than faradays. Faraday's would not imply any direction, Lenz's law takes into account the fact that voltage is above and below the axis.

According to Heineman Faraday's Law is the negative rate of change of flux, so it does predict the signs.
Lenz's law does not predict the magnitude of the voltage, which was a feature of the graph

[Chavi]

Only Faraday's law applied to that case. .

[waack]

for question 8 it did not state which polarity the voltmeter was in i drew above axis for 0-1 and half magnitude under for 2-4 i understand what you mean now partial marks u think??

[Linkage1992]

yeah you'll probably get partial marks.

[DuBistKomisch]

for question 8 it did not state which polarity the voltmeter was in i drew above axis for 0-1 and half magnitude under for 2-4 i understand what you mean now partial marks u think??

I'm starting to think it was pretty ambiguous as we were given no reference polarity for the voltmeter so they may accept either combination of signs as long as they are opposite. Don't take my word for it though.

[m@tty]

for question 8 it did not state which polarity the voltmeter was in i drew above axis for 0-1 and half magnitude under for 2-4 i understand what you mean now partial marks u think??

I'm starting to think it was pretty ambiguous as were given no reference polarity for the voltmeter so they may accept either combination of signs as long as they are opposite. Don't take my word for it though.

This is what I think too.

[SixWinged]

for question 8 it did not state which polarity the voltmeter was in i drew above axis for 0-1 and half magnitude under for 2-4 i understand what you mean now partial marks u think??

I'm starting to think it was pretty ambiguous as were given no reference polarity for the voltmeter so they may accept either combination of signs as long as they are opposite. Don't take my word for it though.

Agreed, if they don't, they'll have to accept Lenz's Law for the following question as it is used to take into account polarity. Seems more correct imo to accept either for the graph than the law.

[waack]

i think u might be right but u never know i even stated at the top assuming positive voltage is p-q lol oh well we shall see

[kyzoo]

=/ I was thinking about whehter you had to put the first section (t=0 to t=1) above or underneath the x-axis. I came to the conclusion that ti doesn't matter. You don't have a power source with a positive/ngative terminal in the diagram, therefore you can't tell which way si positive current and which way is negative.