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OCR PHYSICS B G495~ 18th June 2015 AM ~ A2 Physics

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Reply 80
Original post by RoryV
I need an A* in physics to get into Uni. Based on my coursework raw marks, and how many UMS i think i'll have got in G494, i reckon i need 138UMS in G495.
Anyone know roughly what sort of raw mark 138UMS would translate to? I was thinking maybe around 80/100 but not too sure


JJust to check, you do know to get an A* you need atleast 90% ums on each A2 unit.
(edited 8 years ago)
Original post by Mars256
JJust to check, you do know to get an A* you need atleast 90% ums on each A2 unit.


90% average across both
You could get 95% then 85% and get an A*


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Reply 82
Why is the answer A?
Original post by Mutleybm1996
I hear G484 was horrible this year
With even top students barely scraping 40


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Guess it's just physics in general! And I'm going to be doing it at uni, I enjoy learning about it but I hate the examinations on it...
Original post by Ghengis
Why is the answer A?


Flux is uniform in the whole iron core, it's flux density that is different due to the change in the number of coils on either side of the transformer. It is not flux linkage as there are different numbers of coils on either side of the transformer, hence flux linkage isn't the same.
Reply 85
Original post by RealTime
Flux is uniform in the whole iron core, it's flux density that is different due to the change in the number of coils on either side of the transformer. It is not flux linkage as there are different numbers of coils on either side of the transformer, hence flux linkage isn't the same.


thanks
Original post by urz13
How/why does the emission of secondary electrons from the sample limit the resolution of the electron microscope?
Also, is the resolution essentially determined by the size of the diffraction spots, which theoretically could be as small as the wavelength of the electrons? Can't quite get my head around these ideas which seem to be suggested by the questions on P1, so if someone could clear it up for me that'd be great :smile:


I'm unsure about the secondary electrons thing, I don't want to give you a potentially wrong answer.

However, resolution is the smallest resolvable thing, which in terms of visible light would be half the wavelength of the smallest wavelength of light ~250nm. Diffraction spots are like the maxima and minima in Young's double slit experiment where the separation of maxima and minima corresponds to a factor of the wavelength of light, so if you can work out the wavelength from this then the resolution would be half of this calculated wavelength,
Reply 87
Original post by urz13
How/why does the emission of secondary electrons from the sample limit the resolution of the electron microscope?


What makes you say that?
According to wikipedia, a secondary electron detector in fact maximises the SEM's resolution.
(edited 8 years ago)
Reply 88
What's the formal definition of flux linkage?
Reply 89
Original post by STATER
What makes you say that?
According to wikipedia, a secondary electron detector in fact maximises the SEM's resolution.


This question:
A de Broglie wavelength of 5 pm for 50 kVelectrons suggests that the resolution of an electron microscope isconsiderably better than the best optical microscopes. What factors may limitthe resolution of an electron microscope in practice?

And the answer:
-Electrons all have negative charge so repel one another, limiting minimum spot size
-Electrons may cause local heating of sample that leads to secondary electrons.

It has me a bit confused!
Reply 90
Original post by urz13
This question:
A de Broglie wavelength of 5 pm for 50 kVelectrons suggests that the resolution of an electron microscope isconsiderably better than the best optical microscopes. What factors may limitthe resolution of an electron microscope in practice?

And the answer:
-Electrons all have negative charge so repel one another, limiting minimum spot size
-Electrons may cause local heating of sample that leads to secondary electrons.

It has me a bit confused!


I would just mention relativistic effects tbh. You can say that the it becomes impractical to decrease the de Broglie wavelength of the electron beyond a certain point due to relativistic effects. Seems like much less of a headache.
Reply 91
what is the relationship between no of turns in a coil and the flux density?
Reply 92
Can anyone help with this? Cheers

Use the relationship Bqr= energy of particle/speed of light

to show that the unit of magnetic field strength, the Tesla, is equal to N A^-1 m^-1
(edited 8 years ago)
Don't know about you guys but I'm just winging this exam, haven't done any past papers yet :/
Reply 94
Original post by Mutleybm1996
90% average across both
You could get 95% then 85% and get an A*


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..

just checked the spec, you're right, my bad
Reply 95
Original post by urz13
This question:
A de Broglie wavelength of 5 pm for 50 kVelectrons suggests that the resolution of an electron microscope isconsiderably better than the best optical microscopes. What factors may limitthe resolution of an electron microscope in practice?

And the answer:
-Electrons all have negative charge so repel one another, limiting minimum spot size
-Electrons may cause local heating of sample that leads to secondary electrons.

It has me a bit confused!


I think the problems with the secondary electrons in this case is that they are not being 'released' by the bombardment from the electron beam but by heating of the sample and this therefore interfers with the secondary electrons produced by the beam, which is what you are trying to measure??? this may be possibly the reason?
Original post by Rhetorical Hips
Don't know about you guys but I'm just winging this exam, haven't done any past papers yet :/


oooh :redface: good luck! Hopefully it is a reasonable exam
Original post by Rhetorical Hips
Don't know about you guys but I'm just winging this exam, haven't done any past papers yet :/


oooh :eek: good luck! Hopefully it is a reasonable exam
Reply 98
Original post by Ghengis
what is the relationship between no of turns in a coil and the flux density?


there isn't one, i don't think, flux density is equal to flux divided by cross-sectional area
Original post by urz13
Can anyone help with this? Cheers

Use the relationship Bqr= energy of particle/speed of light

to show that the unit of magnetic field strength, the Tesla, is equal to N A^-1 m^-1


ImageUploadedByStudent Room1434546836.658241.jpg

Gone wrong somewhere


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