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OCR Physics B G495 Field and Particle Pictures June 21st 2011 Exam Thread

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Reply 280
Another version of fleming's left hand rule (that uses the right hand) is: Index finger = direction of length of the wire/ direction of velocity. Middle finger = direction of magnetic field. Thumb = direction of thumb. It comes from the right hand rule when you are using a cross product. The cross porducts I am referring to is F = I(LxB) or F = q(vxB)
Reply 282
Original post by readyforresults
Its typical ocr b to expect you to assume they'll be working "9 till 5"

that's a gd one LOL I wish I were not working after 5 XD
Reply 283
getting a B isn't difficult, but most times you will drop between 5-10 marks depending on how stupid an OCR question is or how many there are, so that makes the A a little more challenging :colondollar:
Reply 284
Original post by Harry-AA
I put of physics for so long because of further maths :/ Kind of regretting it now. I need a B in this exam to get my A overall, that shouldn't be too much of an ask right? I hate this course, it sucks.


SAME HERE..FP2& 3 'S hard n yeah I need A for uni!!
Arh mega stressing out about this! i have an exam every day this week and one on monday too - not cool!
Mad cramming session now!

Anyone got any ideas about the case study?
xxx
All the equations like:

F=kQq/r^2
Electric potential = kQq/r
Electric field strength = kQ/r^2
Potential difference = kQ/r

are they meant to be negatives? ie. F=-kQq/r^2
Guys, has any one uploaded any notes/questions for section C...


Thanks
Original post by 41jms

70 beats per second means that theres one beat every 0.86 seconds. The reolution of the watr clock is essentially one tenth of a pulse beat. So that's 0.86/10 = 0.086
Reply 290
Original post by Rogercbinboy
All the equations like:

F=kQq/r^2
Electric potential = kQq/r
Electric field strength = kQ/r^2
Potential difference = kQ/r

are they meant to be negatives? ie. F=-kQq/r^2


no I think because it can be a repulsive or attractive force depending on whether q is positive or negatively charged. :smile:
Original post by Rogercbinboy
All the equations like:

F=kQq/r^2
Electric potential = kQq/r
Electric field strength = kQ/r^2
Potential difference = kQ/r

are they meant to be negatives? ie. F=-kQq/r^2


Hm, don't think so?
the sign is told in the sign of the charges surely?

Given that: Repulsive charges are positive and Attractive forces are negative (From CGP Revision guide)
e.g. to test:
Repulsive - two negatives make a positive, so overall positive
Attractive - One negative one positive, so overall negative
So I don't think they do need negatives?
xx
Any 1 have the question paper for jan 2011???


i already have the ms ... if any 1 needs give me a shout and ill upload... but its the QS... either upload it link it to me please

thanks
Original post by Rogercbinboy
All the equations like:

F=kQq/r^2
Electric potential = kQq/r
Electric field strength = kQ/r^2
Potential difference = kQ/r

are they meant to be negatives? ie. F=-kQq/r^2


Potential difference is meant to be negative (when the two charges have opposite charges) because we say that the potential of the test charge is zero at infinity, and thus the potential of the test charge at a point in the electric field is negative. BTW KQq/r = electric potential ENERGY. This is also negative (when the two charges are opposite) because we say that the energy of the test charge is zero at infinity. Positive kinetic energy is needed in order for the test charge to do work against the attractive force (in order for the test charge to reach infinity.)

Electric field strength is negative when the charge of the point charge is negative, and positive for a positive point charge :smile:
(edited 12 years ago)
stupid question probably, but how do you use this equation:

emf = -d(NI)/dt

where NI is flux linkage (I don't have a phi key...)

I just get confused where to put the numbers?
Ironic really as I'm a further maths student, oh well haha
Reply 295
Original post by Geekchic17
stupid question probably, but how do you use this equation:

emf = -d(NI)/dt

where NI is flux linkage (I don't have a phi key...)

I just get confused where to put the numbers?
Ironic really as I'm a further maths student, oh well haha


Well...
emf=voltage
N(phi)=flux linkage which is Number of turns*Flux
dt=change in time


Sometimes you are given the rate of change of Flux d(phi)/dt
You then just simply multiply this by N (number of turns) to get the rate of change of flux linkage. :smile:
Reply 296
Original post by Soppe
Any 1 have the question paper for jan 2011???


i already have the ms ... if any 1 needs give me a shout and ill upload... but its the QS... either upload it link it to me please

thanks


heres all the latest g495 papers...

http://www.johnbright.conwy.sch.uk/depts/science/Y13/P-Y13_Past_Papers.html
Reply 297
Which way does current flow, positive to negative? I heard this is what scientists thought ages ago and they didn't bother to change it when it's actually negative to positive?
Original post by Soppe
Any 1 have the question paper for jan 2011???


i already have the ms ... if any 1 needs give me a shout and ill upload... but its the QS... either upload it link it to me please

thanks


Here you go :smile:
Question paper:

http://www.johnbright.conwy.sch.uk/depts/science/Content/Past_Papers/Y13/Physics/G495Jan11.pdf

and for everyone else the mark scheme:

http://www.johnbright.conwy.sch.uk/depts/science/Content/Past_Papers/Y13/Physics/2011JanuaryMSA2.pdf

:smile: xx
Reply 299
Hopefully OCR redeem themselves a little by making G495 easy and simple, and testing us on what we actually learn, and then making you work for your marks in G494.

A man can dream.

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