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

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also, check this out from the spec:

yeh... the earlier unit.... top notch scheduling there OCR

Brilliant, having a bad enough time going through the whole of G495, nice to know I now have to revise for an exam that's in over a week just so I can pass one that's in 3 days. Well played OCR..

Reply 141

Rogercbinboy

Anyone have the June '08 legacy paper? The folder for jun '08 past papers on the OCR website doesn't have it in but I'm pretty sure it exists.

Cheers

Reply 142

Summerdays

Original post by Rogercbinboy

Anyone have the June '08 legacy paper? The folder for jun '08 past papers on the OCR website doesn't have it in but I'm pretty sure it exists.

Cheers

Apparently not :s

http://www.johnbright.conwy.sch.uk/depts/science/Y13/P-Y13_Past_Papers_Legacy.html

How many past papers has everybody done for this unit?

Reply 143

41jms

could somebody please explain motors and generators to me? just magnetism in generally really i find it hard to get my head around! Like whats the sequence of events.. what induces what etc etc? Id be so so grateful!!

Reply 144

Rogercbinboy

Original post by Summerdays

Apparently not :s

http://www.johnbright.conwy.sch.uk/depts/science/Y13/P-Y13_Past_Papers_Legacy.html

How many past papers has everybody done for this unit?

Ah right weird.

Been revising with the majority of legacy papers and past papers over the last few weeks but I'm redoing them today under exam conditions - want to get all the papers from jan08 to jan11 done including the spec'.

Reply 145

Summerdays

Original post by 41jms

For a motor you are using electrical energy to create mechanical energy. For example, in a d.c motor, you have commutators that ensure that the directon of the current changes every half a rotation. In a d.c rototr it is a coil of wire that is rotating in a magnetic field, but only the part of the coil that is perpendicular to the field/ lines. This provides a torque and causes the coil to rotate. As the coil rotates, it begins to cut through field lines. This causes a 'back emf' to be induced that oppose the original voltage. This reduces the current and thus BIL becomes less, thus the torque becomes less, which means that the rotational kinetic energy of the coil (which i what you want) becomes less. It is better to load the motor so that the angular velocity of the rotor becomes less, and thus the amount of back emf induced becomes les.

With a generator you are using mechanical energy to rotate a coil in order for emf to be induced in the coil.

Reply 146

Mikkels88

do we need to know back emf for this exam? i remember hearing about it but its not in CGP or the spec i dont think... :smile:

could you explain it in any more detail? <3

Reply 147

41jms

Original post by Summerdays

i dont really understand, thanks for explaining though. looks like ill fail this exam.. you can all thank me for bringing the grade boundary down!!

Could you explain/answer this Q..

Protons in the beam move at the speed of light c. At this speed energy E is given by E = pc. Show that a maximum magnetic field strength of 8.3T is more than enough to keep protons of energy 7Tev moving in a circle of radius 4.3km.
e= 1.6x10^-19 C
c = 3x10^8 m/s

Reply 148

Mikkels88

I would do it this way, but from the info youve been given it would seem they want a slightly different method... maybe i need to see the rest of the question? anyways, they clearly want you to use E = pc, but im using E = mc^2 , idk if it works out the same lets see... (they are equal by the way as v = c and p=mv but then why would they give that formula? OCR moment maybe?)
anyhoo:

for circular motion, F = mv^2 / r , (where v is c as they are travelling at the speed of light)
mc^2 = 7TeV, as this is their energy, which is 1.12*10-6 Joules
so the centripetal force is F = 1.12*10-6 / 4300 which equals 2.60*10-10 N.
The force on a charge in a field is also given by F = Bev, and so with a field strength of 8.3T, the force on the protons is F = 8.3 * 1.6*10-19 * c which equals 3.98*10-10

you can see that this force is greater than the force required for the circular motion (the 2.6*10-10) and so is more than enough to keep the protons in a circle

=============================

no idea if thats right to be honest but it seems to work :smile:

(edited 12 years ago)

Reply 149

Summerdays

Original post by 41jms

i dont really understand, thanks for explaining though. looks like ill fail this exam.. you can all thank me for bringing the grade boundary down!!

Here's an explanation from my teacher:

Alternators and dynamos (ie generators) are used to generate electric currents. An external source (eg the car engine) is used to turn the rotor, in order to generate an emf to drive a current. The mechanical energy used to turn the rotor is converted into electrical energy. That is the whole point of them. So it is not correct to talk about a 'back emf' for an alternator, because the point of the alternator is to generate an emf to drive a current.

You only talk about a 'back emf' in the case of a motor, when you are driving a current through the rotor by an external emf, and the point is to convert electrical energy into mechanical energy ( the motor driving a washing machine drum, for example). In this case, the rotor turning in the field does generate a back emf, which opposes your external emf (hence the term 'back' emf) and makes it hard to drive the current through the motor. That is why electrical energy needs to be put in (to drive the current against the back emf), which is how you can get mechanical energy out.

Reply 150

41jms

Original post by Mikkels88

probably is, thank you!!
iv learnt more in that 30 seconds of reading what you wrote than trying to make sense of markscheme for the past 4 hours :frown:

thanks again.

Any chance you could explain q12 on this paper for me?
http://www.johnbright.conwy.sch.uk/depts/science/Content/Past_Papers/Y13/Physics/2864January2009.pdf

iv looked at the mark scheme i just dont understand :frown:

(weve had a really rubbish timetable for physics, like 2/3 hours a week and theres only 3 other people in my class.. arhhhh iv left this far too late :frown:

)

Reply 151

Summerdays

Original post by 41jms

mv^2/r = Bev so B = mv/er = p/er

p = 7x10^12x1.6x10^-19/c = 3.73x10^-15

B= 3.73x10^-15/(1.6x10^-19x4.3x10^3) = 5.43T

8.3T > 5.43T so 8.3T is more than enough.

Reply 152

Summerdays

Original post by Mikkels88

do we need to know back emf for this exam? i remember hearing about it but its not in CGP or the spec i dont think... :smile:

could you explain it in any more detail? <3

"Back emf" is a pain because it is opposing the original emf that is used to drive a current through the coil. The resultant emf is less due to the back emf, and thus the current flowing through the coil is less, and thus BIL (which detremines the torque and thus the rotational kinetic energy) is less.

Reply 153

Helsy

Original post by Summerdays

Hang on... "back emf"??? What on earth is that? I've not heard of that, where's it from! :eek:

Reply 154

Summerdays

Original post by Helsy

Hang on... "back emf"??? What on earth is that? I've not heard of that, where's it from! :eek:

It's not something you really need to know :smile:

EDIT: But it is an emf that is induced that works against the original emf.

(edited 12 years ago)

Reply 155

Mikkels88

Original post by 41jms

Any chance you could explain q12 on this paper for me?
http://www.johnbright.conwy.sch.uk/depts/science/Content/Past_Papers/Y13/Physics/2864January2009.pdf

iv looked at the mark scheme i just dont understand :frown:

(weve had a really rubbish timetable for physics, like 2/3 hours a week and theres only 3 other people in my class.. arhhhh iv left this far too late :frown:

)

I have absolutely no idea what this 'grid' is but im assuming its some sort of conducting mesh or something, because they seem to have a circuit set up either side of it...
for
a)i) its simply a uniform field from the 'grid' to the collector plate, and so arrowed lines go from + to - so from grid to collector, must be the same distance apart to show its uniform

ii) as the voltage is twice as high (6V) the arrowed lines must be twice as close together, to give twice the field strength. The + to - direction is grid to cathode so arrowed lines from grid to cathode, equal distance apart (half the distance apart as your 3 on the other side)

b) ok this is seriously confusing, its terribly explained in the question and im not sure i follow but my understanding would be this:
from 0V - 3V : i think because the electrons are accelerated towards the grid, they gain KE and so all their initial EPE transfers to KE. If V < 3V, they wont have enough KE to overcome the repulsion on the other side of the grid (as that side has a voltage of 3V in the other direction as such, so will slow the electrons down)

from 3V - 5V : the electrons are simply gaining more KE by the time they reach the grid (as they start with more EPE) and so more have enough energy to reach the collector

at 5V : this is a little tricky. I had an explanation but after looking at the MS realised i am slightly wrong. My understanding is now:
electrons still reach the collector with less than 5V starting energy, and so they collide elastically with the mercury atoms (just bounce around without losing energy effectively) until they finally get through and reach the detector.
However, at 5V starting energy, very few get through. This is because they are colliding with mercury atoms and transferring all their energy to them, by exciting the mercury electrons up an energy level (remember energy levels are discrete and so only a CERTAIN energy, 5eV here, can excite electrons up a level).
Above 5V, they simply have more EPE to start and so more KE at the grid (like from 3V - 5V) and so more get through. (but alot are still stopped by mercury atoms if they happen to collide when their energy is 5eV)

does that make it a bit clearer for you? Im sorry i dont think I entirely understand it, but its a classic OCR question where they try and verify a simple understanding by asking a stupid bloody question.

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and cheers Summerdays, some good stuff there :smile:

i think i get it from your explanation above in green, lays it out nice and cleary. basically similar to the idea of eddy currents? ie what you are doing is inducing something that is opposing what you are trying to achieve (via energy loss).