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# AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread] watch

1. (Original post by Boop.)
Can someone explain the answer to the rod in the two coils questions?
As the switch is closed a current flows through coil P so a magnetic field "grows" around that coil and passes through coil Q. Since a change in flux always induces an emf, an emf is therefore induced in coil Q. Coil Q is part of a complete circuit so an induced current flows through it, which shows up as a reading or "flick" on the ammeter arrow. (You don't need to state the direction of the flick as we weren't given enough information to determine.)

But as the magnetic field from P settles down, so does the part of that field going through Q. This means that the magnetic flux through Q is no longer changing now, so the reading on coil Q's ammeter goes back to 0.

Now, as the resistance in coil P's circuit is increased, the current through coil P gets smaller. This means a smaller emf is induced and therefore the magnetic field that is set up around P is reduced. Hence the magnetic flux through Q gets smaller. As this happens it moves in the opposite direction to when it originally entered through Q. (At least I think so! -- could be wrong), so there is an change in flux through Q but this time it is smaller than the initial change in flux, because the initial change of flux through Q went from 0 to X (say), but the second one goes from X to Y (say), where 0<Y<X. Therefore the second change in flux is X-Y and the first one is X-0, and X-0>X-Y, hence the second change in flux is smaller, hence a smaller reading on the ammeter. (But the direction of this reading is what I'm unsure about, although I think it would be in the opposite direction than the initial one because Y-X is a negative change of flux, whereas X-0 is positive.)
2. (Original post by NEWT0N)
As the switch is closed a current flows through coil P so a magnetic field "grows" around that coil and passes through coil Q. Since a change in flux always induces an emf, an emf is therefore induced in coil Q. Coil Q is part of a complete circuit so an induced current flows through it, which shows up as a reading or "flick" on the ammeter arrow. (You don't need to state the direction of the flick as we weren't given enough information to determine.)

But as the magnetic field from P settles down, so does the part of that field going through Q. This means that the magnetic flux through Q is no longer changing now, so the reading on coil Q's ammeter goes back to 0.

Now, as the resistance in coil P's circuit is increased, the current through coil P gets smaller. This means a smaller emf is induced and therefore the magnetic field that is set up around P is reduced. Hence the magnetic flux through Q gets smaller. As this happens it moves in the opposite direction to when it originally entered through Q. (At least I think so! -- could be wrong), so there is an change in flux through Q but this time it is smaller than the initial change in flux, because the initial change of flux through Q went from 0 to X (say), but the second one goes from X to Y (say), where 0<Y<X. Therefore the second change in flux is X-Y and the first one is X-0, and X-0>X-Y, hence the second change in flux is smaller, hence a smaller reading on the ammeter. (But the direction of this reading is what I'm unsure about, although I think it would be in the opposite direction than the initial one because Y-X is a negative change of flux, whereas X-0 is positive.)
What if you said when the resistance is higher, the deflection on the meter would be smaller?
3. (Original post by ChiefKeef)
Yooooooo was the first answer weight??
yes pretty sure it was weight
4. Good luck with phys5 guys. Im off for core 3 tomorrow

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5. What was T in Q5?
6. How marks for an A?
7. (Original post by Will177)
yes pretty sure it was weight
f is the rate of change of momentum

and w=mg which is in Newtons

and f is also newtons

8. (Original post by AR_95)
What if you said when the resistance is higher, the deflection on the meter would be smaller?
That would definitely get a mark, but I have a sneaking suspicion they also wanted you to talk about the direction of the arrow for the second mark because while we aren't given enough info to determine the absolute direction, we are given just enough to determine the direction of the arrow in the second part relative to that in the first part.

I am now fairly sure the direction would be opposite, because if we define change in flux as final flux minus initial flux, then by what I said in my previous post the first change in flux is X=X>0, while the second one is Y-X<0 (since Y<X -- see my last post)
9. They were mostly on V = IR, so when there was an increase in resistance there would be a lower current because the voltage remains constant
10. (Original post by NEWT0N)
If I'm honest I would say bang on an A or slightly higher, but a mid A at the most. A lot of people I spoke to thought last year's paper was harder and that was 53 for an A.
Jesus christ i'm screwed then. I think i've dropped 4-6 in multi-choice. 4 on damping question, 1 mark for current graph(unless they scrap that), 3 on magnet question and 2 on last question. Hopefully I can get a mid A* in unit 5 and medical, last year the average ums for unit 5 in our class was 110 while in unit 4 was only 82. Hopefully I can do it
11. Did the ring affect damping? And how? Never done anything like it
Plus I put weight, cos rate of change of momentum = force, and weight is a force
12. If you ticked the wrong box for damping would you get ecf for explanation or fully zero marks?
13. (Original post by StarvingAutist)
Damping affects amplitude, not time period. Time period is irrelevant.
Conservation of momentum means that the ringed bob experiences less damping because it has higher mass so gives the air molecules less velocity (assuming the same surface area).
Light damping does affect time period, but only negligibly. If the magnitude of the damping is proportional to the velocity of the pendulum (it's actually proportional to velocity^2 when in air), with a constant of proportionality p, then the time period is inversely proportional to sqrt(1-p^2). They didn't want to hear about it in the question though.

(Original post by poee)
THIS IS WRONG.

it is not halved.

at least I think it's not,, each plate has Q charge
It depends on what the question asked. I hope it asked for the number of electrons on the plate in total in which case the negative plate has charge -Q and the positive plate +Q. Was this a multiple choice question or a written one? I can't remember...
14. Actually, I'm saying it was an easy paper, but I've never seen questions like the ring, or the steel fibre metal bla thing come up before so maybe it wasnt
15. (Original post by ssargithan)
If you ticked the wrong box for damping would you get ecf for explanation or fully zero marks?
Fully 0 me thinks
16. (Original post by ssargithan)
If you ticked the wrong box for damping would you get ecf for explanation or fully zero marks?
Seeing as if you ticked the wrong box your explanation couldn't be right because what you were explaining was incorrect, I would think 0 unfortunately. Possibly they might award marks if you did manage to make valid points somehow, but I wouldn't expect any ecf :/
17. (Original post by JizzaStanger)
Light damping does affect time period, but only negligibly. If the magnitude of the damping is proportional to the velocity of the pendulum (it's actually proportional to velocity^2 when in air), with a constant of proportionality p, then the time period is inversely proportional to sqrt(1-p^2). They didn't want to hear about it in the question though.
Yeah I see. I didn't write about T in the question anyway, given that the SHM equations are basically irrelevant.
18. (Original post by k9000)
Doesn't one plate have all the electrons?
Yeah, I agree with you. I'm pretty sure that all the electrons build up on one plate, and the electrons leave the other plate, which is what causes the potential difference
19. what order were the questions in? resonance and damping, then was it electric fields? then mag fields then circular motion?
20. (Original post by connorpayne)
Yeah, I agree with you. I'm pretty sure that all the electrons build up on one plate, and the electrons leave the other plate, which is what causes the potential difference
Ikr, if they had equal numbers there would be zero potential

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