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# AQA Physics PHYA4 - 20th June 2016 [Exam Discussion Thread] watch

1. (Original post by Imtheish)
It says calculate the revolution, and to do that you need to find the time first from he graph. The mark scheme says time is 40ms, and I'm asking why it is. It's the June 2013 paper
Isn't 1 revolution of flux linkage just one complete sine curve, a maximum and a minimum which takes 40ms to happen
2. (Original post by Imtheish)
Attachment 552468
Can someone explain why the time is 40. Please
I don't understand what time your referring to, but the answer for the first question is 1500Hz.
1 revolution takes 40ms
in one minute there's 1500 revolutions (60 / 40x10^-3)
3. (Original post by ReeceFraser)
I'm doing June 2014 section B and the capacitance question says that a capicitor is charged at a constant current for 60 s and the p.d becomes 4.4v, the question is to find the resistance after 30s. In the mark scheme is says the p.d after 30s is 2.2v but i thought that capicitors didnt charge uniformly. Is it only uniform becuase the current is constant?
It's being charged by a constant current.
V=Q/C
V=It/C

So V is proportional to t so as t is halved, V will also be halved to 2.2V.

The current is still the same in the new circuit and that is the point of the variable resistor, to allow different resistances to give the same current (4.5uA). So as the capacitor is charging the resistance is decreased to keep I the same as before. A normal discharge/charging circuit will have a constant resistance.

Hope that's simple enough .
4. (Original post by splashywill)
Unit 4 Physics
A* A B
2015 54 49 45
2014 61 53 48
2013 57 53 47
2012 56 50 44
2011 54 48 42

These are all for June papers..... just and interesting look.
Interesting to see that the A* conversion point has been manually adjusted for the last three papers. Usually the difference between A and A* should be equal to the difference between A and B. I have never seen this happen on any OCR unit. Further evidence (after last year's ridiculously low boundaries for the AS units) that AQA are incapable of setting papers which effectively target the full range of grades?
5. (Original post by particlestudent)
It's being charged by a constant current.
V=Q/C
V=It/C

So V is proportional to t so as t is halved, V will also be halved to 2.2V.

The current is still the same in the new circuit and that is the point of the variable resistor, to allow different resistances to give the same current (4.5uA). So as the capacitor is charging the resistance is decreased to keep I the same as before. A normal discharge/charging circuit will have a constant resistance.

Hope that's simple enough .
Cheers man. Hows this for the 6 mark question following:

When the switch is at position 1 electrons will flow from B to Q and build up a negative charge on plate Q. This negative charge will repel electrons from plate P causing it to become positively charged and the p.d across the plates to increase. As the p.d across the plates increases the energy stored in the capcitor will increase and the current flowing through the circuit will decrease. When the capacitor is fully charged the current will become zero and the p.d between across the resistor will be zero.When the switch is at position 2 the capacitor will discharge with electrons flowing from Q to P. Initially the p.d across the resistor will be maximum meaning maximum current will flow. As the charge of the capacitor decreases the p.d will decrease causing the rate of flow to decrease. This will continue until the capicitor has lost all it charge and the current becomes zero.
6. (Original post by ReeceFraser)
Cheers man. Hows this for the 6 mark question following:

When the switch is at position 1 electrons will flow from B to Q and build up a negative charge on plate Q. This negative charge will repel electrons from plate P causing it to become positively charged and the p.d across the plates to increase. As the p.d across the plates increases the energy stored in the capcitor will increase and the current flowing through the circuit will decrease. When the capacitor is fully charged the current will become zero and the p.d between across the resistor will be zero.When the switch is at position 2 the capacitor will discharge with electrons flowing from Q to P. Initially the p.d across the resistor will be maximum meaning maximum current will flow. As the charge of the capacitor decreases the p.d will decrease causing the rate of flow to decrease. This will continue until the capicitor has lost all it charge and the current becomes zero.
That's fine . I remember doing this paper a couple of days ago and some of the points I remember seeing on the mark scheme were just absurd.
7. Any predictions for the 6 marker ?
8. (Original post by rjdoran)
Magnetic flux=BA
Area which the rod moves through= length * distance covered
(where distance covered, d, equals velocity * time taken, from v=s/t)
Magnetic flux equals BLvt, emf induced equals the rate of change of flux density.
Therefore EMF=BLvt/t, the t's cancel, so the emf=BLv.
Thanks for replying, but I still don't understand why the wire has to move to the right and not vertically up?
9. (Original post by cmrol)
Isn't 1 revolution of flux linkage just one complete sine curve, a maximum and a minimum which takes 40ms to happen
I get it now thanks
10. On my 6th PHYA4 paper today... What has my life come to ah😪
Hopefully this year they'll decide we no longer need 6 mark questions.. I always seem to loose marks on them..
Anyone have any predictions on this paper?
I'm guessing at some momentum/impulse/explosion and more Gravitation stuff this year as there was barely any last year.
Hopefully they will not have so much magnetic induction as that seemed to take up most of last years paper.

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11. Also are there any specific prior 2010 papers people would recommend?

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12. Is it right of me to say that potential gradients are parallel to the field lines?
13. (Original post by cowie)
On my 6th PHYA4 paper today... What has my life come to ah😪
Hopefully this year they'll decide we no longer need 6 mark questions.. I always seem to loose marks on them..
Anyone have any predictions on this paper?
I'm guessing at some momentum/impulse/explosion and more Gravitation stuff this year as there was barely any last year.
Hopefully they will not have so much magnetic induction as that seemed to take up most of last years paper.

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More EM induction please, less G fields - I love EM induction, but don't like G fields as much
14. (Original post by kingaaran)
More EM induction please, less G fields - I love EM induction, but don't like G fields as much
Only if it's just the calculations part. I hate the induction explanation questions!
15. (Original post by ¡Muy bien!)
Thanks for replying, but I still don't understand why the wire has to move to the right and not vertically up?
It won't cut field lines if it moves vertically .
16. (Original post by kingaaran)
More EM induction please, less G fields - I love EM induction, but don't like G fields as much
More capacitors please, like a whole question on capacitors will be good!
17. Can someone please Calculate the angle in radians that the Earth spins through in one hour? I got 0.1047 radians if that right ?
18. (Original post by kother2015)
Can someone please Calculate the angle in radians that the Earth spins through in one hour? I got 0.1047 radians if that right ?
Nope, that's incorrect.

It spins 2(pi) radians in 24 hours so in one hour it would spin 1/12 (pi). Divide the 2(pi) by 24 to get it for one hour.

1/12 pi is around 0.262 radians.
19. can someone help with this question I keep getting -14.4J and have no idea what I'm doing wrong :/

Q1. An electron and a proton are 1.0 × 10–10 m apart. In the absence of any other charges, what is the electric potential energy of the electron?
A +2.3 × 10–18J
B –2.3 × 10–18J
C +2.3 × 10–18J
D –2.3 × 10–18J

https://08c6cd28b8bef288858e878e5745...c%20Fields.pdf
20. (Original post by Mentalmirz)
can someone help with this question I keep getting -14.4J and have no idea what I'm doing wrong :/

Q1. An electron and a proton are 1.0 × 10–10 m apart. In the absence of any other charges, what is the electric potential energy of the electron?
A +2.3 × 10–18J
B –2.3 × 10–18J
C +2.3 × 10–18J
D –2.3 × 10–18J

https://08c6cd28b8bef288858e878e5745...c%20Fields.pdf
You are calculating the electric potential rather than the electric potential energy.

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