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

1. (Original post by Protoxylic)
Yes, so a quarter of a sin wave. The EMF varies over that time from 0 to a maximum emf at 0.5s. Prove for yourself that if the flux linkage is ACos(wt) for constant A=BAN. Then EMF=-dN(phi)/dt which is AwSin(wt) the emf is not constant. Hence the need to find the mean emf
No, the emf is not constant. To find the average you'd add the values at each instant up and divide it by the number of instants.
sum of e = S(Awsinwt) dt
= phi(t)
Then divide it by the number of instants:
average e = phi(t)/total time

That'd be 0.12.
2. Does anyone remember what question number the "halving the electrons" question was on the multiple choice?

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3. (Original post by StarvingAutist)
No, the emf is not constant. To find the average you'd add the values at each instant up and divide it by the number of instants.
sum of e = S(Awsinwt) dt
= phi(t)
Then divide it by the number of instants:
average e = phi(t)/total time

That'd be 0.12.
So you tell me why dividing a number you got given, by another value you got given constitutes 2 marks. Contrast this to the previous question where you were asked to multiply 3 numbers together for 1 mark (flux linkage) then asked the units for another mark (wbturns). All the physicists (as in the teachers) in my school agree with the use of the rms value
4. (Original post by Protoxylic)
So you tell me why dividing a number you got given, by another value you got given constitutes 2 marks. Contrast this to the previous question where you were asked to multiply 3 numbers together for 1 mark (flux linkage) then asked the units for another mark (wbturns). All the physicists (as in the teachers) in my school agree with the use of the rms value
Well, what is wrong with my working? Nothing as far as I can see.
5. (Original post by JizzaStanger)
I think I ticked the middle box as well.
awesome we should be OK then
6. (Original post by StarvingAutist)
Well, what is wrong with my working? Nothing as far as I can see.
Can you remember the values of the cross sectional area and flux density? The peak emf would be BANw. Divide that by root 2 and see if you get 0.12V, I doubt it very much. Since the coil rotates pi/2 from perpendicular to the field to parallel to the field, at t=0.5s this would be where peak emf occurs
7. What was the graph then finally what could we draw to get the mark ?

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8. (Original post by Protoxylic)
Average velocity for CONSTANT ACCELERATION. The emf is not freaking constant it varies with sin. How many times do I actually have to explain this to people lol.

Look at it another way, lets say a you move from 0 to 20 ms^-1 in a time of 20 seconds. Is half way in distance at t=10s? Of course it isn't

The difference here is that it's like the discrepancy between velocity and speed.

If you undergo uniform circular motion with period T, then your average velocity between t=0 and t=T would be zero, in T seconds you've returned to your starting position, so your displacement is 0m. Velocity is displacement / time and therefore velocity is 0m/s. Your average SPEED would be non-zero, clearly. This strange result happens because velocity is a vector.

Although emf is scalar, it is the rate of change of the dot product of two vectors (B.A) and so can easily be negative. The same principle as for velocity applies.
9. (Original post by Protoxylic)
Can you remember the values of the cross sectional area and flux density? The peak emf would be BANw. Divide that by root 2 and see if you get 0.12V, I doubt it very much. Since the coil rotates pi/2 from perpendicular to the field to parallel to the field, at t=0.5s this would be where peak emf occurs
I wasn't dividing by root 2. See my working above and point out the flaw in my reasoning. Don't compare it to your idea, actually find the fault.
10. (Original post by davidharris96)
No it was D,
rate of change of momentum, KGms^-2 which is the same as Newtons
change in momentum equal to impulse (Ns)

rate of change of momentum - I/t (Ns/s) = (N) - equivalent to weight =mg
11. (Original post by king cobra)
What was the graph then finally what could we draw to get the mark ?

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Probably:

1 mark for starting at the right point (same as other line)
1 mark for finishing at the right point (half of other line)
1 mark for right gradient which now appears to be either flat line at 0 A or steeper than the other line
12. is there an unofficial mark scheme
13. (Original post by Protoxylic)
Average velocity for CONSTANT ACCELERATION. The emf is not freaking constant it varies with sin. How many times do I actually have to explain this to people lol.

Look at it another way, lets say a you move from 0 to 20 ms^-1 in a time of 20 seconds. Is half way in distance at t=10s? Of course it isn't
Did the question not say "flux is increased uniformly"
14. (Original post by kevincarreira)
Probably:

1 mark for starting at the right point (half of other line)
1 mark for finishing at the right point (half of other line)
1 mark for right gradient which now appears to be either flat line at 0 A or less steep than the other line
Corrections in bold.

Larger resistance means lower current at all times!
15. (Original post by NEWT0N)
Did the question not say "flux is increased uniformly"
I can't remember, but if that is true then the rotation of the coil would vary. Either way if it did say increased uniformly then the emf would be constant meaning there would be no need to ask for the mean emf.
16. (Original post by Protoxylic)
So you tell me why dividing a number you got given, by another value you got given constitutes 2 marks. Contrast this to the previous question where you were asked to multiply 3 numbers together for 1 mark (flux linkage) then asked the units for another mark (wbturns). All the physicists (as in the teachers) in my school agree with the use of the rms value
or 0.06/0.5 = 0.12 v ?? as emf= BAN(linkage)/t - faradays law??
17. (Original post by king cobra)
What was the graph then finally what could we draw to get the mark ?

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From what was said, you get the marks if you have done a straight line at current = 0 (the question as it was written), or if you answered the question as it should have been starting halfway up compared to the other graph, a shallower curve but still of exponential shape.
18. (Original post by Chazley123)
or 0.06/0.5 = 0.12 v ?? as emf= BAN(linkage)/t - faradays law??
You don't get marks for quoting given formulas on the data sheet
19. (Original post by JizzaStanger)
Corrections in bold.

Larger resistance means lower current at all times!
thanks bud! I didn't get it right so i guessed what they might be from what peoples have said
20. http://www.sunnyskyz.com/happy-pictu...rying-her-baby

^unofficial mark scheme

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