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

Reply 4560

Gome44

10

Original post by NEWT0N

Agreed. They said "average emf" so you just subtract the end points and divide by time

Lol I worked out V rms :/

Reply 4561

Kennethm

2

Original post by CD223

Guys. Bad news I'm afraid

ImageUploadedByStudent Room1434032804.520422.jpg

Posted from TSR Mobile

i did the question as if it was 150k as stated earlier meaning i did a curve but it was half way down lol is that right?

Reply 4562

Protoxylic

14

Original post by NEWT0N

Isn't average when you just subtract the two endpoints? Like average velocity

Also I'm fairly sure that there was a past MC question which involved a similar calculation and I remember getting it right by just doing what I did for this one

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

(edited 8 years ago)

Reply 4563

JizzaStanger

4

Original post by aprocrastinator

uhhh ok, I'm sure I put my intended answer at the time lol but I ticked the middle box I believe...?

Actually I'm pretty sure I was asking for the one with the most damping. Cos I remember I ticked the box which was the longest sentence so it would've been 'cone without the ring' ... dear god this is confusing :biggrin:

I think I ticked the middle box as well.

Reply 4564

StarvingAutist

17

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.

Reply 4565

apriori1234

8

Does anyone remember what question number the "halving the electrons" question was on the multiple choice?

Posted from TSR Mobile

Reply 4566

Protoxylic

14

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

Reply 4567

StarvingAutist

17

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.

Reply 4568

aprocrastinator

10

Original post by JizzaStanger

I think I ticked the middle box as well.

awesome we should be OK then :biggrin:

Reply 4569

Protoxylic

14

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

Reply 4570

king cobra

8

What was the graph then finally what could we draw to get the mark ?

Posted from TSR Mobile

Reply 4571

JizzaStanger

4

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.

(edited 8 years ago)

Reply 4572

StarvingAutist

17

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.

Reply 4573

Chazley123

2

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

Reply 4574

kevincarreira

6

Original post by king cobra

What was the graph then finally what could we draw to get the mark ?

Posted from TSR Mobile

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

Reply 4575

AR_95

21

is there an unofficial mark scheme

Reply 4576

JizzaStanger

4

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!

Reply 4577

Protoxylic

14

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.

Reply 4578

Chazley123

2

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??

Reply 4579

Lau14

11

Original post by king cobra

What was the graph then finally what could we draw to get the mark ?

Posted from TSR Mobile

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.

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

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you