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

1. Can anyone show me how to do this? the answer is C

I know I have to use the equation T=2pi(l/g)^1/2 but can't quite get it
2. (Original post by Music With Rocks)

Can anyone show me how to do this? the answer is C

I know I have to use the equation T=2pi(l/g)^1/2 but can't quite get it
Use T=2pi(l/g)^1/2 and make L the subject of formula for each and try again
3. (Original post by Music With Rocks)

Can anyone show me how to do this? the answer is C

I know I have to use the equation T=2pi(l/g)^1/2 but can't quite get it
gonna be 0.37x9.81 on the bottom of the fraction in the equation, which would mean the the time period would would be something like 2.5 times longer, but it's squared rooted, so more like 1.7 ish times longer, and the one that's roughly closest to that is 2.33 seconds
4. (Original post by Watchmen98v2)
Use T=2pi(l/g)^1/2 and make L the subject of formula for each and try again
Ah thank you very much, I have got it now

EDIT: I never thought to rearrange like that for all these types of questions (I am dumb) you are a saviour, this is great haha
5. hi guys

for exam technque is it ok to do section b first then section a because section b has more marks
6. (Original post by civ1)
hi guys

for exam technque is it ok to do section b first then section a because section b has more marks
yeah I always do B first
7. (Original post by alevelstresss)
yeah I always do B first
i am proper scared for tommorow ?? what about you
8. (Original post by civ1)
i am proper scared for tommorow ?? what about you
nope
9. How do you do this?
Attached Images

10. (Original post by doshea1311)
eddy currents are formed by a changing magnetic flux in the conductor of a transformer etc. The changing flux induces and emf in the core and therefore a current in the core.
Hey, just a general question, think it'd be better to begin section B first rather than A? Spend an hour on section B and the rest on A?
11. (Original post by alevelstresss)
nope
Ok mR confident
12. (Original post by xMillnsy)
How do you do this?
C = Q/V
Q = CV

once you've got Q, divide by the charge of one electron to get the total number of electrons i.e. 4.4x10^10
13. (Original post by xMillnsy)
How do you do this?
Work out the charge on the fully charged capacitor using equation Q=CV.

And you know the magnitude of the charge on one electron is 1.6x10^19.

So divide and find the number of electrons it will take to reach this charge.
14. (Original post by xMillnsy)
How do you do this?
15. (Original post by Music With Rocks)

Can anyone show me how to do this? the answer is C

I know I have to use the equation T=2pi(l/g)^1/2 but can't quite get it
no need to make L the subject, just divide 1.42 by root 0.37
16. i'm not too worried about my understanding but i feel like I'm still really rusty and scared of running out of time lol
17. Can anyone help me with his question:

Which of the following statements is true for a rectangular coil rotating in a uniform magnetic field?

A: The magnitude of the induced EMF is greates when the coil is parallel to the field
B: Increasing the frequency of rotation increases the max EMF
C: The induced Emf is independant of the area of the coil
D: A graph of Emf/time would be linear

I know B is definetly true but I also thought A was true? thanks
18. (Original post by ReeceFraser)
Can somebody go through the 6 marker from Jan 2012, i literally have no idea.
(Original post by ReeceFraser)
Can somebody go through the 6 marker from Jan 2012, i literally have no idea.

Why a current will be induced in the ring

This is to do with Faraday's Law which is If the magnetic flux within a circuit changes, an emf is induced in the circuit and this emf is proportional to the rate of change of the flux linkage.

The current supplied to the coil is alternating so the magnetic field is also alternating, meaning there is a change in flux, this causes an emf to be induced (Faraday's Law states this) within the coil and then because it's a complete circuit, inducing emf also induces a current.

Why the ring experiences a force that moves it upwards

This is Lenz's Law - Induced current flows in such a direction as to oppose the magnetic flux increase.

The coil produces one magnetic field but the current produces a different magnetic field that opposes the field of the coil. Since it's opposes it you can treat it as if there are two like poles near each other. Like poles would repel. So they do. The magnetic force the ring experiences is due to this repulsion as the poles interact.

Why the ring reaches a stable position

There comes a point as it floats higher where the component of it's weight acting downwards due to mg is equal to the magnetic force from the coil repelling it upwards. The resultant vertical component is zero and it becomes stationary. The weight of the ring is obviously constant so the magnetic force must be decreasing in magnitude.

I feel like I've done this question before and it might just seem like I'm repeating the mark scheme so sorry if I basically make just as little sense. If you're totally clueless maybe you should revise Faraday's Law and Lenz's Law...they took me ages to get my head around and I'm only like 80% sure I get them
It kind of helps we did this as a practical too...
19. so ****ed for this exam
20. (Original post by Charlie1523)
Can anyone help me with his question:

Which of the following statements is true for a rectangular coil rotating in a uniform magnetic field?

A: The magnitude of the induced EMF is greates when the coil is parallel to the field
B: Increasing the frequency of rotation increases the max EMF
C: The induced Emf is independant of the area of the coil
D: A graph of Emf/time would be linear

I know B is definetly true but I also thought A was true? thanks
I think B is correct. A is wrong because emf is only induced when it CUTS field lines, e.g. when it is perpendicular to the field. So it is actually 0 here. note: magnetic flux linkage through the coil is maximum when parallel to the field lines, as there are max. number of field lines going through the coil.

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