# (New 2017) AQA A-level Physics 7408 [Exam Discussion]

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#1

AQA A-LEVEL PHYSICS 7408 (NEW SPECIFICATION)

PAPER 1 - 7408/1
Duration: 2h
Date: 15 June 2017
am/pm: am

PAPER 2 - 7408/2
Duration: 2h
Date: 21 June 2017
am/pm: am

PAPER 3 - 7408/3
Duration: 2h
Date: 29 June 2017
am/pm: am

________________________________
1
5 years ago
#2
In electromagnetic induction, the rate of change of flux is the change in field lines per second right? is this the idea of Faraday's law? also what factors affect the rate of change of flux and why?
0
5 years ago
#3
(Original post by Nai18)
In electromagnetic induction, the rate of change of flux is the change in field lines per second right? is this the idea of Faraday's law? also what factors affect the rate of change of flux and why?
Field lines aren't really a 'thing', they just represent the direction of a magnetic field. The rate of change of flux is the area A multiplied by the magnetic field B, provided that the direction of the magnetic field is perpendicular to the area.

The rate of change of flux is affected by the magnetic field strength and the area 'swept' per unit time.

Φ = BA
dΦ/dt = d(BA)/dt

since for a uniform magnetic field, B = const, the rate of change of magnetic flux is only really affected by the rate of change of perpendicular area:

dΦ/dt = B dA/dt

just remember that dA/dt could be represented as a length traversed by a speed per unit time.
1
5 years ago
#4
(Original post by testytesttest)
Field lines aren't really a 'thing', they just represent the direction of a magnetic field. The rate of change of flux is the area A multiplied by the magnetic field B, provided that the direction of the magnetic field is perpendicular to the area.

The rate of change of flux is affected by the magnetic field strength and the area 'swept' per unit time.

Φ = BA
dΦ/dt = d(BA)/dt

since for a uniform magnetic field, B = const, the rate of change of magnetic flux is only really affected by the rate of change of perpendicular area:

dΦ/dt = B dA/dt

just remember that dA/dt could be represented as a length traversed by a speed per unit time.
Oh so if area is L x L and the area swept is L x vt , would you sub that into A and you get induced EMF= B x lvt / t and the delta t's cancel out leaving Induced EMF = Blv

(sorry if this looks bad i dont know how to do the delta signs so assume the t is delta t)
0
5 years ago
#5
(Original post by Nai18)
Oh so if area is L x L and the area swept is L x vt , would you sub that into A and you get induced EMF= B x lvt / t and the delta t's cancel out leaving Induced EMF = Blv

(sorry if this looks bad i dont know how to do the delta signs so assume the t is delta t)
Imagine that you have your conductor cutting flux is a wire of length L.

In a time interval dt, it will cut Lvdt, where v is the speed, so yes, the area A swept per unit time is A = Lvt.

You normally shouldn't be able to assume that you can directly substitute this into E = dΦ/dt, but this is A-level, so as long as you're confident that the rate doesn't change for some reason, then it should be fine. So in the case of a wire crossing perpendicular to a uniform magnetic field, E = BLv
0
5 years ago
#6
(Original post by testytesttest)
Imagine that you have your conductor cutting flux is a wire of length L.

In a time interval dt, it will cut Lvdt, where v is the speed, so yes, the area A swept per unit time is A = Lvt.

You normally shouldn't be able to assume that you can directly substitute this into E = dΦ/dt, but this is A-level, so as long as you're confident that the rate doesn't change for some reason, then it should be fine. So in the case of a wire crossing perpendicular to a uniform magnetic field, E = BLv
Ah i see that makes a lot of sense. In lenz law, is it the current induced due to the induced EMF opposes the change in flux?
0
5 years ago
#7
(Original post by Nai18)
Ah i see that makes a lot of sense. In lenz law, is it the current induced due to the induced EMF opposes the change in flux?
0
5 years ago
#8
(Original post by testytesttest)
....The rate of change of flux is the area A multiplied by the magnetic field B, provided that the direction of the magnetic field is perpendicular to the area. ....
I think you meant magnetic flux NOT rate of change of magnetic flux in saying that the area A multiplied by the magnetic field strength B, provided that the direction of the magnetic field is perpendicular to the area.

(Original post by testytesttest)
….just remember that dA/dt could be represented as a length traversed by a speed per unit time.
dA/dt has unit of [m2/s] but the term “a length traversed by a speed per unit time” has a unit of [m2/s2].
0
5 years ago
#9
(Original post by Eimmanuel)
I think you meant magnetic flux NOT rate of change of magnetic flux in saying that the area A multiplied by the magnetic field strength B, provided that the direction of the magnetic field is perpendicular to the area.

dA/dt has unit of [m2/s] but the term “a length traversed by a speed per unit time” has a unit of [m2/s2].
Clearly the second part refers to something which already has one length dimension. It's not a dimensionless 'nothing' that is crossing the magnetic field. That should be implicit. A crossbar of length L is essentially passing across a magnetic field. The area per unit time cut by this gives the dimensions L^2 T^2.

First part was my mistake but looking at the mathematical expressions should rectify this.
0
5 years ago
#10
How are people feeling for this?
0
5 years ago
#11

aqa sucks btw
5
5 years ago
#12
(Original post by testytesttest)
Clearly the second part refers to something which already has one length dimension. It's not a dimensionless 'nothing' that is crossing the magnetic field. That should be implicit. A crossbar of length L is essentially passing across a magnetic field. The area per unit time cut by this gives the dimensions L^2 T^2. ...
I am not sure if you really understand what I am trying to imply.
The term dA/dt has unit of [m2/s] but the term “a length traversed by a speed per unit time” has a unit of [m2/s2] which seems to agree by you (you say dimensions L2 T-2 – typo in your power for the time).

If this is the case, then your following “representation” is not equivalent:

….just remember that dA/dt could be represented as a length traversed by a speed per unit time.
because the two terms have different dimensions or units.

The term dA/dt is just a metal rod of length L moving perpendicular to the magnetic field at a speed of v or a metal rod of length L cutting the magnetic field at a rate of v.
1
5 years ago
#13
(Original post by Eimmanuel)
I am not sure if you really understand what I am trying to imply.
The term dA/dt has unit of [m2/s] but the term “a length traversed by a speed per unit time” has a unit of [m2/s2] which seems to agree by you (you say dimensions L2 T-2 – typo in your power for the time).

If this is the case, then your following “representation” is not equivalent:

because the two terms have different dimensions or units.

The term dA/dt is just a metal rod of length L moving perpendicular to the magnetic field at a speed of v or a metal rod of length L cutting the magnetic field at a rate of v.
I think you're confused by the idea that the thing traversing the magnetic field itself has a dimension. So an object with dimensions [L] traversing at a speed, i.e dimensions [LT^-1] per unit time gives the correct dimensions for this. My OP was unclear but I explained it in my first response.

I suggest you read my lecture notes to further understand this concept:

0
5 years ago
#14
(Original post by testytesttest)
I think you're confused by the idea that the thing traversing the magnetic field itself has a dimension. So an object with dimensions [L] traversing at a speed, i.e dimensions [LT^-1] per unit time gives the correct dimensions for this. My OP was unclear but I explained it in my first response.

I suggest you read my lecture notes to further understand this concept:

I think we are disagreeing on the dimensions for “something per unit time”. As I can see your lecture note said

area ‘sweeped’ per unit time has a dimension of [L2]

I disagreed. It should have the dimensions of [L2/T].

Look at Q22 in the following link

http://theory.caltech.edu/~politzer/QP-Problems.html

Look at the way others use it:

http://slideplayer.com/slide/4514158/

If you google the term area swept out per unit time, it is not hard to find that the dimensions are L2/T instead of L2.

I believe in physics where we say something per unit time, it should have an additional dimension like [something]/[T].

…an object with dimensions [L] traversing at a speed, i.e dimensions [LT^-1] …
I agree that there is no per unit time, the dimensions are good.

But if you add per unit time to the sentence, then the dimension should be L2/T2.

If you want to insist that I am confused. I am ok. (You can PM me to argue about it.)

PS : In Newtonian gravity, if you have studied Kepler’s law, the quantity “area swept out per unit time” is pretty common and it has the dimension of [L2/T]. I am just surprised that when this quantity is used in electromagnetism, the dimensions change!
0
#15

Someone help me out here. Why exactly is the resultant force equal to ? Isn't also downwards?
0
5 years ago
#16
Think the "extendable" bungee rope as a spring, when you stretch the outward, the spring is "stretching" inward.

So the should be upward while weight is the only downward force.
0
#17
(Original post by Eimmanuel)
Think the "extendable" bungee rope as a spring, when you stretch the outward, the spring is "stretching" inward.

So the should be upward while weight is the only downward force.
Is this always the case? Should I always consider F going upwards?
0
5 years ago
#18
(Original post by markschemes)
Is this always the case? Should I always consider F going upwards?
What do you mean is this always the case?

What is the F that I should always consider going upward?

I am sorry that I don't really know what you are asking.
0
#19
(Original post by Eimmanuel)
What do you mean is this always the case?

What is the F that I should always consider going upward?

I am sorry that I don't really know what you are asking.
I always assumed was downwards for a rope/string.
0
5 years ago
#20
Why?
0
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