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

AQA A-LEVEL PHYSICS 7408 (NEW SPECIFICATION)

PAPER 1 - 7408/1
Duration: 2h
Date: 15 June 2017
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PAPER 2 - 7408/2
Duration: 2h
Date: 21 June 2017
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PAPER 3 - 7408/3
Duration: 2h
Date: 29 June 2017
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(edited 6 years ago)

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Reply 1

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?

Reply 2

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.

(edited 6 years ago)

Reply 3

Nai18

Original post by testytesttest

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)

Reply 4

Original post by Nai18

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

Reply 5

Nai18

Original post by testytesttest

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?

Reply 6

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?

Yeah pretty much, it's given in Faraday's Law by the negative sign in E = -dΦ/dt

Reply 7

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. ....

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].

Reply 8

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.

(edited 6 years ago)

Reply 9

Apachai Hopachai

How are people feeling for this?

Reply 10

Apachai Hopachai

ops, wrong thread bye

aqa sucks btw

Reply 11

Original post by testytesttest

….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.

Reply 12

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:

Screen Shot 2017-05-22 at 15.25.41.png

Reply 13

Original post by testytesttest

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:
https://www.physicsforums.com/threads/the-factors-determining-the-induced-emf-in-a-wire.898856/

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!

Reply 14

Someone help me out here. Why exactly is the resultant force equal to

k\Delta L - mg

? Isn't

k\Delta L

also downwards?

Reply 15

Original post by markschemes

Someone help me out here. Why exactly is the resultant force equal to

k\Delta L - mg

? Isn't

k\Delta L

also downwards?

Think the "extendable" bungee rope as a spring, when you stretch the outward, the spring is "stretching" inward.

So the

k\Delta L

should be upward while weight is the only downward force.

Reply 16

Original post by Eimmanuel

Think the "extendable" bungee rope as a spring, when you stretch the outward, the spring is "stretching" inward.

So the

k\Delta L

should be upward while weight is the only downward force.

Is this always the case? Should I always consider F going upwards?

Reply 17

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.

Reply 18