ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: 3Xamz on November 06, 2010, 11:01:51 am
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Q3) Why is the area 0.15*0.3, as per solutions? The length of the wire is 30cm, okay fair enough. However, why do they use the 0.15m to find the area? That's not part of the wire, it's just how far it is in the magnetic field? :S
Q4) Could someone please give me an explanation to this question? The solutions, I do not get =[
[IMG]http://img593.imageshack.us/img593/1655/85567587.jpg[/img]
Uploaded with ImageShack.us
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Also, wouldn't this be correct:
[IMG]http://img259.imageshack.us/img259/3558/001ul.jpg[/img]
Uploaded with ImageShack.us
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Q3) Why is the area 0.15*0.3, as per solutions? The length of the wire is 30cm, okay fair enough. However, why do they use the 0.15m to find the area? That's not part of the wire, it's just how far it is in the magnetic field? :S
The area shouldn't be 0.15x0.30 and I don't know why the solutions indicate it as such. I just ignored the question :P
Q4) Could someone please give me an explanation to this question? The solutions, I do not get =[
I'm assuming you mean Q5 because that's the one that you uploaded.
The way I do it is as follows. Use the Right hand palm rule. The field is coming out of the page so have your fingers pointing upwards; and the rod is being slid to the left so have your palm facing the left side of the page. Your thumb should point towards Y, indicating that if we were to apply a current running from X to Y then we would produce a force to the left. However, in this question we are exerting a physical force on the rod to the left so a current will be induced so as to generate a force which pushes the rod back in the opposite direction (to the right). So all you have to do is reverse the direction of current in your right hand palm rule, i.e. the current runs into your thumb instead of out of your thumb, i.e. from Y to X. Knowing this, you can then quickly test it yourself. Fingers pointing upwards, thumb pointing towards X, you can see that your palm is facing the right side of the page. And since the induced current is running from Y to X, Y has the greater potential (solutions are wrong).
Also, wouldn't this be correct:
Yes
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Thank you d-ea-6 :)
Insight is such a bullshit exam for Physics.
So many mistakes, just ran across another mistake where they said (30/0.2)=600ohms =='
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Insight 2010 has a lot of errors, HOWEVER question 3 is NOT one of them!!!
If you look at the question carefully, it says the wire is passed through two conducting rails. The fact that they are conducting means that when you add the wire which is connected to both ends, if forms an effective circuit. The area of this circuit within the magnetic field will be .15 * .30cm, and thus you work it out as per usual.
edit: see below
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So, this constitutes a circuit?
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Okay, actually scratch that closed circuit thing sorry.
However, there well still be an induced EMF. Remember that EMF can exist without current, but not vice versa. Consider a single wire moving through a magnetic field. There is no current, however by the relationship EMF=vBL we can clearly see that there is an EMF. Similarly in this case, because there is an effective area (as the wire is connected to both rails) there will be an EMF.
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If you read question 3, it says that the magician turned off the external field, changing the magnetic field from 30mT to 0. When this occurred, the rods would've been stationary, wouldn't they? So would emf=Blv even be applicable?
For question 4, however, you would be correct. And the Insight solutions are correct for the most part. They used the right values but didn't correctly state the formula as emf=Blv (they wrote emf= -NBA/t)
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No, I'm not saying to apply EMF = vBL to this particular case, rather just to demonstrate that you don't need a closed circuit to create an EMF, to amend my previous mistake. :)
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Okay :P
So, ultimately, you're still saying that question 3 isn't an error on their part?
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Yes, i'm sure they are right.
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How are they right though?
Even if we use EMF=BLV, isn't it EMF= 30*10^-3 * .3m * 0.05m/s?
This equates to 4.5*10^-4 Volts.
They have 1.35*10^-3 Volts..
EDIT: I still don't see how they got that value for the area? :(
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There are 3 rods.
3x(4.5x10^-4) = 1.35x10^-4
Edit: Referring to question 4.
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No, don't use emf = vbl, i was just using that to demonstrate a point!! sorry for the confusion.
The wire is connected to both the conducting rails, thus it creates an effective closed of area. This area of the wire and rail within the magnetic field will be 0.15 * .30 m, which is 0.045m.
So therefore, when the magnetic field is switched off, there will be a change in flux, as you have an area and you have a magnetic field strength.
Multiplying this change in flux by -3 and then dividing by the time taken will give you the required answer.
Remember though, because it is not a closed circuit, there won't be a current, HOWEVER there will be an EMF as EMF can exist without current in an open circuit.
sorry, i'm not very good at explaining things.
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Okay, just to clear things up, we're talking about question 3 :P
Last query then.
The wire is connected to both the conducting rails, thus it creates an effective closed of area.
Wouldn't the area be as shown then?
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Oh boy, i seem to be digging myself into a deeper hole. i don't know what the hell to think now. ::)
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LOL.
All you have to do, mate, is say that Insight are wrong. And we can put this to rest :D
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yeah, i'll probably talk to my teacher about it. what a fucked up question.
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hey, i get what your saying, but how can you multiply the whole area by 3, if it is only the rod that has extra lengths in it and not the horizontal rails?
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This is true according to iTute;
If a 10 cm conducting wire moves 5 cm horizontally across a magnetic field;
The the change in area is given by 10cm*5cm = 50cm^2
This will provide an induced EMF, and also produces a current according to Lenz's Law.
This supports what Linkage was saying earlier; but not to the part where he said no current is produced. According to iTute there is indeed a current along the moving wire? :S
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This is true according to iTute;
If a 10 cm conducting wire moves 5 cm horizontally across a magnetic field;
The the change in area is given by 10cm*5cm = 50cm^2
This will provide an induced EMF, and also produces a current according to Lenz's Law.
This supports what Linkage was saying earlier; but not to the part where he said no current is produced. According to iTute there is indeed a current along the moving wire? :S
This is only relevant to question 4. Like I said earlier, the rods are stationary in question 3 when the magnetic field is switched off.
But, if you'd like, we can look at question 4 and find that it works as follows:
emf = nBlv = 3 x (30/1000) x (30/100) x (5/100) = 1.35x10^-3 V
Alternatively:
emf = nΔBA/Δt = (3 x (30/1000) x ((15/100)x(30/100))) / 3 = 1.35x10^-3 V (3 is obtained by using t=d/v = 15cm/5cms^-1 = 3seconds)
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Oh yes of course!
So, question 3 is not do-able?
Because its a wire thats not moving, so we don't have a change in flux ?
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Question 3 is not do-able because we don't know the area. The change in flux is 30x10^-3 T x area (as a result of the magnetic field being switched off).
Hypothetically, let's just say that the width of each rod was 2cm. Then it'd just be a standard emf question which you'd answer by doing as such:
emf = nΔBA/Δt = (3 x (30/1000) x ((30/100)x(2/100))) / (6/1000) = however many volts
That's how I see it.
Edit: Yes, my mistake :P
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Yup, that's how I see it too.
Except;
"The change in flux is (30x10^-3)*AREA"
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This supports what Linkage was saying earlier; but not to the part where he said no current is produced. According to iTute there is indeed a current along the moving wire? :S
well that's what i initially thought because it said conducting rails, so i assumed there was a current running along both of them. But then when dea6 started suggesting otherwise, I realised that the rails weren't actually connected to anything so I started doubting myself. It really is an ambiguous question.
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The question is do-able if you make some assumptions. Since there is only one vertical and one horizontal measurement given, it's pretty clear that the area will be the product of those two, even if the diagram doesn't support it.
Anyway, I agree, Insight 2010 was ridiculous, they tried to make it hard and couldn't even do basic maths themselves. So many errors in their solutions.
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LISTEN, point is, the 2010 insight exam is the worst physics exam i have ever come across, and it's my 17th unit 4 practice exam. It is riddled with errors and many of the solutions are ... simply wrong
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LISTEN, point is, the 2010 insight exam is the worst physics exam i have ever come across, and it's my 17th unit 4 practice exam. It is riddled with errors and many of the solutions are ... simply wrong
Are you referring to the last two questions? Cos they are unbelievable. Are the correct answers 150ohm for q17 and 54000J for q18?