Login | Register

Welcome, Guest. Please login or register.

June 04, 2020, 05:27:51 pm

AuthorTopic: Unit 4 Question Thread  (Read 1311 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

#1procrastinator

• Guest
Unit 4 Question Thread
« on: October 12, 2012, 10:43:56 am »
0
1) If you have a wire moving at a constant velocity perpendicular to a magnetic field
(imagine the field lines on this page are pointing down and the wire is moving to the left),
wouldn't the only force on be pushing out of the plane and into for the positive and negative charges respectively?

In the Heinemann test, chapter 10.2 ('Induced EMF: Faraday's Law), it says the force on the positive charges would be along
the wire and out of the page? As I'm typing, I'm thinking maybe they mean push out of against the
side of the wire, not what I initially thought (and what confused me).

What I initially thought was that they mean there's a force acting to the left...sorry, stream of
consciousness.

2) If electrons are the negative charges, what are the positive charges?

3) Why must we have a reference intensity to find the intesity of some sound source? I don't really
get the motivation for this equation $L=10log(\frac{I}{I_o})$

Phantom-II

• Victorian
• Forum Regular
• Posts: 88
• Respect: 0
Re: Unit 4 Question Thread
« Reply #1 on: October 12, 2012, 11:34:36 pm »
+1
1) Yes, but there is no circuit, so there would be a build up of positive charges near your end and negative charges away into the page, polarisation

positive charges are protons! though dont think of conventional current that way, conventional current is the one we use, only because they got it wrong the first time. I dont think the positive particles/protons ACTUAlly travel inter-molecularly

For sound, first you have to understand the human perception is only a small range of intensities, and it is logarithmic in nature. So as you should know from maths, a log graph looks like a curled line. You basically then have to match that set of x-values with y-values,
if you expand that log, I0=10^-12, equation will be: 10logI-10log(10^-12)
which is 10logI+120.
since perceivable intensity are between 0 and 1, we move the graph up 120 units for the yaxis values to be around 0-120dB

oppalovesme

• Guest
Re: Unit 4 Question Thread
« Reply #2 on: October 17, 2012, 09:06:23 pm »
0
I'm missing something really obvious here I think, but could someone help me out with question 3 (AOS1) on the VCAA 2010 exam?
http://www.vcaa.vic.edu.au/Documents/exams/physics/2010physics2-w.pdf
I understand the right hand slap rule and whatnot, but was wondering how to find the direction of the magnetic field within the solenoid - as it appears to be connected to an AC source.
Thanks in advance!

StumbleBum

• Victorian
• Forum Obsessive
• Posts: 280
• Respect: +3
• School: St Joseph's College
• School Grad Year: 2012
Re: Unit 4 Question Thread
« Reply #3 on: October 17, 2012, 09:16:10 pm »
0
I'm missing something really obvious here I think, but could someone help me out with question 3 (AOS1) on the VCAA 2010 exam?
http://www.vcaa.vic.edu.au/Documents/exams/physics/2010physics2-w.pdf
I understand the right hand slap rule and whatnot, but was wondering how to find the direction of the magnetic field within the solenoid - as it appears to be connected to an AC source.
Thanks in advance!

It is a DC source as it is a battery. All batteries that are standalone will be DC.
As for the direction of the magnetic field, the current will flow from the positive terminal to the negative (as always) and will therefore create a magnetic field to the left according to Fleming's Right Hand Rule. Which then, according to the Right Hand Slap rule, with current going from Q to P in the coil and the magnetic field going to the left inside the solenoid, will cause a downwards force on the side PQ.
2011: Mathematical Methods (CAS) [36]

2012: English [35+] | Specialist Mathematics [35+] | Further Mathematics [45+] | Physics [40+] | Accounting [38+] |

oppalovesme

• Guest
Re: Unit 4 Question Thread
« Reply #4 on: October 17, 2012, 09:19:14 pm »
0
It is a DC source as it is a battery. All batteries that are standalone will be DC.
As for the direction of the magnetic field, the current will flow from the positive terminal to the negative (as always) and will therefore create a magnetic field to the left according to Fleming's Right Hand Rule. Which then, according to the Right Hand Slap rule, with current going from Q to P in the coil and the magnetic field going to the left inside the solenoid, will cause a downwards force on the side PQ.
so the solenoid is connected to a DC source..?
it looks like it's the coil connected to the battery in the diagram though :s

StumbleBum

• Victorian
• Forum Obsessive
• Posts: 280
• Respect: +3
• School: St Joseph's College
• School Grad Year: 2012
Re: Unit 4 Question Thread
« Reply #5 on: October 17, 2012, 09:27:38 pm »
+2
so the solenoid is connected to a DC source..?
it looks like it's the coil connected to the battery in the diagram though :s
Oh sorry, didn't explain this part. You'll notice that for question 1, it states that the information is relevant to questions 1-5. This then implies that the solenoid is still connected to the DC battery in figure 2, as it is in figure 1.
2011: Mathematical Methods (CAS) [36]

2012: English [35+] | Specialist Mathematics [35+] | Further Mathematics [45+] | Physics [40+] | Accounting [38+] |

oppalovesme

• Guest
Re: Unit 4 Question Thread
« Reply #6 on: October 17, 2012, 09:31:21 pm »
0
Oh sorry, didn't explain this part. You'll notice that for question 1, it states that the information is relevant to questions 1-5. This then implies that the solenoid is still connected to the DC battery in figure 2, as it is in figure 1.

omggggg.... i think i need to sleep LOL
thank you so much for helping silly me :3

#1procrastinator

• Guest
Re: Unit 4 Question Thread
« Reply #7 on: October 25, 2012, 12:46:15 am »
0
When you find the force of a magnetic field on a wire with a current flowing through it, the force is in the direction of your thumb right (I'm referring to the right hand rule where you have your index in the direction of the current and middle in the direction of the field)?

If the current is the flow of electrons, then why when you use the RHR on the a single electron, it's in the opposite direction? Or is the direction of the current really a flow of protons?

paulsterio

• ATAR Notes Legend
• Posts: 4829
• I <3 2SHAN
• Respect: +427
Re: Unit 4 Question Thread
« Reply #8 on: October 25, 2012, 01:00:38 am »
0
Conventional current is the flow of positive charge (which isn't really the case) but it's opposite to the direction the electrons flow in.

#1procrastinator

• Guest
Re: Unit 4 Question Thread
« Reply #9 on: October 25, 2012, 03:05:32 pm »
0
What do you mean it's not really the case? Are you just saying that the positive charge IS in the direction of the current but it just isn't really moving?

StumbleBum

• Victorian
• Forum Obsessive
• Posts: 280
• Respect: +3
• School: St Joseph's College
• School Grad Year: 2012
Re: Unit 4 Question Thread
« Reply #10 on: October 25, 2012, 04:27:35 pm »
+1
What do you mean it's not really the case? Are you just saying that the positive charge IS in the direction of the current but it just isn't really moving?

The rule is that Conventional Current flows in the opposite direction to that of the electrons.

Conventional current is the flow of positive charge (which isn't really the case)

I think Paul was just saying that you can think of conventional current to be the flow of positive charge, as it will oppose the flow of negative charge (the electron flow). However, conventional current isn't actually the flow of positive charge.
2011: Mathematical Methods (CAS) [36]

2012: English [35+] | Specialist Mathematics [35+] | Further Mathematics [45+] | Physics [40+] | Accounting [38+] |

paulsterio

• ATAR Notes Legend
• Posts: 4829
• I <3 2SHAN
• Respect: +427
Re: Unit 4 Question Thread
« Reply #11 on: October 25, 2012, 04:35:12 pm »
+2
Ah, it's hard to explain what I mean, but yeah SB managed to pretty much hit the nail on the head, but I have to go through a bit of history.

Before we knew what electrons were, we already knew that current was the flow of charge, but we didn't know if positive charges were moving or negative charges were moving, so obviously we had to take a guess and we DEFINED conventional current as the flow of positive charge.

So essentially if we have two points, A and B, current flowing from A to B would mean that a proton flies from A to B and hence A becomes negative and B becomes positive. Well this was the case until it was found out later on that in fact, there were no moving positive charges, it was moving NEGATIVE charges.

But conventions are hard to change, so it was decided to be kept that current is the flow of positive charges, thus, when we say current, we say that the current is flowing from A to B (positive to negative) even though the electrons are flowing from B to A.

#1procrastinator

• Guest
Re: Unit 4 Question Thread
« Reply #12 on: October 25, 2012, 07:00:42 pm »
0
Ok, so if there's a wire

A----------B

and a current flows from A to B, then the direction of the current is positive charges but what's really happening is the electrons are flowing from B to A?

Is that right?

paulsterio

• ATAR Notes Legend
• Posts: 4829
• I <3 2SHAN
• Respect: +427
Re: Unit 4 Question Thread
« Reply #13 on: October 25, 2012, 07:02:25 pm »
0
No there are actually no flow of positive charges, PROTONS DO NOT FLOW.

The direction of conventional current is from A to B, so yes the electrons are flowing from B to A.

#1procrastinator

• Guest
Re: Unit 4 Question Thread
« Reply #14 on: October 25, 2012, 07:28:25 pm »
0
^ Haha, yeah that's what I meant

Thanks a lot, it makes sense now