TangoTangoPapa2 posted this before and i was wondering if anyone had an easy to figure out that 1 and 6 were the solutions and not 4 and 3 or 5,2
Really been bugging my head for a while, without figuring out manually the values each combination wont give is there another way?
Question 11:
https://www2.physics.ox.ac.uk/sites/...2011Paper.pdf
Solution:
http://www.physicsandmathstutor.com/pat/solutions2011/
Thanks in advance!
You are Here:
Home
> Forums
>< Study Help
>< Maths, science and technology academic help
>< Physics
>< Physics Exams

Oxford PAT 2016
Announcements  Posted on  

How helpful is our apprenticeship zone? Have your say with our short survey  02122016 

 Follow
 281
 22082016 13:28
Tagged:Post rating:1 
 Follow
 282
 25082016 10:38
Hey Guys
Applying from Australia here. i tried a few past PAT papers and found the maths section to be alright as our maths syllabus contains very similar questions. i am however having a lot of trouble with the physics section as alot of the content are not taught here in australia
do you guys have any recommendations to good resources/notes i can use to quickly learn all of the laws and formulae that would be in alevels.Post rating:1 
 Follow
 283
 25082016 20:30
(Original post by masterladfrank)
Hey Guys
Applying from Australia here. i tried a few past PAT papers and found the maths section to be alright as our maths syllabus contains very similar questions. i am however having a lot of trouble with the physics sectin as alot of the content are not taught here in australia
do you guys have any recommendations to good resources/notes i can use to quickly learn all of the laws and formulae that would be in alevels.
I used http://www.physicsandmathstutor.com/physicsrevision/ but the documents on the website are not available at the moment.
formulas can be found here : http://www.ocr.org.uk/images/77735d...81tog486.pdf You could google Physics formula booklet/datasheet for other version of the formulae sheets.
The problems in AS challenge and physics section of Oxford PAT are similar. You should attempt those as well. http://www.physics.ox.ac.uk/olympiad/PastPapers.html
!!! Good Luck !!!!!Tagged: 
 Follow
 284
 25082016 21:43
(Original post by NatoHeadshot)
TangoTangoPapa2 posted this before and i was wondering if anyone had an easy to figure out that 1 and 6 were the solutions and not 4 and 3 or 5,2
Really been bugging my head for a while, without figuring out manually the values each combination wont give is there another way?
Question 11:
https://www2.physics.ox.ac.uk/sites/...2011Paper.pdf
Solution:
http://www.physicsandmathstutor.com/pat/solutions2011/
Thanks in advance!Post rating:1 
 Follow
 285
 25082016 22:37
I have a few questions about electromagnetism.
Is there always a magnetic field around a current carrying wire? Even if the current is "steady"? I mean, dI/dt is a constant, not a function of t?
Why are electric fields due to magnetic fields concentric circles? (edit: I have seen the example of putting a copper ring in a uniform magnetic field, and when the flux linkage was increased that induced an emf and hence a current in the ring, and when there's a current, there's an electric field driving the charges, and that electric field would be concentric circles, but the electric field due to changing magnetic field exists without the copper ring as well, or without charges, which I can't understand atm)
ThanksLast edited by lawlieto; 25082016 at 22:44.Post rating:1 
 Follow
 286
 25082016 23:04
(Original post by lawlieto)
I have a few questions about electromagnetism.
Is there always a magnetic field around a current carrying wire? Even if the current is "steady"? I mean, dI/dt is a constant, not a function of t?
Thanks
Every current carrying conductor (Usually we deal with steady current) produces a magnetic field in every point in space around it.
Let us consider a small part of a current carrying conductor of length dl, then magnitude of magnetic field produced about this length at a point P around it is:
 directly proportional to the magnitude of the current
 inversely proportional to the square of the distance between the point and conductor.
 directly proportional to the sine of angle between the length of the conductor and the line joining conductor and point in space.Post rating:1 
 Follow
 287
 25082016 23:17
(Original post by tangotangopapa2)
Yes, there is always a magnetic field around a current carrying wire. Yes even if the current is steady. In fact:
Every current carrying conductor (Usually we deal with steady current) produces a magnetic field in every point in space around it.
Let us consider a small part of a current carrying conductor of length dl, then magnitude of magnetic field produced about this length at a point P around it is:
 directly proportional to the magnitude of the current
 inversely proportional to the square of the distance between the point and conductor.
 directly proportional to the sine of angle between the length of the conductor and the line joining conductor and point in space. 
 Follow
 288
 25082016 23:18
(Original post by lawlieto)
Why are electric fields due to magnetic fields concentric circles? (edit: I have seen the example of putting a copper ring in a uniform magnetic field, and when the flux linkage was increased that induced an emf and hence a current in the ring, and when there's a current, there's an electric field driving the charges, and that electric field would be concentric circles, but the electric field due to changing magnetic field exists without the copper ring as well, or without charges, which I can't understand atm)
Thanks
Hmm, I think there is a better explanation by Dr Kathy Romer. See the following thread, post no. 45.
http://www.thestudentroom.co.uk/show...4164901&page=3 
 Follow
 289
 25082016 23:33
(Original post by lawlieto)
Thank you any ideas about the concentric circle electric fields? I really don't get why those electric field lines are concentric circles around a changing magnetic field and how they exist without charges.
1) When there is change in magnetic flux current is induced in the copper ring. (Direction given by the Lenz's law)
2) Copper ring can be considered as a straight current carrying conductor bent into a circle. And this current induces magnetic field not electric field where field lines are tightly placed parallel circles with centre along the line of ring. There is also induced electric field due to changing magnetic field which is hard to analyse (needs knowledge of vector calculus).
I might be missing something.
Could you please link me to the question, if it is from one?Last edited by tangotangopapa2; 25082016 at 23:48. 
 Follow
 290
 25082016 23:47
(Original post by tangotangopapa2)
I am sorry but this question is not exactly getting into my head. Could you please link me to the question, if it is one?
So there's a magnetic field into the page. The strength of this magnetic field is increased. This will create an electric field, in the form of a set of concentric circles.
(You can imagine that you put a copper ring in the magnetic field. When you increase the strength of the magnetic field, there will be a change in flux linkage through the copper ring which induces an emf, and hence a current in the copper ring. If there's current in the copper ring, an electric field is also present along the ring which drives the charges. However, there will be an electric field present when changing the magnetic field, even if there's no copper ring and the magnetic field is let's say, in vacuum)
So the thing I was wondering about is that, why is the direction of the electric fields is circular? Saying that "if there was a copper ring this and this would happen and then it would be circular" somehow doesn't sound like an explanation to me. (read it in a book written by an authoritative physicist)
EDIT: if I think about it, the magnetic field produced by a current carrying wire is also a "circular" so it makes sense when people say "electricity and magnetism are different sides of the same coin" but there's still a big whyLast edited by lawlieto; 26082016 at 00:02. 
 Follow
 291
 27082016 15:55
Hey guys,
I was doing the new specimen paper from the physmathtut website but I was confused on question 12.
I got the part on the left correct and was able to determine the resistance for each component but why does he use the formula 1/Rt = 1/R1... when calculating the resistance between A and B. I thought you could just add them all together and get 29/10 R? Are they still in parallel or something?
Can someone explain this to me please?
Link to solutions for the paper:
http://www.physicsandmathstutor.com/...2015specimen/
EDIT: Sorry guys, I just realised my mistake... They are in parallel lol.Last edited by hellomynameisr; 27082016 at 16:11. Reason: Mistake found 
 Follow
 292
 27082016 16:56
Try this question if you haven't already. Number c is really tricky. Took about an hour for me to realise.
Hint: It is not necessary to immediately dive into tons of equations from Kirchhoff's laws.

 Follow
 293
 27082016 17:21
(Original post by tangotangopapa2)
Try this question if you haven't already. Number c is really tricky. Took about an hour for me to realise.
Hint: It is not necessary to immediately dive into tons of equations from Kirchhoff's laws.
From the solutions, why does he do what he does in part B? I understand that p=v^2/r but how does he get the values to input them in from
And part C, how does he split up the circuit like that 
 Follow
 294
 27082016 18:13
(Original post by hellomynameisr)
I was able to get part A right but don't understand B and C
From the solutions, why does he do what he does in part B? I understand that p=v^2/r but how does he get the values to input them in from
And part C, how does he split up the circuit like that
Here comes the tricky bit. As resistance in BD is four times larger than resistance AD. Voltage dropped in BD is also 4 times larger than that dropped at AD.
Now it gets easy: The total voltage dropped is V and this is divided in the ratio 1:4. The voltage between BD is 4V/5 and resistance is R. Using the formula v^2/R you get the required answer.
Part c is very tricky and so simple if you use simple trick. Are you sure you want to look at the solution below? Think again!!! :PSpoiler:ShowThe question tries to fool you showing points C and B, but we won't get fooled, will we? OK, enough talking now.
Note that B and C is the same point. Weather you say (R)CD or (R)BD, its the same thing.
So, total resistance is [(BCA//BA) + AD ]//BD.Post rating:1 
 Follow
 295
 27082016 18:36
(Original post by hellomynameisr)
I was able to get part A right but don't understand B and C
From the solutions, why does he do what he does in part B? I understand that p=v^2/r but how does he get the values to input them in from
And part C, how does he split up the circuit like thatPost rating:1 
 Follow
 296
 27082016 19:17
(Original post by tangotangopapa2)
You need to know that voltage between resistors in parallel is same and equal to total voltage. Let's just look at the branch ADB. (R)AD = R/2 and (R)DB = R (fairly simple isn't it?).
Here comes the tricky bit. As resistance in BD is four times larger than resistance AD. Voltage dropped in BD is also 4 times larger than that dropped at AD.
Now it gets easy: The total voltage dropped is V and this is divided in the ratio 1:4. The voltage between BD is 4V/5 and resistance is R. Using the formula v^2/R you get the required answer.
Part c is very tricky and so simple if you use simple trick. Are you sure you want to look at the solution below? Think again!!! :PSpoiler:ShowThe question tries to fool you showing points C and B, but we won't get fooled, will we? OK, enough talking now.
Note that B and C is the same point. Weather you say (R)CD or (R)BD, its the same thing.
So, total resistance is [(BCA//BA) + AD ]//BD.
(Original post by NatoHeadshot)
try redrawing the circuit and you should have solved it pretty easily i hope this helped 
 Follow
 297
 28082016 14:40
Can't believe that summer is nearly over and I learnt almost nothing. Hard time ahead.
But I have a plan. (I always have plans. Only problem is that I never follow them. ). I will start my preparation from today. No, I am tired atm. I will start my preparation tomorrow. . Surely, next week.
How prepared is everyone by now? 
 Follow
 298
 28082016 15:14
(Original post by tangotangopapa2)
Can't believe that summer is nearly over and I learnt almost nothing. Hard time ahead.
But I have a plan. (I always have plans. Only problem is that I never follow them. ). I will start my preparation from today. No, I am tired atm. I will start my preparation tomorrow. . Surely, next week.
How prepared is everyone by now?
I proper feel like I am going to fail this exam and miss my Oxford place
Hopefully not tho lol
I feel okay for the maths section as I've completed the relevant A2 content but its the physics section that has me worried.
However, we all have until Nov 2nd which is plenty of time to get comfortable with everything and do all the past papers + morePost rating:1 
 Follow
 299
 28082016 18:27
For those who have 'Povey's perplexing problems'. Puzzle 1.5, I understand everything apart from the formula used for working out the area of the quadrilateral. Any help would be greatly appreciated.

 Follow
 300
 28082016 18:58
(Original post by hellomynameisr)
Not at all
I proper feel like I am going to fail this exam and miss my Oxford place
Hopefully not tho lol
I feel okay for the maths section as I've completed the relevant A2 content but its the physics section that has me worried.
However, we all have until Nov 2nd which is plenty of time to get comfortable with everything and do all the past papers + more
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: December 4, 2016
Share this discussion:
Tweet
Related discussions:
 2016 Physics Applicants
 Pat 2016 Discussion
 PAT 2016 Thoughts
 Oxford Reach Scholarship for 2016 entry
 Oxford PAT 2015
 The Oxford TSA thread  2016 applicants  4th November ...
 The Oxford Offer Holders Thread: 2016 Entry
 Oxford Offer Holders (2016 Entry)
 Oxford Applicants Stalking Page 2016 Entry
 An "I got into Oxford with these grades" thread
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.