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# Edexcel Unit 4: Physics on the Move 6PH04 (11th June 2015) Watch

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1. I guess maybe its not in the syllabus then.

We should really just get 1 exam board for everyone
2. (Original post by ThatWasHard!)
I guess maybe its not in the syllabus then.

We should really just get 1 exam board for everyone
Ikr. Edexcel is a joke lol

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3. How is revision going?
4. (Original post by Freddy-Francis)
How is revision going?
Got A* in my Unit 4 mock

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5. (Original post by Freddy-Francis)
How is revision going?
I've been neglecting physics so hard. Will start next week Wbu?
6. Can someone tell me if this is correct?
If you need to work out the magnetic flux density if you know n, A and phi then you use phi=BA since magnetic flux isn't affected by the number of turns in the coil (so you don't use phi=nBA)?
7. (Original post by BP_Tranquility)
Can someone tell me if this is correct?
If you need to work out the magnetic flux density if you know n, A and phi then you use phi=BA since magnetic flux isn't affected by the number of turns in the coil (so you don't use phi=nBA)?

Magnetic flux is not affected by the number of coils (φ=ab)

Magnetic flux linkage IS affected by the number of coils (Φ=nab)

Note magnetic flux linkage uses a capital phi and magnetic flux doesn't.

P.s. I am 99% sure this is correct, can someone please tell me if it's wrong

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8. (Original post by cerlohee)
I've been neglecting physics so hard. Will start next week Wbu?
Started this week ^^
9. Anyone know if Baryon numbers actually turn up in exams? Just seen it in my revision guide, but wasn't taught it in class and I haven't seen any exam questions on them yet.
10. (Original post by jay_em)
Anyone know if Baryon numbers actually turn up in exams? Just seen it in my revision guide, but wasn't taught it in class and I haven't seen any exam questions on them yet.
Baryon number is conserved so I guess they could ask about it- just remember all quarks have a Baryon number of +⅓ and everything else is 0.

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11. (Original post by jay_em)
Anyone know if Baryon numbers actually turn up in exams? Just seen it in my revision guide, but wasn't taught it in class and I haven't seen any exam questions on them yet.
I have seen to validate some particle equations and stuff. Where you have to check if everything is conserved

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12. Just saw that lepton number/baryon number/strangeness conservation isn't mentioned in the specification- does make you wonder if we really do need to know about them ...

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13. (Original post by BP_Tranquility)
Just saw that lepton number/baryon number/strangeness consecration isn't mentioned in the specification- does make you wonder if we really do need to know about them ...

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I think we do. Loads of stuff is hidden like that would go under conservation of the collisions part in the spec. Edexcel are really bad with the specs as for physics if you learnt the spec in and out you wont actually know all tje stuff for the questions. Most of the stuff is application they say lol.

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14. (Original post by physicsmaths)
I think we do. Loads of stuff is hidden like that would go under conservation of the collisions part in the spec. Edexcel are really bad with the specs as for physics if you learnt the spec in and out you wont actually know all tje stuff for the questions. Most of the stuff is application they say lol.

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Ah, that is annoying, OCR seem to give a lot more detailed specifications (for chemistry at least) -_-

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15. Can anyone walk me through the thought process in answering this question here?
Attached Images

16. Also, can somebody explain question 14 C part (II) on the june 2013 paper?
17. (Original post by Hectors)
Can anyone walk me through the thought process in answering this question here?
I wouldn't say that electric/magnetic fields were my strongest area but I'll have a stab at it :

So you know that induced emf=-d(nBA)/dt. The greatest change in flux is between the 2 field lines furthermost to the left (due to these 2 field lines giving the biggest gap). The induced emf is therefore at a maximum at these 2 points.

The induced emf is 0V at A and B (since these 2 points have no magnetic field lines passing through them so there's no magnetic flux which means no induced emf!). The induced emf is 0V in the centre of the wire as well (where the field lines are equally spaced) since there's no rate of change in flux here either.

So the graph would initially begin at 0V, and it would curve upwards until it reaches a maximum value. The curve will then sharply go down to 0V at the centre of the wire. The emf would then become negative and decrease down to a minimum value, and again as the length of wire approached B, the curve would go up to 0V again. The reason why the emf becomes negative is because the field direction changes so the direction of induced emf is reversed.

Hope that helped a little
18. (Original post by BP_Tranquility)
I wouldn't say that electric/magnetic fields were my strongest area but I'll have a stab at it :

So you know that induced emf=-d(nBA)/dt. The greatest change in flux is between the 2 field lines furthermost to the left (due to these 2 field lines giving the biggest gap). The induced emf is therefore at a maximum at these 2 points.

The induced emf is 0V at A and B (since these 2 points have no magnetic field lines passing through them so there's no magnetic flux which means no induced emf!). The induced emf is 0V in the centre of the wire as well (where the field lines are equally spaced) since there's no rate of change in flux here either.

So the graph would initially begin at 0V, and it would curve upwards until it reaches a maximum value. The curve will then sharply go down to 0V at the centre of the wire. The emf would then become negative and decrease down to a minimum value, and again as the length of wire approached B, the curve would go up to 0V again. The reason why the emf becomes negative is because the field direction changes so the direction of induced emf is reversed.

Hope that helped a little
Thanks!!
19. (Original post by Hectors)
Also, can somebody explain question 14 C part (II) on the june 2013 paper?
"Initial rate of increase of potential difference" essentially means dV/dt, so draw a tangent to the curve at the beginning and calculate a value for V/t.

Next, we know that C=Q/V and Q=It. So C=It/V. Rearranging this, I=CV/t. Since we know C and we found V/t, we can now find the initial charging current I from the formula
20. (Original post by Navo D.)
"Initial rate of increase of potential difference" essentially means dV/dt, so draw a tangent to the curve at the beginning and calculate a value for V/t.

Next, we know that C=Q/V and Q=It. So C=It/V. Rearranging this, I=CV/t. Since we know C and we found V/t, we can now find the initial charging current I from the formula

Thanks, I've managed to get up to dv/dt but I didn't know what to do with it 😊

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