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Unit 4 Physics Edexcel A2 and Edexcel IAL

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Hey! I want to clear a couple of doubts from this paper. I aplogise in advance since they are a lot but I'm kind of weak in this unit so a minor help will be appreciated! Thanks In Advance!

https://ad57adc4f6eb5ea42b541057f16290e611d4e42b.googledrive.com/host/0B1ZiqBksUHNYcHRSNUJpeFpqNXM/June%202012%20QP%20%20-%20Unit%204%20Edexcel%20Physics.pdf

5- Why is the answer 3.5 s i.e D, how did they come to this answer when no info regarding capacitor and the resistor is given?
13- Why don't we just find the momentum here and then use the formula wavelength=Plank's constant/momentum. Where did E=mc^2 come from?
17(b) onwards. Don't understand properly.
18(c)-I don't get this at all. :tongue: What and how was this supposed to be solved?
For 5 you use the discharge equation.
Ln (0.5)=-t/5 so t=Ln (0.5 )*-5

For 13 λ = h/p is used to find the wavelength associated with a moving particle. Photons are waves, so you cannot use this equation to find their wavelength. Use the photon model equation E = hf to find the photons frequency, then use λ = c (speed of light)/f to find the photon's wavelength.
(edited 7 years ago)
Original post by Mowerharvey
For 5 you use the discharge equation.
Ln (0.5)=-t/5 so t=Ln (0.5 )*-5

For 13 λ = h/p is used to find the wavelength associated with a moving particle. Photons are waves, so you cannot use this equation to find their wavelength. Use the photon model equation E = hf to find the photons frequency, then use λ = c (speed of light)/f to find the photon's wavelength.


Ah okay! Thanks a lot! :smile:
Hi, could anyone explain question 4 to me please?

https://ad57adc4f6eb5ea42b541057f16290e611d4e42b.googledrive.com/host/0B1ZiqBksUHNYcHRSNUJpeFpqNXM/June%202011%20QP%20-%20Unit%204%20Edexcel%20Physics.pdf

Thanks :smile:
(edited 7 years ago)
Original post by candycake


Coil X produces a magnetic field of magnetic flux density 0.002 T, and coil Y is in this magnetic field. The equation for flux linkage is Φ = BAN, so Φ = 0.002 * 0.0004 * 50 = 4x10^-5. You don't need to worry about coil X except for the magnetic field it produces.
Original post by Mowerharvey
Coil X produces a magnetic field of magnetic flux density 0.002 T, and coil Y is in this magnetic field. The equation for flux linkage is Φ = BAN, so Φ = 0.002 * 0.0004 * 50 = 4x10^-5. You don't need to worry about coil X except for the magnetic field it produces.


Thank you!
anyone else find this unit a lot harder than unit 5? Doing past papers I get the lowest grades in this exam out of all my exams.
Original post by klosovic
anyone else find this unit a lot harder than unit 5? Doing past papers I get the lowest grades in this exam out of all my exams.


Yup! Me! The only topic that's a bit of a pain in Unit 5 is Astrophysics. Else, the rest of them is okay. But unit 4 is just depressing! :frown:
Hello! Need help in this paper.
https://ad57adc4f6eb5ea42b541057f16290e611d4e42b.googledrive.com/host/0B1ZiqBksUHNYcHRSNUJpeFpqNXM/January%202012%20QP%20-%20Unit%204%20Edexcel%20Physics.pdf

2. Why is the answer C instead of B?
16(c). What does the time constant have to do with the suitability?
18(bi and biii). How do we even construct the equations here?
Original post by sabahshahed294
Hello! Need help in this paper.
https://ad57adc4f6eb5ea42b541057f16290e611d4e42b.googledrive.com/host/0B1ZiqBksUHNYcHRSNUJpeFpqNXM/January%202012%20QP%20-%20Unit%204%20Edexcel%20Physics.pdf

2. Why is the answer C instead of B?
16(c). What does the time constant have to do with the suitability?
18(bi and biii). How do we even construct the equations here?


For 16(c), you need to prove that the capacitor can charge and discharge during the time period of sound with frequency 20 Hz. So you need to work out the time constant of the capacitor, which is 0.005 s.
The time it takes to charge/discharge is approximately 5T, which in this case is 0.025 s, so the time taken to charge AND discharge is
0.05 s. The time period of sound 20 Hz is 1/20 = 0.05 s, which is the same as the time taken for the capacitor to charge and discharge, therefore the capacitor is suitable.
Original post by Mowerharvey
For 16(c), you need to prove that the capacitor can charge and discharge during the time period of sound with frequency 20 Hz. So you need to work out the time constant of the capacitor, which is 0.005 s.
The time it takes to charge/discharge is approximately 5T, which in this case is 0.025 s, so the time taken to charge AND discharge is
0.05 s. The time period of sound 20 Hz is 1/20 = 0.05 s, which is the same as the time taken for the capacitor to charge and discharge, therefore the capacitor is suitable.


Thanks! :smile: Got it now.
Hi, would any be able to explain questions 13bii and 13biii please?

https://ad57adc4f6eb5ea42b541057f16290e611d4e42b.googledrive.com/host/0B1ZiqBksUHNYcHRSNUJpeFpqNXM/January%202014%20%28IAL%29%20QP%20-%20Unit%204%20Edexcel%20Physics.pdf

Thanks :smile:
Original post by candycake


Yeh so basically, when it is moving through there is a change in flux is what we are looking at. So use this fact to explain bii. For the next prt is a thing about current so we want to know which way the conventional current goes and we can use our hands to figure that out!


Posted from TSR Mobile
Original post by physicsmaths
Yeh so basically, when it is moving through there is a change in flux is what we are looking at. So use this fact to explain bii. For the next prt is a thing about current so we want to know which way the conventional current goes and we can use our hands to figure that out!


I tried using FLHR but I get the wrong answer for bii - presumably the field would be in line with the arrows on the diagram, so which way would the motion be?

Why is there no build up of charge after the plane reaches a constant velocity?

Thanks!
Original post by candycake
I tried using FLHR but I get the wrong answer for bii - presumably the field would be in line with the arrows on the diagram, so which way would the motion be?

Why is there no build up of charge after the plane reaches a constant velocity?

Thanks!


For using FLHR for any conducting object moving in a magnetic field, think of the moving object itself as being a "flow of positive charge", so instead of pointing your thumb in the direction of motion of the plane you point your second finger in that direction instead and use your thumb to work out the direction of force acting on the positive charges and thus which wing will become positively charged.

After the plane reaches a constant velocity, the change in magnetic flux becomes zero as there is no acceleration, so there is no emf induced in the wings and thus no buildup of charge.
Original post by PhysicsIP2016
For using FLHR for any conducting object moving in a magnetic field, think of the moving object itself as being a "flow of positive charge", so instead of pointing your thumb in the direction of motion of the plane you point your second finger in that direction instead and use your thumb to work out the direction of force acting on the positive charges and thus which wing will become positively charged.

After the plane reaches a constant velocity, the change in magnetic flux becomes zero as there is no acceleration, so there is no emf induced in the wings and thus no buildup of charge.


Thank you :smile:
Original post by PhysicsIP2016
For using FLHR for any conducting object moving in a magnetic field, think of the moving object itself as being a "flow of positive charge", so instead of pointing your thumb in the direction of motion of the plane you point your second finger in that direction instead and use your thumb to work out the direction of force acting on the positive charges and thus which wing will become positively charged.

After the plane reaches a constant velocity, the change in magnetic flux becomes zero as there is no acceleration, so there is no emf induced in the wings and thus no buildup of charge.


Just one more thing - For question 13biii, the mark scheme says "The build-up of charge creates an electric fieldThis creates a force in the opposite direction to the magnetic force." Could you please explain this to me?
Original post by PhysicsIP2016
For using FLHR for any conducting object moving in a magnetic field, think of the moving object itself as being a "flow of positive charge", so instead of pointing your thumb in the direction of motion of the plane you point your second finger in that direction instead and use your thumb to work out the direction of force acting on the positive charges and thus which wing will become positively charged.

After the plane reaches a constant velocity, the change in magnetic flux becomes zero as there is no acceleration, so there is no emf induced in the wings and thus no buildup of charge.


After the plane reaches a constant velocity isn't it still cutting lines of flux, and therefore an emf is still induced?
Original post by candycake
Just one more thing - For question 13biii, the mark scheme says "The build-up of charge creates an electric fieldThis creates a force in the opposite direction to the magnetic force." Could you please explain this to me?


As there is a potential difference, there is now a uniform electric field which acts across the wings of the plane from right to left. This exerts a force on the charged particles so now the positive charges are attracted towards the left hand side of the plane where the magnetic field force initially moved them to the right hand wing :smile:
Original post by Mowerharvey
After the plane reaches a constant velocity isn't it still cutting lines of flux, and therefore an emf is still induced?


Sorry yes you're correct, it is cutting lines of flux but as the value for d phi and dt remain the same, the emf is constant so there is no overall increase in the e.m.f.

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