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OCR PHYSICS B G495~ 18th June 2015 AM ~ A2 Physics

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For 10(b) with the motor side length, I said the rate increases. Here's my reasoning:

* F=BIL, and L halves, so the force on the side 'AB' halves.
* However, this means the mass halves too assuming the mass of the wire per unit length is constant, so acceleration on that part of the wire is constant.
* Now I remembered v=ωr. I assumed the same applied to acceleration (turns out it does: a=αr where α is angular acceleration). If all of the side lengths are halved, then AB is half the distance from the axis of rotation, so the angular acceleration on the motor α=a/r is doubled.
* Hence the rate of rotation doubles too.

Not sure if this logic works, but it was only two marks, so I probably hit at least one of the marks.
(edited 8 years ago)
Original post by HennersPD
I did around the point X, is that correct or was it through X?


Through x, centre the dot in the middle


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Original post by RealTime
Oh I don't understand why it would be what you done, since equipotentials cut the field lines at 90 degrees and lies on a line of equal electric field strength, which would be a circle around the point charge


I thought it was the equipotential at x


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Original post by HennersPD
yeah thats true! did you get the last question with the gamma factor? got v = 2.1*10^8 ms-1 and how did you work the resolution out for the flea image?


I got that, yes
I can't remember to be honest
Think I calculated the length it would be in m
Then divided by the 480 pixels


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Original post by Mutleybm1996
I thought it was the equipotential at x


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Yeah it was, did you draw a circle around the point charge that went through X?
Reply 405
Q16 where you had to work out the turns in the solenoid and show that it was 3800, what were the calculations in the next part of the question?
Forgotten what I put as my answer so hopefully Ill remember the working
Original post by Mutleybm1996
I thought it was the equipotential at x


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it was equipotential at X sorry yes i did it through X with the dot at centre! also for the last question I can't seem to work it out now even though i got in the exam... the total energy was: 0.511MeV+(200*10^3 V * 1.6*10-19 C) right? and Erest was 0.511MeV ?
Original post by STATER
Q16 where you had to work out the turns in the solenoid and show that it was 3800, what were the calculations in the next part of the question?
Forgotten what I put as my answer so hopefully Ill remember the working


3800 turns of wire of diameter 0.010m

So length of wire = 3800 * pi * 0.010 = 119m

Since resistance per unit length was 1.6 ohms per meter (given), resistance = 1.6 * 119 = 191 ohms.

Since V = IR and I = 2.0A, V = 2.0 * 191 = 382V ~ 380V
Reply 408
Original post by Greating
3800 turns of wire of diameter 0.010m

So length of wire = 3800 * pi * 0.010 = 119m

Since resistance per unit length was 1.6 ohms per meter (given), resistance = 1.6 * 119 = 191 ohms.

Since V = IR and I = 2.0A, V = 2.0 * 191 = 382V ~ 380V


Thanks. Exactly what I did. Put my mind to rest.
Original post by Greating
3800 turns of wire of diameter 0.010m

So length of wire = 3800 * pi * 0.010 = 119m

Since resistance per unit length was 1.6 ohms per meter (given), resistance = 1.6 * 119 = 191 ohms.

Since V = IR and I = 2.0A, V = 2.0 * 191 = 382V ~ 380V


I used the value of turns as 4000 (given in the question) and the V as 402V. Will I get the marks since it's still the correct method and usually with these there is a range of answers?
Original post by AccountName
It was 81/100 for 135 ums last year (and the year before). Considering that (through the polls linked to this thread and last year's equivalent) this year's paper seems to have been slightly harder than last year's, I'm hoping for a slightly lower 90 percent mark, although I suppose this could be wishful thinking.


That's good then, didn't want to be too optimistic haha
Original post by megatovskimaths
That's good then, didn't want to be too optimistic haha


for the question working out the number of turns and then the V, would you get the marks if you used the 4000 turns from the show that for the previous part which gives V=402V??? just because people used 3800V and I got an answer of 3846 turns
Original post by Greating
3800 turns of wire of diameter 0.010m

So length of wire = 3800 * pi * 0.010 = 119m

Since resistance per unit length was 1.6 ohms per meter (given), resistance = 1.6 * 119 = 191 ohms.

Since V = IR and I = 2.0A, V = 2.0 * 191 = 382V ~ 380V


Mr Greating, why have you multiplied the length of wire by pi? Surely 3800*0.01 gives the total length of the wire in question? Haven't you singly sound the circumference of the wire and multiplied it by 3800?

Looking forward to reading your reply.
Original post by CarlosMartino
Mr Greating, why have you multiplied the length of wire by pi? Surely 3800*0.01 gives the total length of the wire in question? Haven't you singly sound the circumference of the wire and multiplied it by 3800?

Looking forward to reading your reply.


Well it's circular, so you have to calculate the circumeference, which is the length taken for one coil, and hence multiply by the number of coils to find the length of wire
Original post by HennersPD
I used the value of turns as 4000 (given in the question) and the V as 402V. Will I get the marks since it's still the correct method and usually with these there is a range of answers?


I done that too, usually they do give full marks for using the value given in the earlier part of the question
Original post by HennersPD
for the question working out the number of turns and then the V, would you get the marks if you used the 4000 turns from the show that for the previous part which gives V=402V??? just because people used 3800V and I got an answer of 3846 turns


Got exactly the same answer if I remember correctly 😊
Original post by CarlosMartino
Mr Greating, why have you multiplied the length of wire by pi? Surely 3800*0.01 gives the total length of the wire in question? Haven't you singly sound the circumference of the wire and multiplied it by 3800?

Looking forward to reading your reply.


Yeah I got 380V


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Original post by Quackmatic
For 10(b) with the motor side length, I said the rate increases. Here's my reasoning:

* F=BIL, and L halves, so the force on the side 'AB' halves.
* However, this means the mass halves too assuming the mass of the wire per unit length is constant, so acceleration on that part of the wire is constant.
* Now I remembered v=ωr. I assumed the same applied to acceleration (turns out it does: a=αr where α is angular acceleration). If all of the side lengths are halved, then AB is half the distance from the axis of rotation, so the angular acceleration on the motor α=a/r is doubled.
* Hence the rate of rotation doubles too.

Not sure if this logic works, but it was only two marks, so I probably hit at least one of the marks.


I don't think it would be, because the radius of rotation is staying constant, it's just the length of wire in the coil that is halving. I therefore said that the force is halving, so the rate of rotation would halve.
Opinions on grade boundaries people?


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Original post by ryanWales
I don't think it would be, because the radius of rotation is staying constant, it's just the length of wire in the coil that is halving. I therefore said that the force is halving, so the rate of rotation would halve.


Don't forget that all sides of the square coil halved in length, including the one at the back. Therefore, the radius of rotation definitely halves.

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