OCR Physics A G485 - Frontiers of Physics - 18th June 2015

\phi = BAN

, hence you need to pick a certain value of the flux linkage. The value you need to you is the one with the amplitude, hence 2.0\times 10^-^2 Wb-turns . Then just rearrange the equation to find the flux density.

So why is the equation in the formila book phi=BAcosx ?

Reply 401

Tiwa

7

Original post by chem@uni

So why is the equation in the formila book phi=BAcosx ?

That is magnetic flux = area x magnetic flux density. Both Magnetic flux linkage and magnetic flux have the same symbol \phi . Flux linkage = N\phi = BAN . In the CGP book, the unit for flux linkage is capital phi I believe.

Reply 402

chem@uni

9

Original post by Tiwa

That is magnetic flux = area x magnetic flux density. Both Magnetic flux linkage and magnetic flux have the same symbol \phi

. Flux linkage = N\phi = BAN . In the CGP book, the unit for flux linkage is capital phi I believe.

Yes thank you was being a bit of an idiot :smile:

Reply 403

Tiwa

7

Original post by chem@uni

Yes thank you was being a bit of an idiot :smile:

Don't sweat it! It's alright!

Reply 404

chem@uni

9

Original post by Tiwa

Don't sweat it! It's alright!

you couldnt draw the graph for 4 part D just MS is very helpful in regards to "Correct shape"

Reply 405

randlemcmurphy

11

Original post by chem@uni

you couldnt draw the graph for 4 part D just MS is very helpful in regards to "Correct shape"

If you are that stuck sub values of r into the equation to work out F, plot some points then draw the curve. As F is directly proportional to 1/r². The graph of F over r, will decrease but will curve i.e. not a straight line.

Reply 406

Elcor

14

In the spec it says "Describe qualitatively the evolution of the universe

10^{-43}

s after the Big Bang". So that's something like:

Soup of quarks and leptons
Quarks combine to form hadrons
Hadrons combine to form nuclei
Matter and antimatter annihilate with matter left over
Atoms form
Matter and radiation separate - origin of CMB
Stars and galaxies form
Present - Universe at temperature of 3K

But do we need to know the time scales too?

Reply 407

sagar448

4

Original post by Elcor

In the spec it says "Describe qualitatively the evolution of the universe

10^{-43}

s after the Big Bang". So that's something like:

Soup of quarks and leptons
Quarks combine to form hadrons
Hadrons combine to form nuclei
Matter and antimatter annihilate with matter left over
Atoms form
Matter and radiation separate - origin of CMB
Stars and galaxies form
Present - Universe at temperature of 3K

But do we need to know the time scales too?

No.

Reply 408

Hilton184

9

Original post by Elcor

In the spec it says "Describe qualitatively the evolution of the universe

10^{-43}

s after the Big Bang". So that's something like:

Soup of quarks and leptons
Quarks combine to form hadrons
Hadrons combine to form nuclei
Matter and antimatter annihilate with matter left over
Atoms form
Matter and radiation separate - origin of CMB
Stars and galaxies form
Present - Universe at temperature of 3K

But do we need to know the time scales too?

I don't think so. It says 'qualitatively' so I think its trying to indicate that we don't need to know the times.

Reply 409

Elcor

14

Original post by Hilton184

I don't think so. It says 'qualitatively' so I think its trying to indicate that we don't need to know the times.

Original post by sagar448

No.

Cheers. It gives the times in the revision guide so I was wondering if anyone had seen it in an exam Q before.

Reply 410

Elcor

14

When the mark scheme assigns a mark to a formula, do you have to state it or does subbing the values get that mark?

Reply 411

sagar448

4

Original post by Elcor

When the mark scheme assigns a mark to a formula, do you have to state it or does subbing the values get that mark?

I think you have to state it because subbing in the values kind of make it confusing to tell which value is matched for what symbol. So I guess just state it, probably the examiners want quick marking. So they probably just look at the formula and the answer and assign a mark.

What other people think?

Reply 412

Absent Agent

21

Original post by sagar448

I think you have to state it because subbing in the values kind of make it confusing to tell which value is matched for what symbol. So I guess just state it, probably the examiners want quick marking. So they probably just look at the formula and the answer and assign a mark.

What other people think?

Usually If you have the final answer correct you get the full marks even if you haven't written anything down. The marks for formula and workings are for when you don't have the final answer correct

Posted from TSR Mobile

Reply 413

BecauseFP

2

I just want to check that I've got this right, here's my notes for mass spectrometers.

The substance is vaporised and ionised.
It is then accelerated through an electric field.
It then enter a velocity selector, which is an area where a magnetic and electric field exert equal force on the ions.
So BQv=EQ, giving v=E/B.
The selector has a slit positioned such that only particles with right velocity can pass through, then after that the ions pass through the slit, they enter an evacuated circular chamber where only the second magnetic field is present so the ions undergo circular motion.
The ions are then detected by a moveable detector that find the raduis of the path and the relative abudance of the ions.
Then you can do some maths to get the mass (I can't be bothered to do the TEX).
And then using the charge to mass ratio, and relative presence, you can determine their isotopes.

Is this all right? I'm almost worried that it's overkill, but I'm obviously more worried about if it's correct or not.

Reply 414

BrokenS0ulz

3

So the last part of this question seems to contradict a more recent question, maybe I'm just misunderstanding.

So both the spheres are +ly charged and they tell us the force between them is 0.14mN, this should be a repulsive force because they're both +ve. So the bottom sphere should exert a 0.14mN force on the scales, which will exert a 0.14mN force upwards due to newton's 3rd law, so shouldn't the scale decrease in its reading. The answer says an increase in the reading by 0.014 to 8.219g. Also how would you go from a force of 0.14mN to a mass of 0.014g?

Reply 415

randlemcmurphy

11

Original post by BrokenS0ulz

So the last part of this question seems to contradict a more recent question, maybe I'm just misunderstanding.

So both the spheres are +ly charged and they tell us the force between them is 0.14mN, this should be a repulsive force because they're both +ve. So the bottom sphere should exert a 0.14mN force on the scales, which will exert a 0.14mN force upwards due to newton's 3rd law, so shouldn't the scale decrease in its reading. The answer says an increase in the reading by 0.014 to 8.219g. Also how would you go from a force of 0.14mN to a mass of 0.014g?

What recent question does it contradict? Was the actual force 0.14mN, as it does say it is about. Maybe when you divide the actual value by 9.81 you get 0.014g? The force which is being exerted on the bottom sphere is due to the top sphere. And the force the top sphere experiences is due to the bottom sphere. So this means the top sphere experiences an upwards force and the bottom sphere a downwards force. They both don't "feel" an upwards and downwards force.

Reply 416

sagar448

4

Original post by BrokenS0ulz

So the last part of this question seems to contradict a more recent question, maybe I'm just misunderstanding.

So both the spheres are +ly charged and they tell us the force between them is 0.14mN, this should be a repulsive force because they're both +ve. So the bottom sphere should exert a 0.14mN force on the scales, which will exert a 0.14mN force upwards due to newton's 3rd law, so shouldn't the scale decrease in its reading. The answer says an increase in the reading by 0.014 to 8.219g. Also how would you go from a force of 0.14mN to a mass of 0.014g?

Basically the spheres are positively charged we all got that point. So this means that the bottom sphere is pushing upwards on the top sphere and the top sphere is pushing downwards on the bottom sphere. So the magnitude of the force that the top sphere is pushing by is 0.14mN. This can be turned to grams simply by dividing it by 9.81. You may ask that how is this mass related? Well the force has hypothetical "mass". Which is putting weight on the bottom sphere and this is pushing down on the scale therefore adding 0.0142711g to the initial reading.

Also notice that they said the initial reading was 8.205g this means that they have already taken in account of the weight of the sphere sitting on the scale so newtons third law still applies but the upward force from the scale does not need to be taken into account.

:biggrin:

Reply 417

BrokenS0ulz

3

Original post by randlemcmurphy

What recent question does it contradict? Was the actual force 0.14mN, as it does say it is about. Maybe when you divide the actual value by 9.81 you get 0.014g? The force which is being exerted on the bottom sphere is due to the top sphere. And the force the top sphere experiences is due to the bottom sphere. So this means the top sphere experiences an upwards force and the bottom sphere a downwards force. They both don't "feel" an upwards and downwards force.

It doesn't contradict it, I was doing the question wrong :tongue:

thanks for the help. The actual force rounded to 0.14mN. The only thing I don't get now is how we get the 0.014g. I understand the weight of the bottom sphere is effectively increasing by 0.14mN, but dividing this by 9.81 gives 1.43x10^-5 g?

Original post by sagar448

Basically the spheres are positively charged we all got that point. So this means that the bottom sphere is pushing upwards on the top sphere and the top sphere is pushing downwards on the bottom sphere. So the magnitude of the force that the top sphere is pushing by is 0.14mN. This can be turned to grams simply by dividing it by 9.81. You may ask that how is this mass related? Well the force has hypothetical "mass". Which is putting weight on the bottom sphere and this is pushing down on the scale therefore adding 0.0142711g to the initial reading.

Also notice that they said the initial reading was 8.205g this means that they have already taken in account of the weight of the sphere sitting on the scale so newtons third law still applies but the upward force from the scale does not need to be taken into account.

:biggrin:

Thanks! But how does 0.14mN = 0.014g, diving 0.14mN by 9.81 gives 1.43x10^-5?

Reply 418

sagar448

4

Original post by BrokenS0ulz

It doesn't contradict it, I was doing the question wrong :tongue:

thanks for the help. The actual force rounded to 0.14mN. The only thing I don't get now is how we get the 0.014g. I understand the weight of the bottom sphere is effectively increasing by 0.14mN, but dividing this by 9.81 gives 1.43x10^-5 g?

Thanks! But how does 0.14mN = 0.014g, diving 0.14mN by 9.81 gives 1.43x10^-5?