CSSA Physics (1 Viewer)

[umz93]

Funny thing is, I didn't use the torque formula. BECAUSE I read the LENGTH being the given amount. It was not told that it was a square shaped coil


What I did was.

F = mg.

Get the formula for the P on the rotor


Then use F = BIl to get the current.

Yes i did the same thing
Is that also correct

 

[Screamm]

Funny thing is, I didn't use the torque formula. BECAUSE I read the LENGTH being the given amount. It was not told that it was a square shaped coil


Okay here's what was given. The Magnitude of Flux in mT, length of the coil and the mass of the particle on the rotor


I used F = mg to get the force of the mass acting on the rotor


Then I used F = BIl to find the magnitude of the current. Which was 2.2 A or 2.02 A idk.


Then I used the right hand rule to determine the opposing force that must be acted on to counterbalance the weight force, hence the direction of the current was going towards the top.

it did say it was a square shaped coil but, it said a square coil with a side length 0.3.

 

[jamesfirst]

Re: Question on difficulty of CSSA physics

I got F..



I got F..

What youre saying about 2 and 1 is correct although they acted in opposite directions..

So 2 - 1 = 1

Thats if my wokring out was correct though



saaame, hopefully we are right

I dont know what I did, but I used the F/l = blah blah blah formula and just subbed sht in. I canceled out k , I since they were all equal. And somehow I got F.


I don't know I got a feeling that it was 3F, but I dont know which one I chose in the end.

 

[jamesfirst]

Re: Question on difficulty of CSSA physics

I'm pretty sure the 8 marker had little to do with the AC vs. DC argument of Westinghouse and Edison; the stimulus provided did mention it, however the question did not asking anything about it. Hopefully you guys weren't tricked and actually wrote a detailed analysis on how and why AC creates eddy currents (alternating current, changes in flux, induces circular current due to Lenz's Law), the effects of an eddy current (heat), the uses eddy currents have (electronic breaking, induction cooktops), the problems it faces (in reducing efficiency of transformers) and possible methods of resolving these problems (lamination). My friends have said they wrote a good essay about Westinghouse vs. Edison, and I pretty much facepalmed. Hopefully you guys didn't make the same mistake.

I did 1 sentence of AC vs DC wars.

- Transformers
- Eddy currents
- It's problems - power loss = inefficient transformers
- How they are fixed - lamination + ferrites

- 2 uses of eddy currents.


I was very tempted to write about the westinghouse vs edison. But I realised that I had too much shit to write about. I ended up using the extra section below where it says "end of question .xx"

 

[jamesfirst]

it did say it was a square shaped coil but, it said a square coil with a side length 0.3.

Ok, I didnt read that part. But why should it matter?


Why WOULD YOU want to find the torque ?? Unless I didnt read the question carefully and got owned. But I'm pretty sure that I interpreted the question correctly and my working out is correct. I mean the way I did it is logical.

 

[jamesfirst]

Yes i did the same thing
Is that also correct

I'm PRETTY sure it's correct.

 

[Screamm]

Ok, I didnt read that part. But why should it matter?


Why WOULD YOU want to find the torque ?? Unless I didnt read the question carefully and got owned. But I'm pretty sure that I interpreted the question correctly and my working out is correct. I mean the way I did it is logical.

idk... i just saw the word square and assumed that area was a component to be used.... do you think solving for B in terms of letting the force on the mass = the torque will still give marks? it makes me mad not knowing for sure

 

[Qwertyuioplm]

My friend used torque and got 0.2, not 2 as I did. So I think there's a very high possibility that using torque will resolve in you getting the wrong answer (assuming he did the calculations correctly using his method).

 

[jamesfirst]

idk... i just saw the word square and assumed that area was a component to be used.... do you think solving for B in terms of letting the force on the mass = the torque will still give marks? it makes me mad not knowing for sure

torque and force is different tho ...

 

[Screamm]

what about the difference in Ep what did people get for that one?

 

[miniwaybzz]

I'm PRETTY sure it's correct.

i did it that way aswell

 

[jamesfirst]

what about the difference in Ep what did people get for that one?

what's that ?

 

[Screamm]

the work required for the satellite to achieve it's orbit.
this is embarrassing if finding the difference in Ep wasn't the right way to do it :O

 

[jamesfirst]

I did

W = Fs

and replaced F = mg to it and found the work done on the satellite. Because the gravitational potential energy is defined as work being done on an object to move its position from infinite to a point on the field. lol

 

[Screamm]

i did the gpe at the orbit - the gpe at the earths surface to find the difference

i don't feel that confident anymore your method seems better

 

[jamesfirst]

I thought it asked for the work ? lol

 

[Screamm]

yeah it did, but somewhere in my notes i wrote the work for a mass to achieve an orbit is the ∆Ep idk :S

 

[jamesfirst]

I just don't feel confident that I did add the radius of the earth in that work question. I did it on purpose tho, it seemed more appropriate.

 

[Qwertyuioplm]

The work done is equal to the change in gravitational potential energy that the satellite faced, i.e.:

Change in Gravitational Potential Energy = G*mass of satellite*mass of Earth((1 / radius of earth) - (1 / (radius of earth + distance of satellite from radius)))

 

[Qwertyuioplm]

I did

W = Fs

and replaced F = mg to it and found the work done on the satellite. Because the gravitational potential energy is defined as work being done on an object to move its position from infinite to a point on the field. lol

That is wrong because that formula assumes g is a constant value. However as an object moves further away from Earth the value of g changes. Many textbooks are horrible at explaining this concept.