question 9 (1 Viewer)

[vnblueberry]

i was talking to this with my friends and ..
looking at the answers B and C .. the current flowing from Y to X / X to Y
are they not the same thing? i mean .. from x-y .. the current must go from y to x later .. flaw? or is my concept wrong

 

[RingerINC]

i was talking to this with my friends and ..
looking at the answers B and C .. the current flowing from Y to X / X to Y
are they not the same thing? i mean .. from x-y .. the current must go from y to x later .. flaw? or is my concept wrong

your logic is flawed. It says the current through the globe.

Hence one is one direction the other is the opposite direction.

 

[myeewyee]

^^ truth

 

[vnblueberry]

thanks :rofl:

 

[RingerINC]

also, relatively sure that the answer was b...

Because of this

http://www.zamandayolculuk.com/cetinbal/faradaydisk.htm

because they have the same diagram as ours if you flip it around and they labelled electron flow as the opposite to b, hence conventional current is b.

 

[suchet_i]

if u look at eddy currents forming from the point of view of the handle, they will be forming in an anticlockwise direction. such a direction of the current will mean that it travels through to x to the globe and to y and so on .. if you understand what i am saying how can i be wrong?

 

[mlinger]

thing is didnt d say it was an AC current whereas b and c said was DC?.. would this produce an ac current since there are no brushes???

 

[hyparzero]

No, it produces DC.

 

[mithu]

wholly fuck. didn't the start of the question say an AC generator. FUCK. That means no current would be produced

 

[maCe]

As I posted in the Multiple Choice Answers thread:

I put C for Question 9. It's definitely not A - it's a homopolar generator. It's not D either, it can't be AC as there's no change in the polarity - as the metal disc is always rotating in the same direction, relative to the static components (brushes/magnets/indicator). So that leaves B or C.

There's a picture on Page 5 of this thread with a diagram of a homopolar generator, and the accompanying post saying that it indicates the answer is B - might I point out, the orientation of the magnets in that picture is the opposite to whats in the exam paper.

Nah pretty sure it's B

Also, Homopolar motors operate on a similar principle, as in, they don't need a commutator or slip rings etc, and yet operate on DC power. It's easy to make one too, and fun to mess around with

See: http://en.wikipedia.org/wiki/Homopolar_motor

Just because it doesn't use a commutator or appears to use slip rings doesn't automatically mean it's an AC generator. It's a completely different principle for power generation - notice the lack of coils and wires on each side of the armature to connect to said slip rings in 'proper' generators.

As yet though, I think it's too early to say what is wrong and right, until someone provides a fully worked and noted solution to this 1 marker.

(Answer C = X-Y by the way)


Edit for clarity: looking at the above quoted picture again, it is hard to tell which way it is actually turning.. though I would assume the arrow indicates a clockwise direction if you were to look at it from the top down. In which case, as I said, it's opposite to what's in the exam paper, which would lead me to believe the current flows the opposite way, thus Answer C, not B.

 

[fizzwizz]

Hi The correct answer to this q is B.Use Lenz's law to answer it, and we will take answer C to show that it is incorrect.

If current goes through globe from X to Y then a current moves upwards from brush Y to the handle. Now think of this current as a current in a single conductor pointing upwards, passing through the magnetic field which is directed from left to right. Now use the right hand push rule: current up, fingers pointing left to right................this produces a force on the conductor (disc) into the page. But it is already moving into the page (at the bootom) so this force would increase its jmotion, not decrease its motion as required by lenzs law. A force has got to act on the conductor pushing it out of the field, as it enters the field. Using current flow downwards from handle to brush Y gives you this force. So answer is B.