Question I dont understand...

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2710

I am not moving the wire. I am trying to find the force on the wire. Your posts are very confusing. I am not using the Dynamo rule, but the Motor Rule. I do not know why you are using your right hand, unless you was taught a different way.

And then with your next posts, you say that it is pointing right...so, meh, it doesn't matter anymore. I got it.

EDIT:

StoneBridge, can I just ask:

Is the same as this:

Right? (when talking about the force)

Ie: Both move out of the page towards you?

Yes if the 1st one is a wire carrying a conventional current to the left.
The wire will experience a force out of the page.
Both are LHR to find the force on a current in a field. One current conventional (+) the other negative.

Reply 41

rbnphlp

Stonebridge

Yes, and the wire is stationary, but the charged particle is moving.
The LHR is used for a stationary wire carrying current. The wire experiences a force.
If the wire is moving you use the RHR to find the induced emf/current direction.
With the LHR the current is already flowing and the rule finds the force direction.

Yes , that is what Iam saying as well ..In post 29 when I said the former(motion of the wire) uses RHR..

Reply 42

2710

I found another error in this revision guide I think. It's attached.

Sorry its not very clear, but it says:

'emf is zero when the coil is parallel to the magnetic field...

...and greatest when they are perpendicular'

This is wrong right (the bottom graph)

What about the top graph. Its a graph of BAN, ie Flux linkage. I hope you can see the coil

Thanks

Reply 43

2710

Yeh, thats what I thought, rotate them 90 degrees. So If I do that, both graphs will be correct right? Dont want to revise the wrong stuff >__<

Reply 44

Stonebridge

Yes it's wrong. Mind you I have now got crossed eyes from trying to see that diagram. :eek3:

Max emf when coil is parallel to field. Thats when its sides are cutting flux most quickly.
Min emf when the coil is perpendicular to the field.
The little magnet and coil diagrams are wrong.

Reply 45

2710

OK thanks.

I have are system oscillating at 1.5Hz, and I have forced vibrations of 10Hz. How do I find the phase difference? Is there a simple method?

Apparently the answer is out of phase by pi.

Thanks

Reply 46

rbnphlp

2710

draw a sketch ..
for 1.5hz, there 1.5 (1+0.5)oscillations per second..
10 there are 10 osicallitions per sec..
then 10 Hz will be always pi out of phase (1.5=1+0.5, 0.5 oscillation is half a wavelength)

Reply 47

Stonebridge

2710

When they said "find", did they mean calculate? What are you given in the question?
At A Level generally, you would be expected to know that the following is the general way such a system behaves:

•

When the driving force is at a higher frequency than the oscillating system, the sytem is about half a wavelength (pi) behind.

•

When the driving force is at the same frequency as the oscillating system (resonance), the sytem is a quarter of a wavelength behind.

•

When the driving force is at a lower frequency than the oscillating system, the sytem is almost in phase.

It is beyond A Level to prove this.
An example of this is seen in the experiment called "Barton's Pendulums."
A search on YouTube will find loads of vids of this.

Reply 48

sykik

Original post by Stonebridge

Between the parallel plates there is a uniform electric field. That means the value of E (and therefore F on any charge) is the same everywhere. (Don't get confused with the potential at different points between the plates).
For a parallel plate system the field is simply E=V/d (Often called a uniform potential gradient) and is the same everywhere. So F=Eq and E=V/d
It would be the same at the other two points.