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06:05
@ArhamJain Hi Arham. You have probably heard that the field is zero in a conductor, but this does not mean fields cannot pass through a conductor. It just means the fields passing through a conductor must cancel each other to give a net field of zero inside the conductor.
If we have a charged conducting plate with a sheet of charge on both surfaces then the fields from the two sheets of charges do pass through the centre of the conducting sheet but inside the plate the two fields cancel each other.
@Pizza I guess it depends on what exactly the question means. It could mean uₙ makes an angle of 30° with uz. Without having the angle labelled on the figure it's impossible to be sure.
But if θ is the angle you've drawn then I agree with you.
 
2 hours later…
07:43
@JohnRennie Actually, I thought so too initially, but I wasn't able to reconcile this fact with the existence of Faraday Cages
Why are the people in this image not getting shocked then?
07:59
@JohnRennie it says: What changes if the square is oriented so that it forms an angle θ = 30° with respect to ûz. I interpreted it that way
Determine the surface area $S$ of the plates of a parallel-plate capacitor, knowing that the distance between the plates is $d$ and that, when inserted into an RC circuit in series with a resistor of known resistance $R$, it charges exponentially with a time constant $\tau$. (Data: $R = 1 \, \text{k}\Omega; \, \tau = 0.01 \, \text{ms}; \, d = 4 \, \text{mm}
; k = \frac{1}{4 \pi \varepsilon_0} = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2)$
08:30
hi @JohnRennie :-)
08:53
@mo-_- @Pizza Hi :-)
Hi :-)
Hi :-)
@Pizza You have to find the capacitance from the time constant 1/RC then use the parallel plate capacitor equation to find the area.
ah ok so $\tau = R \cdot C$
$C = \tau / R$
Have you studied RC circuits?
08:56
this is actually the only exercise that involves them so I didn't do them very well
thats why i asked here
OK, then just accept τ = RC and we can discuss this in detail if/when you want to do circuits.
ok :-)
@mo-_- What do you want to ask?
$C = \epsilon_0 \cdot S/d \Rightarrow S = Cd/\epsilon_0$
08:59
so thats the solution ?
Yes - simpler than you thought :-)
yes it was quick
@JohnRennie Hi :-)
Hi :-)
(sending diagram ...)
(sending text...)
Consider a long conductor wire carrying a current and three copper rings arranged as shown in the figure. Assuming that the current intensity $i(t)$ increases over time, determine whether an induced current is generated in the three rings and, if so, establish its direction.
09:04
That's quite a nice question :-)
How far have you got with it?
I think we need to consider each ring individually
so ring A ; ring B and ring C
I'll draw a diagram, and we can mark the field from the wire on it ...
B is in the screen on the bottom
and out of the screen on the side above
then perhaps for ring A since the external field is coming out of the screen then the current creates an incoming field, therefore i flows in the opposite direction?
yes :-)
09:12
So is the flux through A changing with time?
I just realised I got the fields the wrong way round :-(
Give me a moment and I'll correct it.
ah yes... I hadn't noticed
That's better :-)
So which rings have an induced current?
the flux varies when the magnetic field (B) varies, so only at the ring B?
We get an EMF round the ring, and therefore an induced current, if the flux through the ring changes with time.
ok
09:19
If you look at ring A, is there a non-zero flux through it?
The flux would actually be quite complicated to calculate as it's a hard integral, but luckily we don't need to calculate it - we just need to know if there is a non-zero flux or not.
ah ok
on the ring A the flux doesn varies
wait
The flux must vary over time for an emf to be generated induced
There is a flux through A because Φ = ∫B dA inside the ring. And since B is non-zero inside the ring we are going to get a non-zero flux.
Yes?
oh right
yes
so its the same for C
Yes, if we consider some small area dA inside the circle at a distance 𝑟 from the wire the field inside that area is B = μ₀i/2𝜋r so dΦ = BdA will be non-zero.
Yes?
yes
09:28
And when we add up the dΦs we will get a non-zero result.
And like you say this will be the same for C. The only difference between A and C is that for A the flux comes out of the screen towards us and for C the flux goes into the screen away from us.
So the next questions is: does the flux through A and C change with time?
What do you think.
yes it varies
So is an EMF, and therefore a current, generated in those rings?
yes
Yes :-)
It was easier than you thought. We go the answer with no calculations needed!
wait
we also have to do ring B
09:32
> And like you say this will be the same for C. The only difference between A and C is that for A the flux comes out of the screen towards us and for C the flux goes into the screen away from us.
Yes?
sorry! i meant B
the ring B
OK, let's look at B. B is a bit special because the wire goes exactly through the middle of the ring, so the area of the ring above the wire is the same as the area of the ring below the wire.
Yes?
yes
So can we use this to help us find the flux through B? What do you think?
yes
09:36
So what is the flux through B?
i think its 0
Yes :-)
The fluxes through the top and bottom halves are equal and opposite so the sum to zero.
So the flux through B is always zero and doesn't change with the current. And therefore there is no induced current in B.
OK, the last bit is to find the directions of the induced currents. Can you see how to do this?
the direction of the current must be such as to oppose the flux variation that generated it
Yes :-)
So what direction is the current in A?
the flux is out of the screen so now it has to be into the screen
so its clockwise ( i dont know if my reasoning is correct )
09:41
Yes :-)
More precisely the increase in the flux through A is out of the screen.
and C is counterclockwise
That is, B is out of the screen and it is getting stronger with time so the change in the external field is towards us.
@mo-_- Yes.
So that's the whole question done!
nice !
It's a good question because it forces you to think about what is happening rather than just reach for an equation. It really tests your understanding.
I have another question of this type (by type I mean that it is not a long exercise, but the topic is different)
09:44
OK, go ahead :-)
@JohnRennie yes, anyway I checked and everything is correct
OK :-)
A circuit consisting of the parallel connection of two resistors (with values $R$ and $2R$ in turn in series with the parallel connection of two capacitors (with values $C$ and $2C$) is driven by a voltage generator. Calculate the time constant of the circuit.
(Data: $R = 50 \, \Omega$; $C = 20 \, \text{nF}$).
what is mean by is driven by a voltage generator
i.e that there is the EMF ?
then I didn't understand how to find time in a circuit
It just means some form of power supply that generates a voltage. The question is being vague because the time constant only depends on the resistors and capacitors so it doesn't matter what voltage is being applied to the circuit.
Are you switching to studying circuits now?
These were questions I had skipped
09:52
The question is quite simple, but if you haven't studied circuits at all you should leave it until you start studying circuits.
Otherwise there will be lots of new concepts we would have to explain.
but I did the part on the circuits, like the current, the resistances, in short these exercises that are done using Kirchoff
but this ask for the time and I didn't know how to do it
Maybe I still have to do it, because I don't remember doing it
Have you studied RC circuits? If not I'll draw a diagram and explain how they work.
no, but I would be interested because there was this question
OK, give me a moment to draw a diagram.
This is a series RC circuit. We call it a series circuit because the resistor and capacitor are in series.
yes
10:00
We start with the switch open so no current is flowing and there is no charge on the capacitor, then at time t = 0 we close the switch and a current starts to flow. But note that this is not a constant current - it changes with time.
Now the voltage across the resistor is just V = IR. Yes?
yes
clear
And to get the voltage across the capacitor we use the capacitor equation Q = CV and rearrange it to get:
V = Q/C
OK so far?
yes
Now if we use Kirchoff we find the voltages across the capacitor and resistor must add up to the battery voltage E, so we get:
IR + Q/C = E
Yes?
yes
10:06
The next step is to note that current is defined as charge per unit time i.e. I = dQ/dt
Then we can substitute for I in the equation and get:
R dQ/dt + Q/C = E
Then we rearrange this to get:
dQ/dt = E/R - Q/RC
Does this all make sense so far?
yes its clear for now
I'll rewrite this slightly as:
dQ/dt = 1/RC (EC - Q)
And now we have a differential equation that we can solve in the usual way.
so i need to put Q on the left
We rearrange it as:
dQ/(EC - Q) = dt/RC
Yes?
@mo-_- Yes :-)
@JohnRennie yes
10:11
So what is the solution?
$\ln|E C - Q| = -\frac{t}{R C} + C$
$|E C - Q| = e^{C} e^{-\frac{t}{R C}}$
$|E C - Q| = K e^{-\frac{t}{R C}}$
Yes :-) and we need to get the constant K from the initial conditions.
but can i remove the | .... | (absolute value)
At the instant the switch is closed t = 0 and since the current only starts flowing at t = 0 the charge Q must be zero as well. Yes?
@mo-_- Yes
$EC - Q = K e^{-\frac{t}{RC}}$
57 secs ago, by John Rennie
At the instant the switch is closed t = 0 and since the current only starts flowing at t = 0 the charge Q must be zero as well. Yes?
K = EC , yes
10:18
Yes, so our final result is:
Q(t) = EC[1 - exp(-t/RC)]
$Q(t) = E C (1 - e^{-\frac{t}{R C}})$
yes :-)
Yes :-)
And we can get I(t) by differentiating if we want to, though we don't need that to discuss the question you originally asked.
Note that RC must have the dimensions of time because t/RC must be dimensionless.
Yes?
yes
We write τ = RC so we can write the equation as:
Q(t) = EC[1 - exp(-t/τ)]
And we call τ the time constant
35 mins ago, by mo-_-
A circuit consisting of the parallel connection of two resistors (with values $R$ and $2R$ in turn in series with the parallel connection of two capacitors (with values $C$ and $2C$) is driven by a voltage generator. Calculate the time constant of the circuit.
(Data: $R = 50 \, \Omega$; $C = 20 \, \text{nF}$).
And your question is asking what the time constant is. Yes?
yes
10:25
Is there a diagram with the question? If not I'll draw one.
no there isnt :(
OK, give me a moment ...
ok :-)
@mo-_- There. I think that is what the question means. Does this look correct to you?
yes i think its correct
ok we can start calulating Req and Ceq
10:33
Yes. Do the calculation and tell me what you get.
1/Req = 1/R + 1/2R = 2/3R
Ceq = C + 2C = 3C
Yes :-)
So we get this. Yes?
so t = Req * Ceq
@JohnRennie yes
Yes. It is as simple as that.
It just needed you to know what the time constant meant for an RC circuit.
16 mins ago, by John Rennie
Note that RC must have the dimensions of time because t/RC must be dimensionless.
Yes?
but is it the same thing in Faraday too?
that is, when we find v(t) = v0 e^(-t/τ)
10:38
Yes. τ must have dimensions of time because the argument to exp() is always dimensionless.
$\tau = \frac{B^2 b^2}{Rm}$
so τ it is in seconds
so it has a dimension
ok :-)
thanks !
10:42
You're welcome :-)
Hi @JohnRennie !
Hi :-)
but in the exercises how do i understand when, for example, i is constant or when it varies over time?
Can you give an example of the sort of question you mean?
Like the moving bar
Which is in a magnetic field
10:48
You just use the basic equations. If we have a bar moving in a field the EMF generated between the ends of the bar is:
E = B𝓁v
Now if the EMF is constant the current will be constant, so we have to ask "does the EMF vary with time?"
Is B the moving bar field?
And E can only vary with time if one or more of B, 𝓁 and v vary with time.
Does this make sense so far?
Yes
So we can always find out whether the current varies with time by just using the equations and looking for any time dependence.
e.g. the question might say the bar is accelerating so v = at. Then we'd get:
E(t) = B 𝓁 a t
And you can immediately see the time dependence.
Mmm
But there shouldn't be t
10:56
I'm not sure what you mean ...
$\epsilon=-\frac{d\phi(B)}{dt}$
If the displacement is linear there should be no dependence on t
It depends on whether the velocity is constant or not.
@Binky Maybe it's better if you send an example exercise
I don't understand how you got E(t)
If the displacement is linear that means the velocity is constant so yes in this case ℰ would be constant
Ah, sorry, I see what you mean. That was a typo :-(
No, wait.
I was correct.
For a bar moving in a magnetic field dΦ/dt is the field swept out per second so it's equal to B𝓁v
So E = B𝓁v
If all of B, 𝓁 and v are constant then E is constant.
But if any of B, 𝓁 or v change with time then E changes with time.
11:02
But B always varies
Because if the bar moves its B changes
Suppose we have an infinite wire with a current I generating a magnetic field around it, and the rod moves parallel to the wire
Then the field at the bar is constant. Yes?
No
Because the bar moves
It induces a magnetic field on the wire
You mean the bar generates a magnetic field so the total field is the (constant) field from the wire plus the changing field from the rod?
11:08
Isn't it the bar that moves the current in the wire?
So does its magnetic field act on it?
Ah, we have been talking about different things.
You're asking about a bar magnet moving past some conductor.
Even open circuits
Though even there when the bar magnet is exactly halfway through the loop there will momentarily be zero EMF generated.
I have an example
OK ... ?
11:18
Suppose i calculate $v_{\infty}$ and the text tells me to calculate the energy dissipated up to that instant, so I write $W=\int_0^{+\infty} \epsilon i$d$t$, so here I have to use i(t) or i constant?
I'd need to see the question as I don't understand what you are asking.
8 mins ago, by Binky
Suppose i calculate $v_{\infty}$ and the text tells me to calculate the energy dissipated up to that instant, so I write $W=\int_0^{+\infty} \epsilon i$d$t$, so here I have to use i(t) or i constant?
Calculate the the energy supplied by the generator
In that case the current varies with time as the speed of the bar changes, so you would use i(t) in your integral.
B is on the screen
Why does it change?
I'm assuming the bar starts at rest at time zero. Since it is at rest the current in the bar is just given by I = Egen/R where R is the resistance of the bar. OK so far?
11:32
but in this case I = (Egen - E) / (R) ?
@Binky
@JohnRennie OK
But as the bar accelerates its velocity 𝑣 increases, and this velocity starts to generate an EMF Ind = B𝓁v that opposes the EMF Egen from the battery. That is the net EMF across the rod is:
Enet = Egen - B𝓁v
And the current is i(t) = Enet(t)/R
So as the velocity increases Enet decreases and the current decreases.
Yes?
Yes
That's why the current decreases with time.
I need to go now. I will be around tomorrow as usual.
Bye :-)
Bye and thanks !
 
7 hours later…
18:49
@JohnRennie Sir, when you are free, could you please also comment on what's wrong with my reasoning with regard to Faraday cages? 🙏

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