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07:18
@mo-_- sry i wasn't online here for few days.... yes i did magnetic field
@JohnRennie Hi
@NOTEBook Hi :-)
I was curious that is burning of coke in presence of oxygen a combination reaction. C forms CO and CO2... so it means combination reactions can form multiple products?
Yes, and combustion of carbon is a good example.
Combustion is usually a complex multistep process.
Coke does not burn directly to CO₂. It burns to CO then the CO burns to CO₂.
You get CO in the reaction products when there is not enough oxygen to burn all the CO to CO₂.
07:25
yes
Thank you
You're welcome :-)
07:58
Hi @JohnRennie :-)
Hi :-)
I did the exercises again and they all worked, only here I dont know how to continue with another method
im gonna ping the exercise again
OK ...
2 days ago, by John Rennie
user image
here i tried to calculate $\Phi$
so i did
$d\Phi = B(x) dA = \frac{\mu_0 i}{2\pi(a+x)} dx$
so $\Phi = \int^L_0 \frac{\mu_0 i}{2\pi(a+x)} dx$
Where dx is a small displacement in the direction of the velocity
@mo-_- OK
08:04
$\Phi = \frac{\mu_0 i}{2\pi} \int^L_0 \frac{1}{(a+x)} dx$
No, wait
No, that's wrong
so $\Phi = \frac{\mu_0 i}{2\pi} \ln(\frac{L+a}{a})$
so here i cant do a = a0 + vt
to do $d\Phi/dt$
because a its not increasing ?
I'll draw a diagram ...
@JohnRennie ah
@JohnRennie ok :-)
wait a is increasing actually maybe i can write a = a0 + vt
The wire is moving upwards i.e. parallel to the wire. Yes?
08:13
yes
So in a short time dt it moves a distance vertically dy = v dt
And it sweeps out an area dA = 𝓁 dy = 𝓁 v dt
Yes?
I don't understand why sometimes we don't do this ... i.e
21 hours ago, by John Rennie
user image
I didn't write dx = v dt here yesterday
but i writed b = b0 + vt
Could you explain to me when the step you took now should be done?
Do you want me to back through this problem using a dx approach? I can do ...
BTW I need to go out for a couple of hours in about 15 mins.
no, I would like to understand if in this exercise that I have written I can now write a = a0 + vt, it is not clear to me
@JohnRennie ok:-)
We are trying to find dΦ/dt because that gives us the EMF. Yes?
08:18
yes
Now there are basically two ways to do this:
1. find Φ then differentiate it to get dΦ/dt
2. consider a short time dt and work out how much Φ changes i.e. the dΦ in the time dt
Then in the second case we just divide dΦ by dt to get dΦ/dt
ok i was doing 1.
Method 2 is almost always faster
I can show you how to use method 2 for that square loop question, but that will have to be after I get back.
ok , but point 1 , how can i continue ?
I'll proceed myself
$\Phi = \frac{\mu_0 i}{2\pi} \ln(\frac{L+a}{a})$
For the rod you cannot use that method because the flux inside the rod is zero.
08:23
ah so for rods I always use method 2
The key difference between a loop and a rod is that the loop encloses some area inside the loop and you can find the flux through that area.
But a rod does not enclose any area so there is no flux through it.
ah ok !
instead for the loop do you recommend method 1 or always method 2?
I recommend using method 2 all the time
It is almost always faster and easier.
so yesterday we used method 1 because that's how I started
Yes. Really I should have immediately told you to stop and use method 2.
08:26
ok :-)
so for the rod and loop method 2 always work ?
Yes.
I'm going now. Shall I ping you when I'm back?
ok :-)
Bye :-)
Will do. Bye :-)
08:44
$d\Phi = \frac{\mu_0 i}{2\pi(a+x)} l \ v \ dt$
$\frac{d\Phi}{dt} = \frac{\mu_0 i}{2\pi(a+x)} l \ v$
up until now it was ok if B was constant, now I don't know what to do because B varies
 
3 hours later…
11:23
@mo-_- I'm back
@JohnRennie Hi :-)
hi :-)
i reached the point above
3 hours ago, by John Rennie
user image
2. consider a short time dt and work out how much Φ changes i.e. the dΦ in the time dt

Then in the second case we just divide dΦ by dt to get dΦ/dt
OK, in a time dt the rod moves a distance vertically dy = v dt
So it sweeps out the area shaded in grey dA = 𝓁 v dt
Yes?
dΦ = B dA = B v dt
@JohnRennie yes
11:26
Except that as you say B is a function of x because it varies with distance away from the wire. B doesn't depend on y, only on x.
yes
dΦ/dt = B(x) 𝓁 v
So what we have to do is divide the grey area into small strips where B is constant, then find the dΦ for each strip, and then integrate to add them up.
Does this make sense so far?
what is the grey area ?
In my diagram I shaded the area that the rod sweeps out in grey.
This grey area. Yes?
yes
11:31
Now consider this strip of the grey area.
ok
The width of this strip is dx and the length of the strip is dy = v dt
yes
So the area of the strip is dA = v dt dx
And the field everywhere inside this strip has the same value B =μ₀i/2𝜋(a+x)
Yes?
wait
we have 2 dA ? dA = 𝓁 v dt and dA = v dt dx
11:35
𝓁 v dt is the area of the whole region shaded in grey
v dt dx is the area of the small strip of the grey area outlined with dashed lines
Yes?
yes
The field varies across the whole grey area, but in the small strip it is constant because everywhere in that strip 𝑥 has the same value. Yes?
i.e. x in that little square?
This region
yes
right
11:41
So the flux through that region is dΦ = B(x) dA = B(x) v dt dx
Yes?
but did you mean that the field varies horizontally from 0 to L?
and x only varies horizontally
The field only varies with 𝑥 i.e. horizontally. It does not vary with 𝑦.
So if we make the red strip very thin, i.e. dx is very small, the field is the same everywhere inside the red strip.
And that field has the value B(x) where 𝑥 is the distance from the end of the rod.
OK so far?
yes
B(x) = μ₀i/2𝜋(a+x)
Yes?
yes
11:46
So we get:
dΦ = μ₀ i v dt/2𝜋 × 1/(a+x)
but so dA =v 𝓁 dt , becomes dA = v dx dt ?
@JohnRennie but isnt a dx missing ?
This time we are choosing our dA to be a small part of the grey area, then we will integrate to find the flux through the whole grey area.
Oops, yes:
dΦ = μ₀ i v dt/2𝜋 × dx/(a+x)
So when we integrate we get:
Φ = μ₀ i v dt/2𝜋 × ∫dx/(a+x)
where we are integrating from x = 0 to x = 𝓁
Yes?
yes
Φ = μ₀ i v dt/2𝜋 ln[(a+𝓁)/𝓁]
Yes?
yes
11:52
And this Φ that we have calculated is the flux that the rod sweeps through in the small time dt
Yes?
yes
So really, although we've called it Φ not dΦ it is a dΦ because it's the change in Φ in a time dt.
so really we have:
dΦ = μ₀ i v dt/2𝜋 ln[(a+𝓁)/𝓁]
But when we wrote dΦ earlier it meant "the change in Φ in a distance dx"
And now it means "the change in dΦ in a time dt"
ah ok
Then divide by dt like we always do:
dΦ/dt = μ₀ i v/2𝜋 ln[(a+𝓁)/𝓁]
yes
11:58
hi
@JohnRennie but what should I do if the bar was still and not moving?
the Area is 0 ?
Φ = 0 so E = 0 ?
Yes. So in that case dΦ/dt = 0
And there is no EMF
There is only an EMF when the rod is moving
i have other 2 questions quick: 1) why when the velocity is constant we can use F = ma ?
maybe when the rod enters in B there is an acceleration?
I'm not sure what you are asking. As I remember the question the rod was being moved with constant velocity, presumably by some external device like a motor.
Yes?
2 days ago, by mo-_-
user image
like here they told me that v is constant but we used F = ma to find the velocity v(t)
ma if v is costant isnt a = 0 ?
12:10
I don't remember that. I think the only time we used F = ma was for that question where the rod was falling downwards due to gravity.
2 days ago, by John Rennie
And yes it is correct to say F = ma
here
we talked about it
Ah, sorry, yes, this was a loop sliding freely not being moved at a constant speed by some external device. Yes?
but the text says: moves without friction on the x-y plane with a constant velocity of 4 × 10⁻² m/s.
I think the idea is that it was moving at constant velocity before it entered the magnetic field.
Then when it entered the magnetic field it started to slow down.
ah ok :-)
thanks, im happy with this
12:16
OK :-)
@JohnRennie Is it too late for a question now?
You can post the question, but I'm busy answering a question in another room right now.
okay :-)
The figure shows a device called a Helmholtz coil. It consists of two coils, each with $N$ turns and radius $R$, separated by a distance equal to $R$. The same current $i$ flows through both coils in the same direction.
Calculate the magnetic field $B$ at the midpoint between the two coils and indicate its direction.
Also, calculate the magnetic field $B$ at the center of each coil and indicate its direction.
$\text{Data}$: $N = 200$ turns; $R = 25 \, \mathrm{cm}$; $i = 12.2 \, \mathrm{mA}$; $\mu_0 = 4 \pi \cdot 10^{-7} \, \mathrm{H/m}$.
i dont know why they used R to indicate the radius and also the distance
@Pizza That's just the way they have set up the question. I would guess the answer comes out simpler if the coil spacing is equal to the coil radius.
12:31
@JohnRennie so I have to use the magnetic field of a circular coil?
Yes. I forget the equation for this but you can quickly Google it.
At the centre of the loop it's B = μ₀i/2R but as we move away along the axis of the loop the equation becomes more complicated.
@JohnRennie so i need to find r at the midpoint
That's easy isn't it? You have a right angled triangle with sides R and ¹⁄₂R so use Pythagoras to find the hypotenuse.
yes
12:39
How do they find where B is pointing?
It's not a terribly exciting question really ...
From the symmetry B has to point along the x axis, so the only question is whether it points left or right. Yes?
yes
And you can use the right hand curl on both coils to find the direction of B
Then sum the two Bs to get the total.
ok thanks :-)
in the end I just have to know how to find B of the circular loop if I can't use the immediate formula
Hi
12:58
Hi :-)
Are you free ?
Yes :-)
What is the value of the flux of the magnetic induction field generated by a coil made up of 50 square turns of side l=3cm struck by the current i=3A through a sphere of radius R which intersects the coil.
But here I have to assume that the sphere all intersects the coil, right?
Is there a diagram?
No
13:06
The magnetic flux through any closed surface, including a sphere, is always zero.
That's because magnetic field lines are always loops so any field line that exits the sphere must re-enter it.
But
Yes ... ?
Does this intersect it all?
I mean the reason
It doesn't matter whether the sphere intersects the coil or not. The magnetic flux though any closed surface is always zero.
Oh okay
13:12
It seems like a trick question ...
But it doesn't tell me that the lines are continuous
With constant verse
Should this be assumed?
I'm not sure what that means ...
Magnetic field lines are always loops. They cannot not be loops.
Oh well, the right answer will be 0 anyway
Thanks very much
Yes
You're welcome :-)

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