Morning @JohnRennie

@sammygerbil Hi. Are you free for problems?

@ayc @blue_eyed_... Questions? If so please post them. I shall be going out soon for about 45 minutes but will look at them when I return.

@sammygerbil did you mean to ping ayc and blueeyed or me?

@Abcd Yes. ayc and blue eyed. They had visited the chatroom.

@sammygerbil i have doubts too ...

@Abcd Well the world is a troubling place, my son ... :)

@sammygerbil what do you mean

@Abcd Just making a joke. What is your question?

@sammygerbil jam (= just a minu6)

Minute*

@sammygerbil what are the current enclosed in a,b and c respecting?

Respectively*

@Abcd Looks like you are learning about Displacement Current. What does the book say about it?

@sammygerbil Book says current is 0 in b and c :'(

But it's i(t) for a

I don't see how

a is like a subset of b and c

If you understand what I mean...

So how can it be 0 for b and c

@Abcd What are the blue regions? Can you post an image of the text?

Sending

@Abcd I think I understand (a) and (b) but not (c).

@sammygerbil ?

@Abcd All 3 diagrams apply Ampere's Circuital Law. Diagrams (a) and (b) are used to find the magnetic field on the rim of the circle of radius r in diagram (a). This circle should really be shaded in blue to show that it is a surface through which current i(t) is penetrating. So the magnetic field on the rim is B where $2\pi rB=\mu_0 i(t)$.

Do you understand so far?

@Abcd In diagram (b) the flat circular surface in (a) has been stretched like a bubble so that it crosses the axis between the plates where the wire no longer penetrates it. So there is no longer any current penetrating this surface S. However, Ampere's Law says the field B on the rim of the circle in (a) is still given by the current which penetrates any surface which has the circle as its rim.

@sammygerbil phh

Ohh*

So the thing us:

Is*

That in a they have considered a full area of the circle as a surface

But in b they have considered the circle as a rim or an opening

There seems to be a contradiction : we have a surface S with no current penetrating it, but we must still have the same field B on the rim of the circle. There must be something missing from Ampere's Law, which says that we should get the same value of B whatever the surface S we choose.

@sammygerbil but still i can see current enclosed in the larger surface too...

See the region between P and capacitor plate

@Abcd That is the point! There appears to be no current penetrating the surface S in diagram (b). So it seems that if we use this surface the field at the rim of the circle should be 0 not B.

@sammygerbil see my new messages

@Abcd The book is saying that there must be an imaginary current flowing between the capacitor plates, called the displacement current.

We know there isn't any flow of charge between the plates, so this displacement current is actually the changing electric field between the plates.

The conclusion is that a changing electric field is equivalent to a steady current. The missing term from Ampere's Law is $dE/dt$. So it should read $$\int_C B.d\ell=\mu_0(I+\frac{dE}{dt})$$

Jam

@Abcd ??

@sammygerbil just a minute

ok

Correction : the equation should read $$\int_C B.d\ell=\mu_0(I+\epsilon_0\frac{d\phi}{dt})$$

@Abcd The current must penetrate the blue surface. In (a) the current in the wire penetrates the blue surface, in (b) the current does not cross the blue surface S.

In (b) the current still crosses the plane area of the circle, but this plane is the blue surface which was used in (a). It is not the blue surface S used in the calculation in (b).

Ampere's Law says the current must penetrate the blue surface, and that this surface can be any shape you like. In (b) we found a surface which has the same rim, but there doesn't seem to be any current penetrating through this surface S!

@sammygerbil ohh

@Abcd I shall go out now. Back in about 45 minutes - I shall ping you. I shall have to go out again this evening, from about 6.45 to 8pm.

Ok

@Abcd Back.

@Abcd You asked about deriving the expression for the intensity of the interference pattern. The amplitudes from the 2 slits are A, 2A and there is a phase difference of $\phi$ between them. Using the Parallelogram Law the squared amplitude of the resultant is $A^2+(2A)^2+2(2A)A\cos\phi=A^2(5+4\cos\phi)$.

The maximum of this expression is $I_0=9A^2$ so $A^2=\frac19 I_0$. Thus we can write the intensity of the interference pattern as $\frac19 I_0(5+4\cos\phi)$. Option 3.

The Parallelogram Law can be obtained from resolving one oscillation of amplitude $E_2$ parallel and perpendicular to the other oscillation of amplitude $E_1$. That is, The resultant x-component is $E_1+E_2\cos\phi)$ and the resultant y component is $E_2\sin\phi$. Then the squared magnitude of the resultant is $$(E_1+E_2\cos\phi)^2+(E_2\sin\phi)^2=E_1^2+E_2^2+2E_1E_2\cos\phi$$ in which I have used $\sin^2x+\cos^2x=1$.

If still not sure draw a vector diagram showing the two oscillations as vectors (phasors) with amplitudes $E_1, E_2$ and angle $\phi$ between them.

@sammygerbil nope, done it in maths. It follows easily from the vector dot product .

$(\vec a + \vec b). (\vec a + \vec b) = |a+b|^2 = |a|^2 + |b|^2 + 2|\vec a||\vec b| \cos\theta$

@sammygerbil one thing, Intensity = $kA^2$ right?

Not directly $A^2$

But you seem to have used that @sammygerbil

i think there'll be k's and they'll cancel ...

thanks @sammygerbil

@Abcd You can insert constant $k$. Then you will get the same answer.

@sammygerbil will you be here at 1:30 am IST ?

That's right $k$ cancels because it is in each term.

@Abcd Not sure. I have been going to bed very late so probably I shall be awake at that time.

@sammygerbil when do you sleep in UK time?

This morning I slept from about 7.30am to 11.30am!

What? 7:30 am is considered morning in India not night

@sammygerbil why do you sleep so late and less ...

@Abcd Some troubling things happening in my life recently which have kept me awake thinking about them.

@sammygerbil ohh

@Blue any nice books on electronic devices, such as transistors and all ?

@EshaManideep Sedra and Smith. Boylestad

Razavi as well.

@sammygerbil are you here

@Abcd yes

jam

@sammygerbil In $\text{Strain} = \dfrac{\Delta L}{L}$

Is L in denominator initial or final length?

@Abcd Initial.

@sammygerbil hmm as I guessed.

Attempt is:

$\dfrac{v_1}{v_2} =\sqrt{\dfrac{T_1}{T_2}}$

then...

:

$Stress = Y \text{Strain}$

$\dfrac{T_1}{T_2} = \dfrac{1}{4}= \dfrac{\text{Initial extension from 2L}}{\text{Final extension from 2L}}$

$v_1/v_2 = 1/2$

$v_2 = 2v_1 = \text{option D}$

But answer given is option C @sammygerbil !!

@Abcd You have assumed that the mass per unit length is not affected by the extension of the elastic string. This assumption works quite well for very stiff strings and ropes for which the extension is a very small fraction of the original length. However, in this case the extensions are large fractions of the original length, so mass per unit length is changed.

@sammygerbil Oh got it :(

@Abcd :)

@sammygerbil please c

@Abcd Any ideas?

@sammygerbil $\ce{n -> p^+ + \beta } + \bar\nu $

@sammygerbil $\ce{p -> n + \beta+ }+ \nu $

@sammygerbil i think it should be B

but thats wrong

@Abcd For what reason?

dont know

well $E = \Delta m c^2$

Thats what option B gives

@sammygerbil i have an exam in the morning(5 hours from now). please make it fast if possible. i have to go to sleep

@Abcd Think about what charges remain on the nucleus after the decay and how many orbital electrons there will be in the atom when it becomes neutral again.

@sammygerbil aren't my decay equations correct?

This probably takes some careful thinking so perhaps best to tackle it in the morning after you have slept.

oh I see. I will do it after returning from the exam.

@sammygerbil is it a tricky question?

K electron capture?

@Abcd Yes your equations are correct but there is more going on than just these equations. The atom has to be neutral again after the decay.

No not K capture.

That is a separate reaction.

there is excess negative in Q2 i see

so i think

one electron should be lost?

@sammygerbil ^

For beta plus decay the nuclear charge decreases so one electron must also be lost from the orbitals.

thats what I said above

Altogether 2 electrons are lost for beta plus decay.

@sammygerbil i still dont understand why there should be -2m_e

I think there should be -1 m_e

Whereas for beta minus decay nuclear charge increases so one electron is lost from nucleus and one is gained in the orbitals. Therefore zero electrons lost altogether.

hmm

Correct answer is A?

hmm

this one also quickly

Atttempt is:

$\dfrac{1}{2}k\dfrac{l^2}4 + \dfrac 12 4k \dfrac{l^2}4 = \dfrac 12 mw^2 A^2$

$A = l/2$

So $\omega^2 = 5k/m$

So I got option D as the answer

but thats wrong

My next question is

Period is $2\pi\sqrt{m/k_{eff}}$ where $k_{eff}$ is effective spring constant.

@sammygerbil please tell the mistake in my method and the way to find the effective spring constant here. Is it series circuit?

Attempt for (18) is:

The springs are in parallel so $k_{eff}=k+4K$.

@sammygerbil you get the same answer as me then lol

Work in air frame.

General formula is:

Hmm. I don't see how we eliminate $k$.

$f_{app} = \dfrac{v_s + v_o }{v_s - v_{soruce}}f$

@sammygerbil i think thats just a typo. Both will be capital.

Air frame:

$v_a = 20$

$v_b = 30$

so $v_{source} = 30$

$v_o = 20$

$f_{apparent} = \dfrac{35}{30}f_{actual}$

But that wont take me to right answer.

Please tell me my mistake.

@Abcd Sorry I shall have to think about this one also. Best for you to get some sleep for your exam. Good luck.

ok good night

@sammygerbil please send the equivalent of voltage in spring circuits

@Abcd ??

@sammygerbil night, do you have some free time?

@JD_PM yes

It is about the rotating disk's problem I told you a couple of days ago: a rotating disk with a point particle in its centre has an initial angular velocity and it moves radially to R.

Well, I got the work done by the centripetal force and also an expression that proves that the rotational kinetic energy when the point particle is at radius R is less than the rotational kinetic energy when the point particle is in the centre. Now I am asked for combining both results to get an expression like this: imgur.com/a/f0PbXqo

I guess I have to modify my integral for the work done by the centripetal force but I am not getting this expression, any idea? :)

@sammygerbil ^

@JD_PM Sorry I don't understand what you are trying to do.

@sammygerbil this is the entire question: imgur.com/a/eqScXzx Note there is a mistake on c); it should be $I$ instead of $I^2$

I am trying to solve d)

@JD_PM Sorry I don't understand this either. Why do you need results of (b) and (c) to prove (d)? It is a straightforward integration problem isn't it? It is completely independent of the rest of the question as far as I can see.

@sammygerbil yes I agree. For me it seems an independent calculation as well. I was trying to think how could I relate both results to get d) but did not see the way to do so.

I will ask my teacher next week and see what I am told. If there is something interesting to get out of it I will tell you.

As always thank you for your help :)

Have a good night

@JD_PM Good night.