Problem Solving Strategies

RishiNandha Vanchi

2:34 AM

I have a weird doubt

Law of em induction

When we talk only about emf we have dphi/dt = -e

ie, the derivation is infact for integral E dl and not exactly for a circuit

so my doubt is: when a circuit is taken, a dI/dt term will appear?

I being current in circuit

RishiNandha Vanchi

2:48 AM

I ran into this when I was trying to go backwards from Biot Savart to Lenz (like how we did historically before maxwell’s eqns became the fundamental description)

Ashish Ahuja

4:31 AM

@JohnRennie hi

I have a couple of questions to ask, are you free?

John Rennie

@AshishAhuja hi :-)

@AshishAhuja yes I'm free.

Ashish Ahuja

Q: Given an equilateral triangle made up of three rods each of length $l$, find the electric field strength at the centroid of the triangle. The linear charge density on two of the rods is $+\lambda$, and the third is $-\lambda$.

Here's what I thought

By the superposition principle the magnitude of the electric field will be $2E$, where $E$ is the electric field exerted by any one of the rods at the centroid.

Now we simply need to integrate for one of the rods and find $E$

John Rennie

Yes

Ashish Ahuja

However, I ended up struggling with the integration since it got rather complicated and I got the wrong answer. I tried both integrating with the angle and the distance from the midpoint of the rod. Could you write down the integral and then I'll try solving it?

The results I found online seemed even more complicated than what I had gotten...

John Rennie

It's a standard result i.e. the field due to a finite rod. In this case we are halfway along the rod so the component parallel to the rod vanishes and we need to consider only the perpendicular component. Yes?

Ashish Ahuja

4:39 AM

yes

John Rennie

I think you can do it either using the angle or just using distance along the rod. I'll draw a diagram:

Ashish Ahuja

Okay.

I know the basic principle, if we do it based on angle we would take a small charge dQ at some distance which we can relate to the angle and length, find the perpendicular component of this dE due to dQ and integrate.

So if you'd like you can skip straight to the integral but anything is fine.

yes

Can I use MathJax?

4:44 AM

yes

John Rennie

The horizontal component is: $$ dE = \frac{k dq}{r^2}\cos\theta $$

Ashish Ahuja

yes

John Rennie

And $\cos\theta = y/r = y/\sqrt{x^2 + y^2}$

Ashish Ahuja

yes

John Rennie

And $dq = \rho dx$

So we're going to end up with a somewhat messy equation:

$$ dE = k\rho y \frac{dx}{(x^2 + y^2)^{3/2}} $$

Ashish Ahuja

4:48 AM

yeah this is exactly what I got too.

Then the indefinite integral becomes complicated and we get a large term with natural log in it

John Rennie

I think that's a standard integral. I'd have to look it up as I'm out of practice doing integrals.

Ashish Ahuja

yeah even I don't know how to do such integrals, I looked it up. It was rather complicated. But the answer is

$$\frac{\lambda}{2 \pi \epsilon l}$$

John Rennie

That seems a surprisingly simple result for a horrible integral :-)

Ashish Ahuja

yeah lol

John Rennie

Where $\ell$ is the distance from the rod?

Ashish Ahuja

4:53 AM

no l is the length of the rod

John Rennie

But unless the rod in infinite the result must depend on the distance from the rod.

Ashish Ahuja

yes but the distance from the rod can be related to $l$. it is an equilateral triangle

John Rennie

Ah, sorry, you meant the result in this case, not the general result for the field of a finite rod.

Ashish Ahuja

ah yes

John Rennie

OK. You seem to have this under control, except for the integral which no-one normal can do anyway.

Ashish Ahuja

4:56 AM

well I haven't studied integration formally in maths yet, who knows jee might have such weird integrals as well :O

John Rennie

No, you wouldn't be expected to integrate that in the JEE. Not even advanced.

Ashish Ahuja

oh ok

I have a couple more questions, I'll go ahead?

John Rennie

Yes

Ashish Ahuja

I was able to solve it about the center of mass.

But my first approach was to try doing it about the instantaneous axis of rotation, since that generally makes stuff much easier. How would we go about doing that?

John Rennie

I don't think using the IAOR is going to be easy.

Ashish Ahuja

5:03 AM

because calculating torque about IAOR is messy?

John Rennie

When you use the centre of the ring you have some symmetry to help you but that isn't present if you find the torque about the IAOR.

Ashish Ahuja

ah ok, that's what I thought too but was wondering if there was a way.

Last question:

The question is rather easy to solve once you take the surface tension forces as $\sigma a \sqrt2$

because two forces of $\sigma a$ in perpendicular directions

John Rennie

You mean taking the component of the force due to the surface tension along the diagonals.

Ashish Ahuja

yes, which would also be the net force due to surface tension. It gave me the correct answer, but I'm not sure why the surface tension force would be so

John Rennie

Yes, I agree.

The surface tension is just a force normal to the edges of the square, and it's perfectly reasonable to divide that into two components at 45° to the rods.

Shall I draw a diagram?

Ashish Ahuja

5:10 AM

to my understanding the net surface tension force over the whole edge should be $\sigma a$, not necessarily over the point charge as well?

@JohnRennie ok

John Rennie

Actually virtual work might be the easiest way to do this ...

Ashish Ahuja

I've never actually used virtual work before

John Rennie

The idea is that if the system is at equilibrium then its potential energy must be at a minimum or a maximum (minimum for stable equilibrium and maximum for unstable equilibrium).

Ashish Ahuja

this is unstable right?

John Rennie

So for some parameter x that describes the system dU/dx = 0.

@AshishAhuja yes, this will be unstable, though in fact we don't care whether it's stable or unstable. In both cases dU/dx = 0, and that means if we move the system an infinitesimal amount the PE shouldn't change.

Ashish Ahuja

5:16 AM

oh right

John Rennie

Suppose you imagine displacing the sides outwards a distance dx, then the work done against the surface tension on each side is just $a\sigma dx$.

Ashish Ahuja

yes

John Rennie

And the work done by the electrostatic repulsion is $Eq\sqrt{2}dx$

And for the total PE not to change we require that the two works cancel out i.e. $a\sigma dx = Eq\sqrt{2}dx$

Ashish Ahuja

@JohnRennie on a single point charge?

John Rennie

@AshishAhuja I've considered a single edge and a single charge. Multiply both sides by 4 to get the total work, but the factor of 4 cancels anyway.

Ashish Ahuja

5:21 AM

wait what have you defined E as? The net electric field on a point charge due to the other three?

John Rennie

Yes

It out to give the same result, because it is effectively just balancing the forces at the charges just like the first calculation.

Ashish Ahuja

yup got it. I'll try it out now...

John Rennie

Actually it's no faster is it?

In both cases you're basically just calculating the same forces.

Ashish Ahuja

yes it looks like. For me forces would probably be faster since this is something I need to stop and think about.

John Rennie

Yes. Virtual work can be easier in some cases, but here your original approach is just as fast and probably easier to understand.

Ashish Ahuja

5:25 AM

I'm getting a different answer from this method?

I'll type it out, maybe I'll spot the mistake.

$$E = \frac{kq}{a^2} \times (\frac{1}{2} + \sqrt2)$$

7 mins ago, by John Rennie

And for the total PE not to change we require that the two works cancel out i.e. $a\sigma dx = Eq\sqrt{2}dx$

$$a = \frac{Eq \sqrt2}{\sigma}$$

$$a = \frac{kq}{a^2} \times (\frac{1}{2} + \sqrt2) \times \frac{q \sqrt2}{\sigma}$$

$$a^3 = \frac{kq^2}{\sigma} \times (\frac{1}{ \sqrt2 } + 2)$$

given answer is at the bottom

hopefully I didn't mess up the algebra..?

John Rennie

Hmm, I got $$ a^3 = \frac{kq^2}{\sigma} \times (\frac{1}{ 2 } + \sqrt{2}) $$ but i did it in a hurry and could easily have missed a factor of $\sqrt{2}$ somewhere.

Ashish Ahuja

@JohnRennie looks like you missed multiplying the $\sqrt2$ here

John Rennie

Oops, yes :-)

Then I get the correct answer

Didiving by that extra $\sqrt{2}$ gives the answer in your book.

Ashish Ahuja

don't you need to multiply by the $\sqrt2$?

John Rennie

No, wait, I need to multiply by $\sqrt{2}$ ...

Hmm, so we're a factor of 2 away from the correct result.

Ashish Ahuja

5:35 AM

yes

John Rennie

I'll have to think about this ...

It must work because it's balancing the same forces in both cases.

Ashish Ahuja

Ok, np. Would it be possible for you to explain why the force on the point charge will be $\sigma a \sqrt2$? That would answer my original query completely, since that is what I am not convinced of.

(force due to surface tension)

John Rennie

I must admit it seems obvious to me. I'm not sure how to approach this ...

Ashish Ahuja

Okay, I'll explain exactly what's bugging me.

The force on each edge is $\sigma a$.

if we had a rod say on this edge

the force on each small section of this rod would not be $\sigma a$, since all the small sections must have the same acceleration as the whole rod but have a smaller mass

(assuming small acceleration for time dt)

I just don't get why the net force on the whole edge must be the same as the net force on the point charge.

John Rennie

For each element of the rod take the force $\sigma dx$ and write it as the sum of two components at 45° to the rod.

Actually let me take a step back.

Ashish Ahuja

5:42 AM

ok. To make it simpler you can just take one edge with one rod if you'd like, we can ignore the other surface.

John Rennie

There are two forces acting on the rod. There is the force in towards the centre and there is a compressive force acting along the length of the rod. yes?

Ashish Ahuja

compressive force? due to what?

John Rennie

Ignore the charges for now and just consider a frame made from four rigid rods with the soap film in the centre.

Ashish Ahuja

ok

John Rennie

Now imagine making the rods soft, then the surface tension is going to compress them and shrink the size of the square. Yes?

Ashish Ahuja

5:45 AM

yeah I got the reason of the compression force, I was originally thinking of another situation.

John Rennie

OK. So the point is that at each point we have the two forces at right angles.

Ashish Ahuja

yes

John Rennie

Let me draw a diagram:

If you consider a small element of the rod you have these forces acing on it. The red arrows are the force due to the compression and the blue force is the surface tension force.

Ashish Ahuja

yes

John Rennie

And you can add pairs of forces to rewrite the total force as two components at 45°:

Ashish Ahuja

5:52 AM

ok

John Rennie

So add up all the elements and you'll get the total force on the edge as a pair of components at 45° i.e. in the direction of the purple arrows.

Ashish Ahuja

yes

John Rennie

And since the rods are rigid we can make these forces act at any point on the rod. So we make them act at the ends of the rod.

That gives the diagonal forces at the corners that balance out the diagonal forces due to the charges.

Ashish Ahuja

hmm but at the corner (say top left), there will be no compression force from the top right?

John Rennie

Remember that four rods exert forces on each other.

So at the corners we have the forces from the other two rods that this rod is connected to.

Ashish Ahuja

5:57 AM

hmm ok

By doing this haven't you just proven that the force on the corners will be two components each of magnitude $$\frac{\sigma dx}{\sqrt2}$$

John Rennie

We integrated didn't we, so $dx$ integrates to $a$

so you get $2 \times \sigma a/\sqrt{2}$

Ashish Ahuja

There is something I'm missing in my thought process, I guess I'll come to realize it sooner or later.

Thank you

RishiNandha Vanchi

6:14 AM

> https://i.stack.imgur.com/5VlmO.jpg
@JohnRennie if you are free sir?

Ill send the msg again im not able to figure out how to link it

I have a weird doubt
Law of em induction
When we talk only about emf we have dphi/dt = -e
ie, the derivation is infact for integral E dl and not exactly for a circuit
so my doubt is: when a circuit is taken, a dI/dt term will appear?
I being current in circuit

I ran into this when I was trying to go backwards from Biot Savart to Lenz (like how we did historically before maxwell’s eqns became the fundamental description)

John Rennie

I'm busy for a while. I'll ping you when I'm free.

RishiNandha Vanchi

Wokay :D

John Rennie

6:45 AM

@RishiNandhaVanchi In a circuit you typically have a resistance. If you consider a loop and at time zero introduce a changing magnetic field then this will induce an EMF and dI/dt will transiently be non-zero. But as the resistance increases the voltage drop due to the resistance increases until it is equal to the induced EMF and that point dI/dt becomes zero.

Or you could look at this another way. If you have a perfect inductor and apply an EMF then both the current and field increase linearly with time i.e. dI/dt and dB/dt are constant. In a real inductor we have resistance and the current becomes constant once the voltage drop due to resistance balances the applied EMF.

RishiNandha Vanchi

@JohnRennie so the expression i ended up is right and we normally neglect dI/dt / assume steady state in our analysis at highschool level?

In the picture i mean

John Rennie

Well we always get a steady state. The hypothetical zero resistance circuit causes various problems when we try to do calculations with it.

RishiNandha Vanchi

No im taking a loop with resistance

John Rennie

In a loop with resistance we get a steady state.

RishiNandha Vanchi

And conserving energy, I end up with lenz law only if dI/dt = 0, doubt is if I went wrong with the energy conservation or if we normally assume dI/dt = 0

At highschool level

@JohnRennie ohhhh

So in reality there is dI/dt transient even if we assume self inductance = 0?

like i got in the expression

John Rennie

6:58 AM

Well there is no current before you start changing the field and a constant current once the voltage drop across the resistance balances out dΦ/dt so there must be a transient current increase in between.

Srijan M.T

7:12 AM

@JohnRennie Sir , can I say data is a raw fact in computer science subject,

John Rennie

I'm not sure what a raw fact means. It is not a term I have seen before.

Srijan M.T

Ok. Np

RishiNandha Vanchi

8:33 AM

@JohnRennie So my steps were right?

Srijan M.T

@JohnRennie V quick. I got 2s component of 10 = 10. Is it right ?

John Rennie

Do you mean two's complement?

Two's complement

Two's complement is a mathematical operation on binary numbers, and is an example of a radix complement. It is used in computing as a method of signed number representation. The two's complement of an N-bit number is defined as its complement with respect to 2N; the sum of a number and its two's complement is 2N. For instance, for the three-bit number 010, the two's complement is 110, because 010 + 110 = 8 which is equal to 23. The two's complement is calculated by inverting the bits and adding one. Two's complement is the most common method of representing signed integers on computers, and more...

Srijan M.T

Yes sir

@JohnRennie

I am getting wrong for 10. Don’t know why