Problem Solving Strategies

NOTE Book

05:39

@JohnRennie Hi... just a doubt regarding language... like sphere with common centre are called concentric sphere, cylinders on top of each is called concentric or coaxial cylinders. It shows same image online when i search for both

John Rennie

Hi :-)

I would probably say coaxial but I think concentric is probably also OK.

With cylinders you usually talk about their axis rather than their centre.

NOTE Book

yes

ok

Thank you

NOTE Book

07:44

@JohnRennie Hi

John Rennie

Hi :-)

NOTE Book

i am getting an additional negative sign when i am trying to find potential energy due to point charge

John Rennie

Can you give some details?

NOTE Book

PE=-W_CONSERVATIVE= ∫F.ds from infinite to a point right?

John Rennie

Yes

But there is a common confusion with the sign of work because you can have work done by a system and work done on the system.

It can be surprising hard to figure out which is correct.

NOTE Book

07:48

F= kqQ/r^2 at final point away from center of charge and ds vector should be towards charge

if i am considering work done by electrostatic force i.e. work by charge

than angle between F and ds is 180 right?

John Rennie

Are you starting at infinity then pushing the charge q inwards towards the fixed charge Q?

NOTE Book

yes

then PE=-∫fdscos180= ∫FdS

since F= kqQ/r^2 integrating it gives -kqQ/r after putting limits

but potential at that point should be kqQ/r

wait

i am getting an extra negative sign but can't figure out how

John Rennie

I get the same problem when I do these problems. I usually just figure out the sign using common sense.

NOTE Book

like it is obvious that PE should increase and be more than at infinite?

John Rennie

08:04

Well suppose you are pushing the charge q inwards from infinity towards the charge Q. Then obviously you are doing work. Yes?

NOTE Book

yes

John Rennie

When you are doing work energy is transferred from you to the system you are doing work on i.e. your energy is decreasing and the system's energy is increasing.

NOTE Book

yes

John Rennie

That must mean that the PE of the system made up from the two charges Q and q is increasing i.e. if the PE starts at zero at infinity then the PE at distance r < ∞ must be positive.

NOTE Book

yes

John Rennie

08:07

That's the sort of reasoning I do to get the sign.

NOTE Book

yes i understand it conceptually but i always get wrong sign for point charge.

someone asked this same question on the site and i saw it before this chapter but i can't find it now :(

John Rennie

I'd have to go away and think about it to figure out where the unwanted minus sign is coming from, but to be honest I don't care because I always get the sign from common sense anyway.

NOTE Book

wait just 1 min

Point charge potential (sign problem) this name problem i found it

this is the post

4

Q: Point charge potential (sign problem)

I'm a bit embarrassed, but I'm not able to compute the electric potential at point $P$ (at a distance $R$ from the origin) generated by a positive unitary point charge in the origin with the right sign. Simply use the definition $$V(P) = -\int_\infty^P \vec E\cdot d\vec l,$$ forgetting the consta...

John Rennie

OK

NOTE Book

according to accepted answer my integration is wrong

thank you very much

John Rennie

08:14

👍

Pizza

09:10

Hi @JohnRennie :)

John Rennie

09:26

@Pizza Hi :-)

Pizza

@JohnRennie are you free?

John Rennie

Yes

Pizza

Ah ok, I had done an exercise but I don't know if I did it really well, now I'll send the text

A square loop of side $l$, with resistance $R$, moves at a uniform velocity $v$ along the $x$-axis, in a magnetic field $B$ directed along the $z$-axis, pointing into the page. The loop lies in the $xy$-plane. At a certain moment, the loop exits the region where the magnetic field is present. Calculate the induced e.m.f., the induced current, and determine the direction of the current. Also, calculate the power dissipated due to the magnetic force exerted on the circuit.

uscente is outgoing in english i think

Shall I try to say what I did?

John Rennie

09:42

OK, go ahead :-)

Pizza

so initially I said, I have to write $\Phi = B \cdot A = A \cdot l^2$

John Rennie

OK ...

(B . 𝓁²)

Pizza

but I noticed that there is a part where the loop has not entered, the field, is it a mistake to write like this?

$\epsilon = -\frac{\Delta \Phi}{\Delta t}$

John Rennie

The reason there is an EMF induced in the square loop is because the flux through the loop is changing with time. Yes?

Pizza

yes

so I wanted to find the time the loop took to exit the field

John Rennie

09:47

The flux through the loop is the field B times the area A = 𝓁a

where 𝑎 is the length of the loop in the field.

OK so far?

Pizza

ah ok, I thought about it, but I wasn't sure, anyway

@JohnRennie yes

John Rennie

Now suppose the front edge of the loop leaves the field at time t = 0, then the length 𝑎 is given by:
a = 𝓁 - vt

Yes?

Pizza

Yes

John Rennie

So the flux as a function of time is given by:
φ(t) = B𝓁(𝓁 - vt)

Pizza

so we have to find time for the loop to exit the field?

or not?

John Rennie

09:51

No, you do not need to find the time for the loop to exit the field.

Pizza

ah ok, then I would have done it like this now :

John Rennie

We now have an expression for φ(t), so we can simply use:
E = -dφ/dt

Yes?

Pizza

$\Delta \Phi = \Phi_f - \Phi_i = 0 - B𝓁(𝓁 - vt)$

that is, I assumed that eventually the loop would exit the field completely

as the loop moves into the field

John Rennie

@Pizza Your approach will work because you are assuming dφ/dt = Δφ/Δt

Yes?

Pizza

yes

John Rennie

09:54

and in fact this is true because the velocity is constant so yes you can do it that way.

Pizza

I don't know how to find $\Delta t$, I would have written t = l/v, but that's not possible here

John Rennie

But a better way is to find the expression for φ(t) because then you can differentiate it to get E, and that will work even if v is not constant.

That's the method I started explaining above

Pizza

ah ok

John Rennie

If the front edge of the loop leaves the field at time t = 0 then the distance it has travelled after time t is just s = vt.

Yes?

Pizza

yes

t = s/v ?

John Rennie

09:58

So the length of the loop left inside the field is a = 𝓁 - vt
Yes?

Pizza

@JohnRennie yes

John Rennie

And the area of the loop left inside the field is A = 𝓁a = 𝓁(𝓁 - vt)

Yes?

Pizza

ah okok yes

John Rennie

Do you see where this is going now?

Pizza

φ(t) = B𝓁(𝓁 - vt)

John Rennie

10:01

Yes :-)

So what is the EMF?

Pizza

do we need to derive this for t ?

E = -dφ/dt

John Rennie

Yes, E = -B𝓁v

Does this all make sense so far?

Pizza

yes

John Rennie

So can you complete the problem using this approach?

Pizza

let me see, then I can calculate the induced current using Ohm's law

$V = IR$

V = induced EMF

I = -B𝓁v/R

yes ?

John Rennie

10:06

Yes :-)

Pizza

To determine the direction I have to use Lenz's law, right?

John Rennie

You could do, but I would do this a slightly different way.

Pizza

how?

John Rennie

Pizza

but if the magnetic field enters the sheet then the current circulates clockwise, right?

John Rennie

10:09

The EMF is being generated by the left edge of the loop i.e. the edge I have coloured blue. The equation you got E = B𝓁v is just the equation for a conductor of length 𝓁 moving at speed v in a field B.

Pizza

yes

John Rennie

Now suppose you have a charge q (positive charge because conventional current is the flow of positive charge) somewhere on that blue edge.

Pizza

ok

John Rennie

Then the force on that charge will be given by the Lorentz force:

F = q v × B

Yes?

Pizza

yes

can I use the right hand rule?

to find v × B

John Rennie

10:12

@Pizza Yes you can!

So what direction do you get for the force?

Pizza

so index finger points in the direction of v

B in the sheet

I find that it must form a 90 degree angle towards the top

John Rennie

Pizza

F = qvB

John Rennie

Yes, it's like the picture above.

Pizza

yes like this

John Rennie

10:15

So the force is upwards. Yes?

Pizza

yes

John Rennie

So the (positive) charges in the blue edge will move upwards, and that means the current is flowing clockwise round the loop.

Does this all make sense?

Pizza

yes

John Rennie

OK :-)

Now can you do the last part?

Pizza

@JohnRennie i have to use this right?

John Rennie

10:18

For which part?

Pizza

curl my fingers in the direction of the current

John Rennie

I'm not sure what calculation you are doing now ...

Pizza

my thumb gives the direction of the field
So in this case a clockwise current produces a field going into the page , so we did correct

John Rennie

@Pizza Yes

Pizza

ah ok :)

@JohnRennie i can use $P = R \cdot I^2$ ?

John Rennie

10:21

Yes. The electrical power being dissipated by the current in the loop must be the same as the mechanical power being produced by the force as you pull the loop.

That has to be true because energy is always conserved.

And here mechanical energy is being converted into electrical energy.

OK so far?

Pizza

yes

John Rennie

So can you do the last part of the question?

Pizza

P = R * (-B𝓁v/R)^2

John Rennie

Yes

Pizza

so that's it

the first part was the hardest

John Rennie

10:27

Yes. The mechanical power is Fv, so we get:
Fv = R(B𝓁v/R)²

And just divide through by v

Pizza

yes

ah ok

but so if I wanted to write Δφ/Δt

how could i find Δt

John Rennie

Suppose the front edge of the square leaves the field at time t = 0, then the time taken for the flux to fall to zero is the time the trailing edge of the square leaves the field.
Yes?

Pizza

because I have seen that in some exercises it is calculated the time for the loop to enter the field

John Rennie

Isn't that exactly the same as the time the loop takes to leave the field, just in reverse?

Pizza

yes

@JohnRennie yes

John Rennie

10:32

So can you calculate that time?

Pizza

t = a/v

John Rennie

We've chosen the time t = 0 to be the time at which the front edge leaves the field, so at this moment the length of the square left in the field is 𝓁.
Yes?

Pizza

@JohnRennie yes

John Rennie

So the time for the square to leave the field is ... ?

Pizza

t = l/v

John Rennie

10:39

Yes.
At time t = 0 the whole square is in the field so φ = B𝓁²
And at time t = 𝓁/v the whole square has left the field so the flux is zero

So we have Δφ = B𝓁² and Δt = 𝓁/v

Pizza

I didn't understand, at time t = 0, isn't there that part ahead that isn't in the field?

John Rennie

I'll draw a diagram

Pizza

I'm getting confused because before the area was different from this: 𝓁²

John Rennie

Pizza

but therefore at t = 0 the entire loop is in the magnetic field (even the part that is outside in the drawing)?

John Rennie

10:48

The diagram in your question shows a time when part of the loop is outside the field i.e. 0 ≤ t ≤ 𝓁/v

Pizza

yes

John Rennie

So is it clear now?

Pizza

I didn't understand whether it should be written t = l/v or t = (l-a)/v, could I consider both cases?

John Rennie

You could do the calculation either way, but it seems simplest to start at the instant when the leading edge is at the edge of the field because that's simple.

Why would you start when the loop is part way out of the field?

Pizza

because it was shown like that in the image

John Rennie

10:53

But why does that matter? You are free to do the calculation any way you want, so you should choose the simplest way.

We can do it that way if you want to show we get the same answer ...

Pizza

you mean writing t = (l-a)/v or t = l/v ?

John Rennie

Pizza

using t = l/v i get the same answer

@JohnRennie ok

so t = (l-a)/v

John Rennie

If we start from here, i.e. we define t = 0 at this moment, then the time taken for the square to leave the field is Δt = a/v.
Yes?

Pizza

yes

John Rennie

10:57

And now the initial flux is φ = B𝓁a and the final flux is zero so:
Δφ = B𝓁a

Yes?

Pizza

yes

John Rennie

So Δφ/Δt = B𝓁v

Yes?

Pizza

yes

John Rennie

Which is the same answer as we got before.

Pizza

yes it would be -B𝓁v

A thousand thanks !

John Rennie

11:00

OK :-)

Binky

Hi

John Rennie

Hi :-)

Binky

Can I ask a quick question?

John Rennie

Yes, go ahead :-)

Binky

en.m.wikipedia.org/wiki/Energy_conversion_efficiency

But here we mean to calculate W_output/W_input ?

John Rennie

11:05

It's not a very good question as it's rather unclear.

If you have a 10kg box on a (frictionless) 10° ramp then the force you need to exert to hold the box stationary is given by:
F = Mg sinθ = 100 sin(10°) ≈ 17.4N

OK so far?

Binky

Ok

John Rennie

But in the question the box is being pushed with a force of 200N. So the box is accelerating up the ramp i.e. its KE is being increased as well as its PE.

Binky

Yes

John Rennie

So if your aim is just to raise the box vertically, i.e. just increase its PE, then the extra force is wasted.

That is, if you just used a 100N force to raise the box vertically upwards you would use a lot less energy.

And I would guess that difference is what the question is about, though I'm not totally sure as the question is not clear.

mo-_-

@JohnRennie Hi sir

John Rennie

11:11

Hi :-)

Binky

@JohnRennie So 17.4 Newtons was the minimum force to take it to the top?

John Rennie

Yes

Binky

Mmm

John Rennie

That would push the box up the ramp without changing its velocity

Binky

Could the efficiency be F_min/F?

John Rennie

11:17

I would guess the efficiency is the ratio of the work done.

Satwik

@JohnRennie Hi! I used to discuss questions with you guys a few years back. I'm glad to see you and this room still active. Cheers!

John Rennie

Work = force times distance, and since the distance is the same (3.5m) then yes the efficiency is just the ratio of the forces.

@Satwik Hi :-) As long as the JEE exists this room will stay busy :-)

Binky

Thanks very much

John Rennie

You're welcome :-)

mo-_-

@Satwik i am new here! :)

Satwik

11:19

@JohnRennie haha spot on

@mo-_- Hi

mo-_-

@JohnRennie are you still here?

John Rennie

Yes

mo-_-

@Satwik Hi :)

@JohnRennie I have a rather long exercise, can I send it?

John Rennie

Yes, go ahead

mo-_-

Four point charges of equal absolute value q and signs as shown in the figure are placed at the vertices of a rectangle with larger side b and smaller side a, lying in the x,y plane.

1. Determine the electric field vector at the point (0, -a/2) in the reference system shown in the figure.
2. Determine the potential at the point (0, -a/2), assuming the potential at infinity is zero.
3. Determine the work needed to move the four charges away (without swapping their positions) so they are placed at the vertices of a rectangle still centered at O but with sides 2b and 4a.

John Rennie

11:31

Have you got the diagram?

mo-_-

yes wait a sec

sorry

imgur.com/S56M13b

here

John Rennie

How far have you got with this?

mo-_-

@JohnRennie I don't know how to do point 3

John Rennie

You do it indirectly.

Suppose you calculate the total PE when the sides are a and b. then you recalculate the PE when the sides are 2a and 2b.

Then the difference must be the work done to move from the first configuration to the second.
Yes?

mo-_-

Yes

John Rennie

11:39

So can you calculate the PE for the initial state i.e. when the edges are a and b?

mo-_-

yes

John Rennie

What do you get?

mo-_-

U(i) = (2 / (4 π ε₀)) * (q² / a) + (2 / (4 π ε₀)) * (q² / b) + (1 / (4 π ε₀)) * (q² / √(a² + b²))

i did also the final one

John Rennie

Yes, though you can write this more succinctly as:
U = 2kq²(1/a + 1/b + 1/√a²+b²)

mo-_-

oh okay

John Rennie

11:46

Wait, I got the signs wrong ...

To find the PE we join up all the charges, so we get the six red lines, then we add up the PE for each line.
Yes?

mo-_-

Yes

John Rennie

For the top line and the bottom line the charges have the same sign, so the PE for each line is +kq²/b making +2kq²/b in total.

For the left line and the right line the charges are different, so the PE is -kq²/a for both lines making -2kq²/a in total.

mo-_-

U(i) = -(2 / (4 π ε₀)) * (q² / a) + (2 / (4 π ε₀)) * (q² / b) - (1 / (4 π ε₀)) * (q² / √(a² + b²))

John Rennie

And for the two slanted lines the charges are different so we get -2kq²/√a²+b²

U(i) = -(2 / (4 π ε₀)) * (q² / a) + (2 / (4 π ε₀)) * (q² / b) - (1 / (4 π ε₀)) * (q² / √(a² + b²))
                                                                 ^
                                                             should be 2

You missed the factor of 2 in the term for the slanted lines.

mo-_-

true

John Rennie

11:52

U = 2kq²(-1/a + 1/b - 1/√a²+b²)

So can you do the same when the sides are 2a and 2b?

mo-_-

yes

U = 2kq²(-1/2a + 1/2b - 1/√4a²+4b²)

John Rennie

Yes :-)

And we can write √4a²+4b² as 2√a²+b² by taking that factor of 4 outside the square too.
Yes?

mo-_-

Yes

John Rennie

So we get:
U' = 2kq²(-1/2a + 1/2b - 1/2√a²+b²)

And then we can cancel the 2s to get:
U' = kq²(-1/a + 1/b - 1/√a²+b²)

Yes?

mo-_-

Yes

John Rennie

11:58

And we know for any system the change in PE is the same as the work done, so now you can find the answer to part 3. What do you get?

mo-_-

I get 0

John Rennie

Really?

U = 2kq²(-1/a + 1/b - 1/√a²+b²)
U' = kq²(-1/a + 1/b - 1/√a²+b²)

Yes?

mo-_-

@JohnRennie no wait i checked

I made a mistake in picking

@JohnRennie yes

John Rennie

Where U is the PE when the sides are a and b, and U' is the PE when the sides are 2a and 2b.

And we know the change in PE is the same as the work done, so what is the work done?

mo-_-

im calculating it

W = -[q² / (4 π ε₀) * (1/a - 1/b + 1/√(a² + b²))]

John Rennie

12:07

Yes :-)

mo-_-

force modulus, the sign is given by the direction of the force right?

anyway thanks sir :)

John Rennie

You're welcome :-)

mo-_-

@JohnRennie are you here on Sunday too?

John Rennie

Yes, I'm here every morning from about 5 a.m. to 1 p.m. UK time.

mo-_-

oh okay thanks for the information

Transcript for

Problem Solving Strategies