Problem Solving Strategies

Hema

02:20

@sammygerbil I understood now thank you!

Abcd

08:04

@JohnRennie Are you there

John Rennie

@Abcd morning

Abcd

@JohnRennie magnetic force on a current carrying loop in magnetic field is 0.

Then how can it have potential energy $-\vec{\mu} \cdot \vec{ B}$

@JohnRennie Hmm, I think that energy is associated with torque coz torque is non zero.

John Rennie

Ah, yes, that's the dipole energy

$\vec\mu$ is presumably the dipole moment

Abcd

yes\

Abcd

08:28

@JohnRennie Why have we taken $\cos \theta = 1$ in this derivation, line 7

John Rennie

In the first integral, what are you integrating on the right hand side? I can't make out what the symbol is.

Abcd

@JohnRennie $$dW = \int \vec{\tau}. d\vec{\theta}$$

John Rennie

What is $\tau$?

Abcd

external torque

John Rennie

Ah

You mean here, where the $\cos\theta$ disappears?

I assume that's because we are rotating the dipole about an axis normal to it, so $\theta = \pi/2$

No, wait

Abcd

08:38

@JohnRennie yes

John Rennie

Ah, they are treating $\theta$ as an vector with the direction along the axis of rotation i.e. the axis that the angular velocity would have. So the torque and angle directions are parallel so $\theta = 0$ hence $\cos\theta=1$

Abcd

@JohnRennie Oh thanks.

@JohnRennie Magnetic field energy outside a solenoid is 0 because net field (one above the solenoid and other below it) is 0 ?

John Rennie

Is the magnetic field energy outside a solenoid zero?

Abcd

i dont know.

John Rennie

9

Q: fraction of magnetic energy stored outside a solenoid

If I have a long solenoid, e.g. length $l$ and radius $r$ with $l = kr$, where k >> 1, with a nonpermeable (e.g. air) core, how much of the magnetic energy is stored outside as compared to inside? If I go by the Wikipedia article on solenoids, $B = \frac{\mu Ni}{l} \to H = \frac{Ni}{l}$ inside t...

According to that the magnetic field energy outside the solenoid is never zero, but in the limit of $L \to \infty$ the ratio of the energy outside to the energy inside tends to zero.

Abcd

08:51

@JohnRennie In the previous derivation why have they taken $\theta = 90^\circ$ as 0 potential energy reference?

John Rennie

@Abcd There is quite a nice intuitive argument here

@Abcd Taking $\theta = 90º$ means the energy function is nicely symmetric i.e. $U(0) = -U(\pi)$ and I'd guess that's why it has been done.

But of course you can take the zero of the PE anywhere you want.

Abcd

@JohnRennie But for $\theta= 90^\circ$ potential energy is maximum as torque is maximum

But 0 is not the maximum that $\vec{\mu}.\vec{B}$

John Rennie

Remember that PE is the integral of the torque

It isn't maximised where the torque is maximised.

Abcd

oh

@JohnRennie Is there any reason/logic (not symmetry) for it to be maximum when theta = pi?

John Rennie

Since you get the torque by differentiating the PE the maxima/minima of the PE are where the torque is zero.

I'm not sure what other reason you need. It's like asking why the PE is a maximum at the top of a hill ...

To increase $\theta$ from zero to $\pi$ takes work. Then as you increase the angle from $\pi$ to $2\pi$ you get work back.

Abcd

09:10

@JohnRennie Why have they defined: $\vec H = \dfrac{\vec B}{\mu_o}- \vec I$

John Rennie

I've never really understood the difference between $\vec H$ and $\vec B$

Abcd

11 hours ago, by Abcd

@JohnRennie Please see this then.

John Rennie

Ah right. When you have $\mu_r \ne 1$ you use $H$ to mean the bit of the field not due to the polarisation of the material i.e. $H = B/\mu_0$.

No, wait ...

$H = B/\mu_0 - M$ where $M$ is the field due to the magnetisation of the material.

So in your original equation presumably $\vec I$ is the field due to the magnetisation of the medium.

Abcd

@JohnRennie I just dont understand what this equation is. What is its significance? How is it defined?

John Rennie

Where did you see it? What's the context?

Abcd

09:18

@JohnRennie I saw it in my book only. But the explanation was vague$^\infty$

The book didnt even properly explain what $\vec M$ is.

John Rennie

In a vacuum you get $B = \mu_0 H$ - so far so good.

Abcd

no.

What is H?

John Rennie

It's just the magnetic field with the factor of $\mu_0$ taken out.

Abcd

@IceInkberry @AvnishKabaj @Jasmine Did you all study Maxwell's Law of induction (induced magnetic fields) and do questions on it like the magnetic field inside a capacitor being charged?

@JohnRennie Whats the purpose of it?

John Rennie

In a vacuum $H$ and $B$ differ only by a constant factor so there's no need to use $H$.

But in a magnetic medium there are two fields present:

1. the external field applied to the medium

2. the internal field due to the magnetisation of the medium

Abcd

09:22

yes sure

then?

John Rennie

For example in an iron transformer core the external field would be the field due to the current in the windings,

while the internal field would be due to the magnetisation of the core.

In that case you divide the total field into two parts. One is the external field and the other is the magnetisation field.

And $H$ is the external field.

So the total field would be something like: $\vec B = \mu_0 \vec H + \vec M$

Abcd

@JohnRennie M??

John Rennie

$\vec M$ is the field due to the magnetisation of the medium

Abcd

Oh okay.

Then?

John Rennie

That's all there is to it.

Abcd

09:28

@JohnRennie Please relate this to the question.

John Rennie

If the coil didn't have the magnetic material in the middle then the field inside it would just be the usual $\mu_0 N I/L$

With the core the field is $\mu NI/L$ and we can split this into $B = \mu_0 NI/L + (\mu-\mu_0)NI/L$

So from above we get $H = \mu_0 NI/L$ and $M = (\mu-\mu_0)NI/L$

Length. I'm using $N$ for the total number of turns so the number of turns per unit length is $n = N/L$

To be honest I don't really understand what the question is asking.

Abcd

@JohnRennie Electric field is increasing. How did he get the direction of induced magnetic field?

John Rennie

What is that diagram?

Abcd

@JohnRennie Any idea what's the use of $\mu$ in the question?

John Rennie

@Abcd if you put (for example) an iron core into a solenoid you increase the field by a factor of $\mu/\mu_0$. That's why you're told what $\mu$ is.

I wonder if the question means $\mu_r = 1400$. That would make more sense.

Avnish Kabaj

09:43

@Abcd yes

Abcd

@AvnishKabaj When?

@JohnRennie Please see for more context^^^

Avnish Kabaj

@Abcd while studying emi

Like 3 weeks ago

Abcd

@JohnRennie I am asking that electric field is increasing so how did he determine the direction of induced magnetic field

@AvnishKabaj is it in syllabus. (please see carefully that I am talking about Induced magnetic fields)

John Rennie

@Abcd that comes from Maxwell's equations.

Abcd

@JohnRennie That's just: $$\int_C B. ds = \mu_o\epsilon_o \dfrac{d\phi_E}{dt}$$

Avnish Kabaj

09:45

@Abcd ¯\_(ツ)_/¯

John Rennie

Curl of B is proportional to $d\vec E/dt$

Abcd

@JohnRennie hmm so?

How did he get that direction

John Rennie

Assuming the current is zero $$\nabla \times \mathbf B = \mu_0\epsilon_0 \frac{d\vec E}{dt} $$

\nabla

It's the vector $(d/dx, d/dy, d/dz)$

Abcd

@JohnRennie I dont know it.

John Rennie

And $\times$ is the cross product.

Abcd

09:48

@AvnishKabaj do you know how to find the direction of induced magnetic field.

John Rennie

$\nabla \times$ is the curl of the vector it's operating on.

Abcd

@JohnRennie Please explain in some way like Lenz's law

Qualitative treatment.

John Rennie

I don't think it can be explained in any more simple way ...

Anonymous

@Abcd Qualitatively curl describes the manner in which a vector field (in this case the magnetic field) rotates at a certain point

Abcd

@Blue I want to know how to determine the direction of induced magnetic field using elementary procedures like Lenz's law

Anonymous

09:51

The direction of rotation is same as the direction of change of $\vec{E}$

Abcd

@Blue Please dont explain that.

Not in syllabus. And I have never used it.

John Rennie

Use the right hand screw rule

Abcd

@JohnRennie how?

John Rennie

Abcd

I see that the direction of induced magnetic field should support the change in flux

@JohnRennie I didnt mean that "how"

I meant how is it applicable here

John Rennie

09:53

The thumb points in the direction of $d\vec E/dt$ and the fingers show the direction of the field.

Anonymous

@Abcd I'm trying to explain how you can determine the direction of induced magnetic field using Maxwell's equation. If you don't want to know it, fine. But then the only other alternative left to you is memorization.

Abcd

43 secs ago, by Abcd

I see that the direction of induced magnetic field should support the change in flux

John Rennie

I'm not sure that helps

Anonymous

There are of course tons of tricks to memorize it which you'll find in textbooks

Abcd

@JohnRennie Can this line of thought be extended to find the direction^

Avnish Kabaj

09:53

Wait no I haven't done magnetic

Field

Why are you studying out of syllabus stuff

It's a waste of time

Abcd

@AvnishKabaj I thought its in syllabus

Thats why I asked you

Avnish Kabaj

I thought you were saying electric Field

My bad

Anonymous

Eh, they're complementary

Abcd

@AvnishKabaj Btw displacement current and all is there in the chapter EM waves. '

1 min ago, by Abcd

@JohnRennie Can this line of thought be extended to find the direction^

@JohnRennie ^^

John Rennie

No, or at least I don't see how.

Abcd

09:57

Okay, Thanks for helping.

I will ask this part on main.

Anonymous

10:07

@Abcd Dude, if you were a little more patient then the direction of the magnetic field would pretty intuitive from part (b) of the image you posted. You'd just need to keep the two Maxwell equations in mind in order to convince yourself that it's true but that's not necessary once you build the intuition.

Anonymous

Anyhow, I don't feel like helping anymore.

John Rennie

10:25

@Abcd hi

Abcd

@JohnRennie Why does Blue say that the direction is pretty intuitive from diagram b?

I was basically looking for an explanation using the "support the change" part.

John Rennie

@Abcd I suspect he just means you get used to what direction the field rotates in from experience. I can't see any reason why it's intuitive other than you've done it before.

Abcd

Okay.

John Rennie

@Abcd you could argue that the changing field has the same effect as a current in the same direction ...

In Maxwell's equations that is the case. The full equation is:

$$\nabla \times \mathbf B = \mu_0\left( \vec J + \epsilon_0 \frac{d\vec E}{dt} \right) $$

So a current and a changing field in the same direction induce the same field. You can imagine an infinite wire with current flowing in the direction of the changing field.

Anonymous

@Abcd I don't know what the principle is, but we surely didn't study anything under the name of Maxwell's law of induction.

John Rennie

10:38

I have to go now. I'll be around for a bit in a few hours, then again tomorrow morning.

Jasmine

12:56

@Abcd no

Abcd

15:23

@JohnRennie Are you there?

John Rennie

@Abcd hi

Abcd

@JohnRennie Please let me know the error in my attempt.

3 mins ago, by Abcd

@JohnRennie Please let me know the error in my attempt.

@JohnRennie Thats the coordinate axis in the first diagram. Also, initially left end of the log was kept at (0,0).

John Rennie

Hmm, calculating the centre of mass is the correct way to do the problem ...

I would take the origin to be at the centre of the plank, then $mx$ is always zero for the plank. In that case the position of the COM is just $mx/(M+m)$ where $x$ is the diatsnce of the man from the centre.

Abcd

@JohnRennie Not getting the right answer :((

@JohnRennie But my coordinate axis should work too.

John Rennie

15:40

Hmm, I get $mx/(M+m)$ for (a) and $2mx/(M+m)$ for (b). What answer is given?

Abcd

@JohnRennie What is your x?

John Rennie

Oops sorry, $x=L/2$

Abcd

@JohnRennie your answer's correct.

2 mins ago, by Abcd

@JohnRennie But my coordinate axis should work too.

John Rennie

I don't understand your method ...

Abcd

@JohnRennie Can you see my coordinate axis at the top left corner?

John Rennie

15:43

No

You're measuring the distances from the end of the rod?

Abcd

No see:

@JohnRennie Can you now see my coordinate axes?

John Rennie

No.

Abcd

@JohnRennie Now?

John Rennie

Are you calculating the position of the COM using the usual expression: $\frac{1}{M} \Sigma mx_i$

Abcd

First please tell me if you can see my axes :( @JohnRennie

John Rennie

15:46

I can see the red lines, but what I can't see is what they are intended to mean.

Abcd

Okay, let me show more clear diagram

@JohnRennie Arrow denotes man.

@JohnRennie x is distance of man from origin.

y is the distance that the log moves behind

@JohnRennie Now is my diagram clear so that I can elaborate more?

John Rennie

OK I think I see what you're doing. The origin is the point where the end of the plank is initially. As the man walks right the plank moves left

Abcd

@JohnRennie ya exactly

18 mins ago, by Abcd

John Rennie

And $x$ is the distance to the centre of mass.

Abcd

@JohnRennie Please see if the rest of the steps make sense

@JohnRennie Capital $X$

Small x is distance of man from origin

John Rennie

15:53

OK

When the man is halfway (part a) then the distance to both the man and the plank COM is just $x$ ...

Abcd

@JohnRennie ??

@JohnRennie I am deriving for general situation

For part a, isnt x+y = L/2 ???

John Rennie

To be honest I see no benefit in pursuing this

Abcd

Never mind Then.

Transcript for

Problem Solving Strategies