Why the magnetic field doesn't have to describe a circle?

[JD_PM]

Homework Statement

Imagine an infinite straight wire pointing at you (thus, the magnetic field curls counterclockwise from your perspective). Such a magnetic field equals to:

$$B = \frac{\mu I}{2 \pi s} \hat{\phi}$$

I want to calculate the line integral of $B$ around the circular path of radius $s$

Homework Equations

The Attempt at a Solution

I know the solution:

But I have been thinking on how could we prove that it does not have to be circle. The line integral over the closed path should always yield the perimeter of a circle for the answer not to change. However, this seems contradictory; I actually want to proof that it does not depend on the path.

What am I missing here?

Thanks

 

[Master1022]

But I have been thinking on how could we prove that it does not have to be circle.
Thanks

Ampere's Law states \oint_{C} \vec H \cdot d \vec l = I_{enclosed} for a given path C. This is true for any given path (e.g. circle, ellipse, random shape, etc). Therefore, we can start from this and then derive a formula for B (that way our choice of path will be incorporated into the formula)

The line integral over the closed path should always yield the perimeter of a circle for the answer not to change. However, this seems contradictory; I actually want to proof that it does not depend on the path.
Thanks

I believe this confusion arises from the fact that we have defined B using the cylindrical symmetry. We are starting with a pre-defined B with a circular path. However, if we were to start from principles and use Ampere's law, we could define B for any other shaped path and it would yield: \vec B = \frac{\mu_{0} I}{Perimeter}

Then, we substitute it back into the integral, the perimeter in the denominator will cancel with the perimeter term formed by the loop integral.

I hope I have written in a way so as to not confuse you too much. To try and summarise:
- To prove that it does not have to be a circle: Derive a formula for B from Ampere's law, noting that any path can be chosen (as long as it doesn't pass through the wire)
- Substituting this new result into the integral again will yield the result \mu_{0}I as before, which is independent of the path taken.

 

[Orodruin]

The integral form of Ampere’s law is related to the differential form through the curl theorem.

The curl theorem let's you deform the path of integration as long as the curl of the field you are integrating is always zero where the path is. The curl of your B field is zero everywhere except on the wire.

 

[JD_PM]

Ampere's Law states \oint_{C} \vec H \cdot d \vec l = I_{enclosed} for a given path C. This is true for any given path (e.g. circle, ellipse, random shape, etc). Therefore, we can start from this and then derive a formula for B (that way our choice of path will be incorporated into the formula)
I believe this confusion arises from the fact that we have defined B using the cylindrical symmetry. We are starting with a pre-defined B with a circular path. However, if we were to start from principles and use Ampere's law, we could define B for any other shaped path and it would yield: \vec B = \frac{\mu_{0} I}{Perimeter}

Then, we substitute it back into the integral, the perimeter in the denominator will cancel with the perimeter term formed by the loop integral.

OK I get what you're aiming at. However, Griffiths writes about the integral of B around a circular path before introducing the Ampere's Law (4th Edition; 3.1; Straight-Line Currents). That is why I thought I should figure out why the path the magnetic field takes around the infinite straight wire doesn't have to be a circle without using Ampere's Law.

 

[JD_PM]

The integral form of Ampere’s law is related to the differential form through the curl theorem.

The curl theorem let's you deform the path of integration as long as the curl of the field you are integrating is always zero where the path is. The curl of your B field is zero everywhere except on the wire.

Is there any other way of justifying the irrelevance of the loop's shape without using Ampere's Law?

 

[Orodruin]

Is there any other way of justifying the irrelevance of the loop's shape without using Ampere's Law?

No. The fact that you can change the integration contour is predicated on the curl of the magnetic field being zero and the differential form of Ampere's law is $\nabla\times \vec B = \vec J$ (up to some constants).

Of course you could take an arbitrary field and see if it has curl zero - but in order to compute the magnetic field at all to verify that it is the magnetic field for your particular situation you will need to apply Ampere's law. The point is that the magnetic field looks the way it does because it needs to satisfy Ampere's law.

 

[JD_PM]

No. The fact that you can change the integration contour is predicated on the curl of the magnetic field being zero and the differential form of Ampere's law is $\nabla\times \vec B = \vec J$ (up to some constants).

Of course you could take an arbitrary field and see if it has curl zero - but in order to compute the magnetic field at all to verify that it is the magnetic field for your particular situation you will need to apply Ampere's law. The point is that the magnetic field looks the way it does because it needs to satisfy Ampere's law.

It seems to me that Griffiths introduced the example of a circle; if we were to calculate the line integral of the magnetic field around a square we would get: $\frac{2\mu I}{ \pi }$

Right?

 

[Orodruin]

Not right. The integral would be the same. In addition, the result of the integral is not a vector.

Edit: It is also the case that $\hat \phi$ has no meaning as an integrated quantity. It is a local basis vector that varies from point to point.

 

[JD_PM]

Not right. The integral would be the same. In addition, the result of the integral is not a vector.

That’s where I get confused. My understanding is that the perimeter should change because we changed the shape...

 

[Orodruin]

That’s where I get confused. My understanding is that the perimeter should change because we changed the shape...

The perimeter has nothing to do with it. The curl of the field is zero, therefore the integral does not change when you change the integration curve as long as you go the same number of turns around the wire. This follows directly from the curl theorem.

The point of the integration around the circle is that the integrand becomes constant and you can move everything except $d\ell$ outside of the integral. This is not the case for the square. I suggest that you try to compute the integral along one of the sides of your square manually to see what you get (the full result is 4 times that). For example, you can use the line $x = x_0$, $y = t$ from $t = -x_0$ to $+x_0$. Note that you will have to express $\hat \phi$ in Cartesian coordinates to do this.

 

[PeroK]

It seems to me that Griffiths introduced the example of a circle; if we were to calculate the line integral of the magnetic field around a square we would get: $\frac{2\mu I}{ \pi }$

Right?

For this particular field, that depends only on $\hat{\phi}$ and not on $\hat{s}$, you can parameterise any curve in terms of $s, \phi$ and the integral reduces to the simple integral for $d\phi$.

 

[Master1022]

The perimeter has nothing to do with it. The curl of the field is zero, therefore the integral does not change when you change the integration curve as long as you go the same number of turns around the wire. This follows directly from the curl theorem.

The point of the integration around the circle is that the integrand becomes constant and you can move everything except $d\ell$ outside of the integral. This is not the case for the square. I suggest that you try to compute the integral along one of the sides of your square manually to see what you get (the full result is 4 times that). For example, you can use the line $x = x_0$, $y = t$ from $t = -x_0$ to $+x_0$. Note that you will have to express $\hat \phi$ in Cartesian coordinates to do this.

I see where I made a mistake when talking about the integral. Surely, if we defined the path in terms of other parameters, the form of B would change, while still outputting the same field?

Thanks

 

[Master1022]

That’s where I get confused. My understanding is that the perimeter should change because we changed the shape...

Yes, I apologise for my first post. We can only treat the perimeter as a result if we can take the \vec B out of the integral (which can only happen if \vec B \cdot d \vec l = 1. If they are not parallel, will be other terms to deal with.

 

[PeroK]

Yes, I apologise for my first post. We can only treat the perimeter as a result if we can take the \vec B out of the integral (which can only happen if \vec B \cdot d \vec l = 1. If they are not parallel, will be other terms to deal with.

The field is $\frac{1}{s} \hat{\phi}$, which is a special case, because the polar line element is $dl = sd\phi \hat{\phi} + ds\hat{s}$, so the integrand reduces to $d\phi$.

 

[JD_PM]

The field is $\frac{1}{s} \hat{\phi}$, which is a special case, because the polar line element is $dl = sd\phi \hat{\phi} + ds\hat{s}$, so the integrand reduces to $d\phi$.

You suggest the following approach. ; using cylindrical coordinates (the current is flowing through the z axis; that is why $B = \frac{\mu I}{2 \pi s} \hat{\phi}$):

I get this method.

Now I am thinking about the cartesian coordinates approach and Orodruin's #10th comment

 

[JD_PM]

Yes, I apologise for my first post. We can only treat the perimeter as a result if we can take the \vec B out of the integral (which can only happen if \vec B \cdot d \vec l = 1. If they are not parallel, will be other terms to deal with.

We can take B out of the integral in this case; it is always equal to the magnetic field due to an infinite straight wire

 

[PeroK]

You suggest the following approach. ; using cylindrical coordinates (the current is flowing through the z axis; that is why $B = \frac{\mu I}{2 \pi s} \hat{\phi}$):

View attachment 240297

I get this method.

Now I am thinking about the cartesian coordinates approach and Orodruin's #10th comment

That's it exactly. Why worry about Cartesian coordinates?

 

[JD_PM]

That's it exactly. Why worry about Cartesian coordinates?

Because I want to understand why the shape of the path doesn't matter in cartesian coordinates.

Once I get something I will post it

 

[Orodruin]

Because I want to understand why the shape of the path doesn't matter in cartesian coordinates.

The path integral is coordinate independent. It does not matter in Cartesian coordinates because it does not matter in polar coordinates.

 

[kuruman]

Here is something to consider. Start from the line integral over a loop, any kind of loop.$$\oint \vec B \cdot d\vec l=\mu_0~I_{enc.}$$Now consider the tangential field averaged over the loop. This is $$\langle B_t \rangle=\frac{\oint \vec B \cdot d\vec l}{\oint d l}=\frac{\mu_0~I_{enc.}}{\oint d l}.$$This says that the averaged tangential field does not change if you distort the shape of the contour loop as long as (a) you enclose the same amount of current and (b) you don't change the length of the perimeter $\oint dl.$

 

[JD_PM]

The perimeter has nothing to do with it. The curl of the field is zero, therefore the integral does not change when you change the integration curve as long as you go the same number of turns around the wire. This follows directly from the curl theorem.

I understand your reasoning, which is based on the fact that $B$ is a conservative field (AKA $\nabla$ x $B = 0$ ).

 

[JD_PM]

The perimeter has nothing to do with it.

This is still shocking me to be honest. $\oint_0^L ds$ yields the perimeter; $L$ is the length of the path.

Griffiths calculates the line integral around the circular path (as an example) and then derives Ampère's Law (though it flaws, as he says, due to the fact that is based on an infinite straight wire).

Till here everything is fine but then is when it's asserted that "the path doesn't have to be a circle".

After kuruman's comment I changed my initial thought; I am convinced that the perimeter cannot change if we want Ampère's Law to be satisfied. My intuition tells me that what's happening is that you distort the circumference of the loop but the length of the path is still $2\pi s$

It's like taking a straight rope of length $L$; Although I shake it up and down, the length will remain $L$.

This explanation seems reasonable for me, what do you all think?

 

[kuruman]

... I am convinced that the perimeter cannot change if we want Ampère's Law to be satisfied.

I am not sure what you mean by this. Ampere's law is a law of nature that is always true. The loop-averaged tangential field is always proportional to the enclosed current and inversely proportional to the perimeter.

 

[JD_PM]

I am not sure what you mean by this. Ampere's law is a law of nature that is always true. The loop-averaged tangential field is always proportional to the enclosed current and inversely proportional to the perimeter.

What I mean is that the perimeter of an Amperian loop doesn't change.

 

[Orodruin]

This is still shocking me to be honest. $\oint_0^L ds$ yields the perimeter; $L$ is the length of the path.

Yes, but unless the path is a circle you cannot move everything else out of the integral to get that integral.

Griffiths calculates the line integral around the circular path (as an example) and then derives Ampère's Law (though it flaws, as he says, due to the fact that is based on an infinite straight wire).

Till here everything is fine but then is when it's asserted that "the path doesn't have to be a circle".

After kuruman's comment I changed my initial thought; I am convinced that the perimeter cannot change if we want Ampère's Law to be satisfied. My intuition tells me that what's happening is that you distort the circumference of the loop but the length of the path is still $2\pi s$

It's like taking a straight rope of length $L$; Although I shake it up and down, the length will remain $L$.

This explanation seems reasonable for me, what do you all think?

No. As has already been explained in this thread.

 

[JD_PM]

No. As has already been explained in this thread.

Why am I wrong? If the length of the path is $2\pi s$ and we 'shake' it, the perimeter does not change (which is indeed true).

 

[Orodruin]

Why am I wrong? If the length of the path is $2\pi s$ and we 'shake' it, the perimeter does not change (which is indeed true).

Because the perimeter has absolutely nothing to do with it. You have already seen that the perimeter has absolutely no influence on the integral for the result of the integral for the circle and you seem to argue that it will have something to do with the final result for a shape different from the circle, yet it should give you the same as the circle. You are simply wrong and the reason as has been pointed out repeatedly is that it is only for the circle that you get the integral $\oint d\ell$ at all.

 

[JD_PM]

Because the perimeter has absolutely nothing to do with it. You have already seen that the perimeter has absolutely no influence on the integral for the result of the integral for the circle and you seem to argue that it will have something to do with the final result for a shape different from the circle, yet it should give you the same as the circle. You are simply wrong and the reason as has been pointed out repeatedly is that it is only for the circle that you get the integral $\oint d\ell$ at all.

So if I am not mistaken, if the path were to be different from a circle, the Ampère's Law'd perfectly hold:

$$\oint \vec B \cdot d\vec l=\mu_0~I_{enc.}$$

The only difference is that now $B$ is not decreasing at the same rate as the non-circular loop increases, which means that $B$ is not constant and cannot be taken out of the integral. However, if we were to compute $\oint \vec B \cdot d\vec l$ we would indeed get the same result that in circle's case!

WoW, this is impressive...

I'd like to test this, do you know about any famous example of non-circular Amperian Loop?

 

[kuruman]

WoW, this is impressive...

Gauss's law is just as impressive except that it involves a surface integral instead of a line integral.

I'd like to test this, do you know about any famous example of non-circular Amperian Loop?

You can test this theoretically by yourself. Consider the usual circle of radius $R$ with a current-carrying current $I$ at its center. Now consider a square Amperian loop of side $R\sqrt{2}~$inscribed in the circle. Calculate $\oint \vec B\cdot d\vec l$ around the square loop and show that it is equal to $\mu_0~I$. Hint: Exploit the symmetry of the problem. You only need to find $\int \vec B\cdot d\vec l$ along one side of the square and multiply the result by 4.

 

[Orodruin]

You can test this theoretically by yourself. Consider the usual circle of radius $R$ with a current-carrying current $I$ at its center. Now consider a square Amperian loop of side $R\sqrt{2}~$inscribed in the circle. Calculate $\oint \vec B\cdot d\vec l$ around the square loop and show that it is equal to $\mu_0~I$. Hint: Exploit the symmetry of the problem. You only need to find $\int \vec B\cdot d\vec l$ along one side of the square and multiply the result by 4.

See #10 and #11 ...

 

[kuruman]

See #10 and #11 ...

Oops, I replied without rereading all the posts to check for duplicates.

 

[JD_PM]

You can test this theoretically by yourself. Consider the usual circle of radius $R$ with a current-carrying current $I$ at its center. Now consider a square Amperian loop of side $R\sqrt{2}~$inscribed in the circle. Calculate $\oint \vec B\cdot d\vec l$ around the square loop and show that it is equal to $\mu_0~I$. Hint: Exploit the symmetry of the problem. You only need to find $\int \vec B\cdot d\vec l$ along one side of the square and multiply the result by 4.

Let's focus on getting the magnetic field due to one of the four edges, which is a straight wire.

Such a magnetic field is:

$$B = \frac{\mu I}{4\pi R'}(sin \theta_2 - sin \theta_1)$$

Where

$$R' = \frac{R\sqrt{2}}{2}$$

This yields:

$$B = \frac{\mu I}{2\pi R}$$

The magnetic field due to the entire square is:

$$B = \frac{2\mu I}{\pi R}$$

But I should get $\mu I$. I must be missing something...

 

[JD_PM]

See #10 and #11 ...

Sorry, I did not think about it.

 

[kuruman]

But I should get μI\mu I. I must be missing something...

You are missing that you you are not looking for the "magnetic field due to the entire square". You need to do the line integral $\oint \vec B \cdot d\vec l$ around the square knowing that $\vec B = \frac{\mu_0~I}{2\pi r}\hat {\theta}$. Note that along the side of the square $\vec B$ varies both in magnitude and direction.

 

[JD_PM]

You are missing that you you are not looking for the "magnetic field due to the entire square". You need to do the line integral $\oint \vec B \cdot d\vec l$ around the square knowing that $\vec B = \frac{\mu_0~I}{2\pi r}\hat {\theta}$. Note that along the side of the square $\vec B$ varies both in magnitude and direction.

But isn't $B = \frac{\mu I}{4\pi R'}(sin \theta_2 - sin \theta_1)$? If I am not mistaken, $\vec B = \frac{\mu_0~I}{2\pi r}\hat {\theta}$ is the magnetic field due to an infinite straight wire (which is not the case here; we're dealing with a finite straight wire).

To calculate $\oint \vec B \cdot d\vec l$ I'd tend to do the following:

$$B = \oint \vec B \cdot d\vec l = \oint \frac{\mu I}{4\pi R'}(sin \theta_2 - sin \theta_1)\cdot d\vec l = \frac{\mu I}{2}(sin \theta_2 - sin \theta_1)$$

Is this what you were asking for (this is just for one edge of the square)?

I'd say that $\oint d\ell$ is still a circular loop; the square's case with field's point in the center of the square is like four times straight wire's case with field's point above the middle point of the straight line.

If this is not what you were asking I may better continue studying magnetostatics' chapter and then come back

 

[kuruman]

You misinterpreted what you were asked to do. Here is a re-statement of the problem as seen in post #29

An infinite wire carries current $I$. Calculate the integral $\oint \vec B\cdot d\vec l$ around a square Amperian loop of side $a=R\sqrt{2}$ oriented so that the wire goes through the center of the square and is perpendicular to its plane.

Maybe this will make it clearer what you need to do.

 

[JD_PM]

Oh, I see. So the drawing'd be something like this:

By symmetry, the magnitude of $B$ is constant around the Amperian square loop. Then, taking just one of the square's edges I calculate $B$ having the field point in the square's center:

$$\oint \vec B \cdot d\vec l = B \oint dl = B R\sqrt{2} = \mu_0~I$$

Note that both $B$ and $dl$ are parallel; we can drop the dot product.

Solving for B:

$$B = \frac{\mu_0~I}{R\sqrt{2}}$$

This is consistent with Ampere's law; for one edge:

$$\oint \vec B \cdot d\vec l= \frac{\mu_0~I}{R\sqrt{2}}R\sqrt{2}=\mu_0~I_{enc.}$$

For the entire square:

$$\oint \vec B \cdot d\vec l= 4\frac{\mu_0~I}{R\sqrt{2}}R\sqrt{2}=4\mu_0~I_{enc.}$$

Am I right? If not, please let me know and I'll keep trying.

 

[Orodruin]

By symmetry, the magnitude of $B$ is constant around the Amperian square loop.

No it isn't.

 

[kuruman]

Am I right? If not, please let me know and I'll keep trying.

You are not right. Look at the drawing below that shows the situation. The magnetic field has constant magnitude on the dashed circle but not on the square because the distance to the center of the square varies as you move along the side of the square. You need to find an expression for vector $\vec B$ at an arbitrary point on the side shown, find $d\vec l$ (for this side it is $dx~\hat x$), find the dot product $\vec B \cdot d\vec l$ (note that $\vec B$ and $d \vec l$ are not parallel) and add all such dot products continuously (i.e. integrate) from one corner of the square to the other. Then and only then you multiply the result by 4. Note: The wire carrying current into the screen and the circle of radius $R$ enclosing the square are not shown for clarity.

 

[JD_PM]

You are not right. Look at the drawing below that shows the situation. The magnetic field has constant magnitude on the dashed circle but not on the square because the distance to the center of the square varies as you move along the side of the square. You need to find an expression for vector $\vec B$ at an arbitrary point on the side shown, find $d\vec l$ (for this side it is $dx~\hat x$), find the dot product $\vec B \cdot d\vec l$ (note that $\vec B$ and $d \vec l$ are not parallel) and add all such dot products continuously (i.e. integrate) from one corner of the square to the other. Then and only then you multiply the result by 4. Note: The wire carrying current into the screen and the circle of radius $R$ enclosing the square are not shown for clarity.

View attachment 240643

Thank you, enlightening explanation and drawing (may you tell what app did you use to make it up? :) )

You need to find an expression for vector $\vec B$ at an arbitrary point on the side shown

My bad, the magnetic field isn't constant; I'm unsure whether the following is correct:

I'd say we are interested in the component of the magnetic field lying on the x direction (it must be perpendicular to the current) So our magnetic field depends on $\alpha$:

$$\oint \vec B \cdot d\vec l= \oint \frac{\mu_0~I}{2 \pi r} cos \alpha dx$$

Once here I feel like the idea is good; it is just about figuring out how $r$ depends on $x$ so that I can solve the integral.

Am I going right?

 

[robphy]

This might help:
https://www.glowscript.org/#/user/m...older/matterandinteractions/program/13-fields (matterandinteractions/program/13-fields)
Select Magnetic.
Measurement type: Ampere's Law.

Draw a square-like loop... (release the mouse to complete the loop).
Drag a current-carrying wire into the center of the loop.
The circulation \oint \vec B\cdot d\vec l (essentially, the sum of tangential-components of \vec B times the perimeter) is shown.

Note how \vec B\cdot \Delta\vec l varies along the Amperian loop.
The circulation is the total around the loop.
As you reposition the wire [the loop can't be repositioned in this software],
\vec B\cdot \Delta\vec l varies,
but the circulation is unchanged as long as the current is enclosed by the loop.

You should play with different configurations to get a feel for the calculations.

 

[kuruman]

Thank you, enlightening explanation and drawing (may you tell what app did you use to make it up? :) )

I used PowerPoint and saved the figure as a .PNG file.

Once here I feel like the idea is good; it is just about figuring out how rr depends on xx so that I can solve the integral.

Am I going right?

You're doing fine. Now you have to make sure that there is only one variable under the integral sign. For the time being you have three, $\alpha$, $x$ and $r$. Do the trig.

 

[JD_PM]

Do the trig.

I've been thinking in using the law of cosines to express $r$:

$$r^2 = \gamma^2 + (z + \frac{R\sqrt{2}}{2})^2 - 2\gamma(z + \frac{R\sqrt{2}}{2})cos\beta$$

But this over-complicates things...

I've been trying to play with $\alpha$, $x$ and $r$ so that I get an expression in function of just one of these, but got nothing of interest...

May you give me a hint?

 

[kuruman]

Forget the upper triangle. It's not needed and I am not sure what $z$ is all about and why it is there. You need to calculate the $\vec B \cdot d\vec l$ along the horizontal side of the bottom right triangle. Label the angles of the bottom triangle, they have to be related to $\alpha$ and $\beta$. Also, don't worry about the law of cosines; you have a right triangle so the Pythagorean theorem works just as well to relate the square of one side to the squares of the other two.

 

[JD_PM]

Forget the upper triangle. It's not needed and I am not sure what $z$ is all about and why it is there. You need to calculate the $\vec B \cdot d\vec l$ along the horizontal side of the bottom right triangle. Label the angles of the bottom triangle, they have to be related to $\alpha$ and $\beta$. Also, don't worry about the law of cosines; you have a right triangle so the Pythagorean theorem works just as well to relate the square of one side to the squares of the other two.

OK Let's see now.

I was dealing with the line integral:

$$\oint \vec B \cdot d\vec l= \oint \frac{\mu_0~I}{2 \pi r} cos (\alpha) dx $$

I had to do the trig:

We just need to focus on an infinitesimal path of the line integral (just on one of the four sides), so doing the trig just with one of these triangles suffices:

$$\tan (\alpha) = \frac{x}{s}$$

$$x = s \tan (\alpha)$$

$$dx = s \sec^2 (\alpha) d\alpha$$

We also know that:

$$r = s \sec (\alpha)$$

We end up with:

$$\frac{\mu_0~I}{2 \pi}\int_{-\pi/4}^{\pi/4}d\alpha = \frac{\mu_0~I}{4}$$

But this is the line integral of just one side of the square amperian loop. If we multiply per 4 we indeed get:

$$\oint \vec B \cdot d\vec l = \mu_0~I$$

How do you see it now?

Sorry for the late reply.

 

[kuruman]

Looks good. Don't worry about the late reply. I'm glad you didn't give up and saw it through the end. I hope you understand line integrals a tad better now.

 

[JD_PM]

Thank you for you patience hahaha

I am glad of being part of PF community.