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

#### JD_PM

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

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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

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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

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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

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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...

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#### kuruman

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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

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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.

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#### Orodruin

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By symmetry, the magnitude of $B$ is constant around the Amperian square loop.
No it isn't.

#### kuruman

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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.

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#### 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

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This might help:
https://www.glowscript.org/#/user/matterandinteractions/folder/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

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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

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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?

#### kuruman

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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.

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

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