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

In summary, the magnetic field around the infinite straight wire equals: B = \frac{\mu I}{2 \pi s} \hat{\phi}
  • #36
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.
 
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  • #37
Oh, I see. So the drawing'd be something like this:

Screenshot (248).png


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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  • #38
JD_PM said:
By symmetry, the magnitude of ##B## is constant around the Amperian square loop.
No it isn't.
 
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  • #39
JD_PM said:
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.

AmperianSquare.png
 

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  • #40
kuruman said:
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? :) )

kuruman said:
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##:

Captura de pantalla (535).png


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

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?
 
  • #41
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 [itex]\oint \vec B\cdot d\vec l[/itex] (essentially, the sum of tangential-components of [itex]\vec B[/itex] times the perimeter) is shown.

Note how [itex]\vec B\cdot \Delta\vec l[/itex] 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],
[itex]\vec B\cdot \Delta\vec l[/itex] varies,
but the circulation is unchanged as long as the current is enclosed by the loop.
240779

240780

240781

You should play with different configurations to get a feel for the calculations.
 
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  • #42
JD_PM said:
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.
JD_PM said:
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.
 
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  • #43
kuruman said:
Do the trig.

Captura de pantalla (547).png

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?
 
  • #44
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.
 
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  • #45
kuruman said:
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:

TRIG.png


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.
 
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  • #46
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. :oldsmile:
 
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  • #47
Thank you for you patience hahaha

I am glad of being part of PF community.
 
<h2>1. Why doesn't the magnetic field have to describe a circle?</h2><p>The magnetic field is a vector quantity that describes the direction and strength of the magnetic force at any given point. It does not have to describe a circle because the magnetic field lines can take on various shapes depending on the orientation and magnitude of the magnetic field.</p><h2>2. What determines the shape of the magnetic field?</h2><p>The shape of the magnetic field is determined by the arrangement and movement of electric charges. The direction and strength of the magnetic field can also be influenced by the presence of magnetic materials.</p><h2>3. Can the magnetic field ever form a circle?</h2><p>Yes, in certain situations, the magnetic field lines can form a circular shape. This usually occurs when there is a current-carrying loop or a circular magnet. However, this is not always the case and the magnetic field can take on other shapes as well.</p><h2>4. How does the magnetic field affect charged particles?</h2><p>The magnetic field exerts a force on charged particles that are moving through it. This force is perpendicular to both the direction of motion and the direction of the magnetic field. The strength of the force depends on the charge of the particle, its velocity, and the strength of the magnetic field.</p><h2>5. Why is it important to understand the shape of the magnetic field?</h2><p>Understanding the shape of the magnetic field is crucial in many fields of science and technology. It helps us understand how magnetic materials behave, how charged particles are affected, and how to control and manipulate magnetic fields for various applications such as in motors, generators, and medical devices.</p>

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

The magnetic field is a vector quantity that describes the direction and strength of the magnetic force at any given point. It does not have to describe a circle because the magnetic field lines can take on various shapes depending on the orientation and magnitude of the magnetic field.

2. What determines the shape of the magnetic field?

The shape of the magnetic field is determined by the arrangement and movement of electric charges. The direction and strength of the magnetic field can also be influenced by the presence of magnetic materials.

3. Can the magnetic field ever form a circle?

Yes, in certain situations, the magnetic field lines can form a circular shape. This usually occurs when there is a current-carrying loop or a circular magnet. However, this is not always the case and the magnetic field can take on other shapes as well.

4. How does the magnetic field affect charged particles?

The magnetic field exerts a force on charged particles that are moving through it. This force is perpendicular to both the direction of motion and the direction of the magnetic field. The strength of the force depends on the charge of the particle, its velocity, and the strength of the magnetic field.

5. Why is it important to understand the shape of the magnetic field?

Understanding the shape of the magnetic field is crucial in many fields of science and technology. It helps us understand how magnetic materials behave, how charged particles are affected, and how to control and manipulate magnetic fields for various applications such as in motors, generators, and medical devices.

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