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We know that from Oersted's experience the lines of force of the magnetic field generated by a wire crossed by current $I$, are concentric circumferences of radius $r$ (variable) where the centre is a point of the wire.

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To prove that these are actually circumferences I have followed this path. The law of the magnetic field of a current crosses the wire is:

$$B(r)=2k_m\frac{I}{r}, \quad \tag{1}$$

where $k_m=k_e/c^2$. Now from $(1)$ we have also:

$$r=2k_m\frac{I}{B(r)}, \quad \tag{2}$$

If we indicate with

$$K=2k_m\frac{I}{B(r)}, \quad \tag{3}$$ we have:

$$r=K, \quad \tag{4}$$

But if $r=\sqrt{x^2+y^2}\,$ ($2-$dimensions plane) then from $(4)$, $$x^2+y^2=K^2$$ which is exactly a circumference with center in in the origin (a point of the thread where we have fixed a Cartesian orthogonal reference system).

If $r=\sqrt{x^2+y^2+z^2}\,$ ($3-$dimensions space) should I draw spheres with a thread point in the middle or are they always circumferences with the center of a point of the wire?

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    $\begingroup$ Well, in this formula, the $r$ is the one we use with the cylindrical coordinates (because of the cylindrical symmetry of the field), so we can't really "switch" to a spherical $r$. $\endgroup$ – Syrocco Jun 26 at 22:19
  • $\begingroup$ @Syrocco You're right. I had already thought about this but being a distance I thought about finding a relationship that would give me circumferences. I understood your comment. Could you kindly give me an exhaustive answer on how I should proceed mathematically? These are subjects that I dealt with at least 26 years ago. Thank you. $\endgroup$ – Sebastiano Jun 26 at 22:27
  • $\begingroup$ I'm not sure I understand what you're asking me to do. You've already found the circumference: $x^2+y^2=K^2$. But you can't generalize this to $x^2+y^2+z^2=K^2$ because $r$ isn't a radius (it's the orthogonal projection of $r$ onto the plane orthogonal to the wire). I'm sorry if I misunderstood your question. $\endgroup$ – Syrocco Jun 26 at 22:41
  • $\begingroup$ @Syrocco Don't worry about it. Could you give me a detailed mathematical explanation as an answer? Greetings. $\endgroup$ – Sebastiano Jun 26 at 22:46
  • $\begingroup$ You have written B(r) as a scalar when in reality it is a vector. Also I is not a constant but there is a current distribution. The reason there is no z is that we have reduced dimensionality by exploiting symmetry. Else you would have to sum/integrate B for each current element in the wire. $\endgroup$ – Paul Childs Jun 26 at 23:59
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You said:

If $ r = \sqrt{x^2+y^2+z^2} $ (3−dimensions space) should I draw spheres with a thread point in the middle or are they always circumferences with the center of a point of the wire?

But it simply isn't, in that formula $r$ is defined as the distance between the point in which you evaluate the field and the wire, in the same manner as you define the distance between a point and a straight line.

I really don't understand what you are doing in your calculation.

In the first place, I want to clarify a point, what I think you are trying to find with your calculation is the set of points in which the field has the same intensity, but this is not the definition of lines of force, maybe you are making some confusion.

Moreover, even if you want to find the set of points in which the field has the same intensity, let me redo that in a cleaner way.

You have to start from the assumption of costant intensity of the field:

$$ B(r) = C $$

where C is a costant. Then, for the law you mentioned:

$$ 2k\frac{I}{r} = C $$

So:

$$ r = 2k\frac{I}{C} $$

If one further assumes that I is costant, then r is costant too:

$$ r = C' $$

Being $r$ defined as $\sqrt{x^2 +y^2}$ then you have:

$$ \sqrt{x^2 + y^2} = C' $$

That is the equation of a circumpherence.

So, this is a correct process to get the set of points in which the field has the same intensity, but I stress again that this are not the lines of force.

The lines of force associated to a vector field are defined (in an elementary way) as those lines such that the vector field is always tangent to them, in this case this two distinct concepts coincide, but it is a mere coincidence.

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  • $\begingroup$ In the meantime, I thank you for your elegant response. I had a doubt simply. If there is something that is not correct you can edit my question. $\endgroup$ – Sebastiano Jun 27 at 10:49
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    $\begingroup$ Thank you, this is my first answer, and also I'm not a native english speaker, so I apologize if my english is not clear sometimes. $\endgroup$ – Fabio Di Nocera Jun 27 at 11:47
  • $\begingroup$ Welcome for me and my best regards from Sicily :-) $\endgroup$ – Sebastiano Jun 27 at 13:41

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