Magnetic field produced by a solenoid

In summary: So if the winding density is standardized, then the length of the coil doesn't really matter. However, if you have a lot of variation in the winding density, then the length of the coil will play a bigger role. So, in summary, the magnetic field of a solenoid is determined by the length of the coil, the specific wire you're using, and the voltage applied.
  • #1
Nabeshin
Science Advisor
2,207
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Ok, so anyone who has studied magnetism knows that the magnetic field due to a solenoid is given by the equation [tex]B=\mu_oNI[/tex] Where N is the number of turns in a given length. Well, in thinking about these, I tend not to make measurements and would rather like to predict the magnetic field of the solenoid, so I tried to "simplify" the formula. Here goes:

[tex]B=\mu_oNI[/tex]
N is turns, which I call n per length, which I will call [tex]L_s[/tex]
[tex]B=\frac{\mu_onI}{L_s}[/tex]
From ohms law, I = V/R
[tex]B=\frac{\mu_onV}{L_sR}[/tex]
And R is equal to [tex]\frac{\rho L_w}{A}[/tex] Where [tex]L_w[/tex] is the length of wire and A is its cross-sectional area.
[tex]B=\frac{\mu_onVA}{\rho L_s L_w}[/tex]

Up till here I'm confident, but after this I'm not so sure. Now, I tried to model the helical nature of the wire wrap by creating a vector-valued function:
[tex]r(t)=sin(t)i-cos(t)j+\frac{r_w}{\pi}t k\left[/tex]
Now in doing this I assume that the coils are wrapped as tight as possible. By this I mean that the horizontal spacing between two loops is equal to 2r, or the diameter of the wire. So this function should move a distance of 2r up for every turn (2*pi radians).

Now, arc length is given by the formula: [tex]s(t)=\int||r'(t)||dt[/tex] So...

[tex]s(t)=\int\sqrt{sin^{2}(t)+cos^{2}(t)+(\frac{r_w}{\pi})^{2}}dt[/tex]
Now, in one turn of the wire we rotate 2pi radians, so let's evaluate from 0 to 2pi..
[tex]s(t)=\int^{2\pi}_{0}\sqrt{sin^{2}(t)+cos^{2}(t)+(\frac{r_w}{\pi})^{2}}dt=2\sqrt{\pi^{2}+r^{2}}[/tex]
So this is the length of wire per one turn, so we can say:

[tex]\frac{Length\:eek:f\:wire}{Turn\:eek:f\:wire}=2\sqrt{\pi^{2}+r^{2}}[/tex]

And turns of wire is n, so we finally arrive at:

[tex]L_w = 2n\sqrt{\pi^{2}+r^{2}}[/tex]

Back to the main mission! Substituting in for [tex]L_w[/tex] produces:

[tex]B=\frac{\mu_o A V}{2\rho L_s\sqrt{\pi^{2}+r^{2}}}[/tex]

So I gather that the magnetic field of a solenoid depends on three things: How long the solenoid is, the specific wire you're using (both size and material), and the voltage applied. Well, does this make sense to anyone? :rofl:
 
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  • #2
Nabeshin said:
So I gather that the magnetic field of a solenoid depends on three things: How long the solenoid is, the specific wire you're using (both size and material), and the voltage applied. Well, does this make sense to anyone? :rofl:
Accept that you left out the fourth thing. Strengthen the magnetic field is a function of "field density." And the density of the field is determined by the number of turns in the coil per inch of links along the coil. So, the field strengths does not only depend on the length of the coil, but also the density of the windings. Unless, of course, you're talking about a standardized winding density, in which case length of the coil is all that matters.
 
  • #3
LURCH said:
Accept that you left out the fourth thing. Strengthen the magnetic field is a function of "field density." And the density of the field is determined by the number of turns in the coil per inch of links along the coil. So, the field strengths does not only depend on the length of the coil, but also the density of the windings. Unless, of course, you're talking about a standardized winding density, in which case length of the coil is all that matters.

Well, I assumed the coils were right next to each other which is as tight as they can get, which I think is what you mean by standardized winding density.
 

What is a solenoid?

A solenoid is a coil of wire that is tightly wound in a helical shape. It is commonly used in electronic devices to create a controlled magnetic field.

How is a magnetic field produced by a solenoid?

A magnetic field is produced by a solenoid when an electric current flows through the coil of wire. The magnetic field is created by the alignment of the individual magnetic fields of the electrons within the wire.

What factors affect the strength of the magnetic field produced by a solenoid?

The strength of the magnetic field produced by a solenoid is affected by the number of turns in the coil, the current flowing through the wire, and the material of the core inside the coil. Increasing any of these factors will result in a stronger magnetic field.

How can the direction of the magnetic field be determined in a solenoid?

The direction of the magnetic field in a solenoid can be determined by using the right-hand rule. Point your right thumb in the direction of the current flow in the coil and your fingers will curl in the direction of the magnetic field.

What are some practical applications of the magnetic field produced by a solenoid?

The magnetic field produced by a solenoid is used in a variety of applications, including electromagnets, motors, generators, and speakers. It is also used in medical imaging technology such as MRI machines.

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