Magnetic field of circular loops and solenoid

[al33]

I don't understand something. At the center of N circular loops, the magnetic field is μ_0NI/2a. And that for a solenoid is μ_0nI. Why are they not the same when the number of loops is large and the length for the solenoid is long?

 

[Charles Link]

It's two very different geometries that you are trying to compare. Note: $n=\frac{N}{L}$, where $L$ is the length of the solenoid.$\\$ In the first case, $a$ is the radius of the ring(s), and it has a very short length. Essentially, $a>> L$. $\\$ For the second case,=the solenoid, its radius doesn't matter, so long as it is fairly long compared to its radius. For the solenoid formula to be accurate, $L >> a$.

 

[al33]

If we place many rings side by side, it looks just like a solenoid, right? And if we apply Ampere’s law on both cases, aren’t we supposed to get the same result? If not, how come? There must be some point that I haven’t figured out.

 

[Charles Link]

If we place many rings side by side, it looks just like a solenoid, right? And if we apply Ampere’s law on both cases, aren’t we supposed to get the same result? If not, how come? There must be some point that I haven’t figured out.

For the first case, $a>> L$. The first case does not work once $L$ starts to get large enough to make a short solenoid. $\\$ Meanwhile, Ampere's law only works for the long solenoid geometry. Biot-Savart works for any geometry. Biot-Savart can readily be computed on-axis for the solenoid of medium length. Let me see if I can find the result in a google and give you a "link": https://notes.tyrocity.com/magnetic-field-along-axis-of-solenoid/ This "link" really needs a figure to show what the angles $\Phi_1$ and $\Phi_2$ are, but perhaps it is somewhat apparent. Here is a "link" with a diagram. See p.2. The angles are called $\theta_1$ and $\theta_2$ in this diagram. http://www.pas.rochester.edu/~dmw/phy217/Lectures/Lect_27b.pdf And see the formula at the bottom of p.6. This second "link" is using cgs units, so a couple conversion factors are necessary to get to the MKS result. $\\$ Editing: You can even use the formula $B=\frac{\mu_o nI}{2}( \cos(\Phi_1)-\cos(\Phi_2))$ to work the case with $a>>L$, and you do get the formula $B=\frac{\mu_o NI}{2a}$ that you presented above. (You let $n=\frac{N}{\Delta}$, (with $L=\Delta$), and $\Phi_1=\frac{\pi }{2}-\frac{\Delta}{2a}$, and $\Phi_2=\frac{\pi}{2} +\frac{\Delta}{2a}$. In the limit $\Delta \rightarrow 0$, you get the first formula above).

 

[al33]

Wow, thanks for the link and the editing part. I should and sould not have posted this thread. I posted so that I could see all of these great derivations. I should not because I am afraid that I have wasted some of your time. I made a mistake interpreting the result for the N loops. That’s for the geometry when you have N loops in the same plane but not for the case by placing loops side by side. Of course the first case cannot use Ampere due to the bad symmetry.

Btw, I couldn’t agree more with your motto~