# I Magnetic field of circular loops and solenoid

1. Feb 13, 2018

### 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?

2. Feb 13, 2018

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, $L >> a$.

Last edited: Feb 13, 2018
3. Feb 13, 2018

### 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.

4. Feb 13, 2018

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).

Last edited: Feb 13, 2018
5. Feb 13, 2018

### 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~