# Law for magnetic field due to solenoid

1. Jun 12, 2015

### NooDota

1. The problem statement, all variables and given/known data

Just a quick clarification about the law: B = 4pi * 10^-7 * n *I/L

Does n refer to total number of loops, or just the number of loops in a single layer?

Like, If I have a solenoid that's made up of 4 layers and 1000 total loops, do I plug in 1000 or 250?

2. Relevant equations

3. The attempt at a solution

2. Jun 12, 2015

### Hesch

Let me rewrite it:

B = μ0*H

It looks like an attempt to calculate the B-Field in the center of a solenoide. Thus:

By "n/L", turns per length is meant, so you could write it:

Bcenter = μ0 * I * ( ΔN / ΔL ).

n = ΔN , L = ΔL ( if you understand what I mean )

Last edited: Jun 12, 2015
3. Jun 12, 2015

### Hesch

4. Jun 12, 2015

### rude man

1000.
n refers to the total number of loops in that formula.
However, conventionally n is used to mean "loops per unit length"; in SI that would be loops per meter.
It doesn't matter whether loops are in 1st or 2nd or 3rd or 4th layer, or any other layers.

Last edited: Jun 12, 2015
5. Jun 12, 2015

### NooDota

Okay, thank you all.

6. Jun 13, 2015

### Hesch

I don't think you are right here. It is not just a convention, but a premise. The intension of the formula is to calculate the H or B field at a distinct location ( the center ) in the solenoide.

Of symmetrical reasons you can only find the H and B fields at the center: Amperes law just tells you what the mean values will be, following the circulation path. Now calling the lower left corner of the rectangle ( circulation path ) in the figure in #3, A, then clockwise the other corners B, C, D. If AB and CD are very close to each other, the H fields parallel to AB and CD will be zero, because the H-field will be perpendicular to AB and CD due to symmetry. Making BC and DA longer ( or not centered ), the H-field will no longer be perpendicular to AB and CD. The lengths of AB and CD are assumed infinite, so that the H-field along BC is zero.

Therefore the formula can only find the H-field at the center of the solenoid, and thus DA must be kept very small. That's why n = ΔN / ΔL, ( not N/L ).

I know that the result will be the same, no matter if you use n or N. I'm just speaking of the premises as for the formula, and that the H-field is calculated at the center of the solenoid.

Last edited by a moderator: May 7, 2017
7. Jun 13, 2015

### rude man

As before, I am confused by your riposte, feeling it to be inapposite.

Last edited by a moderator: May 7, 2017
8. Jun 13, 2015

### NooDota

Ahh, I don't know any of the stuff you're talking about. I've only studied magnetic fields in my school and they just gave us direct laws to calculate them. (Straight line, solenoid, and a circular circuit or whatever you call it)

9. Jun 13, 2015

### rude man

I wouldn't worry about it! (BTW your 'circular circuit' is probably a one-turn loop or a toroid).

10. Jun 13, 2015

### Hesch

Why?

It's important to understand, that the H-field is not constant along a solenoid ( the H-field has most strength at the center ). As for other locations in/nearby the solenoid, Biot-Savart law must be used.

I'm just specifying that, as it is not mentioned in post #1.

How come that you are confused about that? Why is it inapposite?

This is the "wrong" sketch:

I've picked out the sketch in #6.

Last edited: Jun 13, 2015
11. Jun 13, 2015

### rude man

Because you did not address the OP's question (post #1). Th OP's formula is the prima facie formula for axial B in a finite-length solenoid. The main purpose of its derivation is to show the multifarious utility of Ampere's law, and is always accompanied by a statement that the so computed field is approximate. I don't think the OP wanted to go beyond that point. I pointed out that the formula does not discriminate among windings from different layers. Thus, my answer of "1000" which is all he asked for.

For a more accurate computation of the axial B field inside or outside the solenoid one invokes the Biot-Savart law, applied to each winding individually and then summed for all windings. The ensuing summation is more simply replaced by the appropriate integral. Theodoros.mihos has recently posted a site which shows one way of accomplishing this:

Last edited: Jun 13, 2015
12. Jun 13, 2015

### rude man

That's why n = ΔN / ΔL, ( not N/L ).
Same thing, unless the spacings are uneven!

13. Jun 14, 2015

### Hesch

Its the same result, but not the same "thing".
I've written this of educational reasons, so that the OP ( what does that mean? ) wont have to remember the formula. The OP has just to remember the priciple in where the formula has derived from. Having understood that, and if the OP can somehow place a symmetric circulation path through a toroid, the OP can see through how the formula must be as for a toroid. ( H = N * I / 2πr ). I think it's more constructive to explain the idea ( how to find the H-field in the center of a toroid ) in the formula, than to reply with a "1000".

My english is not excellent, but I've found help here:

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html

( Same idea ).