# Solenoid, How Many Layers?

Hello,
The problem I am struggling with is this

You are designing an electromagnet capable of producing a magnetic induction field of 1 T
by winding a solenoid around a long cylinder. A solenoid is typically wound by starting
at one end of the cylinder and placing loop after loop directly next to each other. Once
the other end of the cylinder is reached the first winding layer of the solenoid is complete
and the next layer is started by again placing loop after loop next to each other. You are
using $300 \mu m$ diameter wire and the current needed to reach 1 T should not exceed 10 A.
What is the minimum number of winding layers required?

$B=\mu \frac{N}{l} I$​

It seems that there is information missing in the question, such as the material of the wire and the material of the core so I suppose I should just choose which as it doesn't matter really as they will just be values which can be plugged in later. From the information in the question I am guessing I need to use the diameter of the wire to find the length $l$ of the solenoid. I think the length of the solenoid is the number of loops multiplied by the diameter of the wire, $N$ x $d=l$ but substituting this into the Ampere's law gives

$B=\mu \frac{1}{d} I$​

I'm not sure that I've gone the right way here as in this expression there isn't anything I don't know. I've tried using the expression for resistance of a wire which includes the length of the wire and the area but since I only have the current in the Ampere's law expression using the resistance expression will introduce a voltage which I have not been given, eg
$R=\frac{\rho L}{A}=\frac{4 \rho L}{\pi d^2}$​
Then substituting this in to ampere's law
$B=\mu \frac{N}{l} \frac{V \pi d^2}{4 \rho L}$​
Getting the voltage in my expression doesn't seem to help my cause much so I am out of ideas. Any help with how I could proceed or where I may be going wrong would be greatly appreciated.

Hi.
You don't need to concern yourself with wire material, voltage or resistivity here since the current you need to work with is a given: 10A. Also, the magnetic field in the solenoid doesn't really depend on the length of the solenoid but on the density N/L which, for a single layer, the value of the wire's diameter allows you to determine as you did (1/d). As for permeability, if you need a numerical result and it's not specified otherwise i'd simply use the vacuum one, but you can indeed plug that in the end...

Hi,

Thanks for the response. So do you mean that from this equation;
$B=\mu \frac{1}{d} I$​

I would find $d$ when $I=10$ and $B=1$ and take that value of $d$ to mean the depth of wire in that situation, then divide by the diameter of the wire (as given in the question) to give the number of layers?

If that is it then the problem I have is that $d$ for the overall solenoid comes out as smaller than the $d$ for the wire quoted in the question and since it would be several layers thick it should be several multiples of the wire's diameter.

Thanks again

TSny
Homework Helper
Gold Member
I think the length of the solenoid is the number of loops multiplied by the diameter of the wire, $N$ x $d=l$

Wouldn't the length of the solenoid be the number of loops in one layer times the diameter of the wire?

rude man
Homework Helper
Gold Member
Hi,

Thanks for the response. So do you mean that from this equation;
$B=\mu \frac{1}{d} I$​

That is the equation for the B field for one layer. But you want k layers. k can be non-integer.

In which case

$kB=1 \small{\text{ Tesla}}$ where $B$ is for one layer​
so

$k=\frac{1}{B}=\frac{d}{\mu I}=\frac{300 \text{ x }10^{-6}}{1.26 \text{ x }10^{-6}\text{ x }10}=23.8095 \small{\text{ layers}}$​

Does this look right? 23 layers seems a big number but then again 1 tesla is also big.

rude man
Homework Helper
Gold Member
In which case

$kB=1 \small{\text{ Tesla}}$ where $B$ is for one layer​
so

$k=\frac{1}{B}=\frac{d}{\mu I}=\frac{300 \text{ x }10^{-6}}{1.26 \text{ x }10^{-6}\text{ x }10}=23.8095 \small{\text{ layers}}$​

Does this look right? 23 layers seems a big number but then again 1 tesla is also big.

Looking good! I got exactly the same number. And your reasoning is sound. 1 Tesla from a solenoid without a high-permeability (large μ) core is indeed a very large B field.

Excellent, thanks a lot to everyone.