# Why do magnetic coils consist of many thin wires?

Magnetic coils are usually made of thin wire turned into spring shape:

Magnetic field which is generated by this solenoid is proportional to the number of turns N.

But why? What changes if we melt all wires together and make copper tube? Of course we will need a gap. But finally we'll get coil with 1 turn of very thick wire.

Will such coil generate smaller magnetic field with the same current? And if yes, then why?

I think I should think on current density and probably with "melted" wire I should make the same current density, i.e. N times larger current then in normal coil.

But why is it possible to avoid the nesessity of bigger current by just splitting wire into thin channels and looping? Why does this trick work?

UPDATE

So, regard two coils on two pictures. The number of loops is the same and equals N, but in case (1) loops are connected serially and in case (2) loops are connected parallel. The movement of electrons is identical in both cases, so the current in case (2) is N times bigger than in case (1).

Magnetic field is identical.

The question IS: why do we prefer case (1)? Why is it easier to run electrons in serial connection? Why isn't these devices just technically equivalent?

UPDATE 2

Suppose, we have one "sick" turn, which has resistance of

$R_0$

If we connect the voltage $U$ to it, then we'll get current of

$I = {U \over R_0}$

Magnetic effects are proportional to (1) current and (2) number of turns, because effects of all turns sum to each other (approximately).

So, we wish to keep the product $NI$ constant

$NI = J_0 = const$

Suppose we split the wire into $N$ turns, remaining the total cross-section of copper the same.

The resistance of each turn will be

$r = {R_0 N}$

because it's cross section is now Nth part of total initial cross-section and $1/R$ is proportional to cross-section.

The total resistance of the coil will be

$R = r N = R_0 N^2$

because turns are connected in series. I.e. current should pass thinner wire and longer length of this wire.

If we apply the voltage of $U$ we will get current of

$I = \frac{U}{R_0 N^2}$

and $J$ of

$J = \frac{U}{R_0 N^2} \cdot N = \frac{U}{R_0 N}$

In other words, to have $J=J_0$ we should power our coil with

$U = J R_0 N$

and we will have current of

$I = \frac{J}{N}$

in the case.

This obviously means, that the more turns we have, the higher voltage should be applied. Hence, the less turns the coil has, the better magnetic properties it has.

Then why use coils with many turns?

• One can do the same thing with one turn. The only difference is the voltage/current at which the coil will operate and how fast it can switch. The more turns a coil has, the harder it is to switch it fast because of its inductance, so a fast switching coil will operate on a higher current and lower voltage with fewer turns and a DC coil will usually be made to operate on a larger voltage with more turns. – CuriousOne Jul 24 '16 at 21:17
• ELI5 answer: If you have one loop carrying 1 amp, you get a certain magnetic field. If you have two loops carrying 1 amp each, you get a stronger magnetic field because the two fields combine. – user253751 Jul 24 '16 at 21:30
• @CuriousOne see my second picture. It has the same number of turns, but they are just connected differently. – Dims Jul 25 '16 at 9:19
• In the normal coil, the loops are in serial connection. To send $1\,\mathrm A$ through each loop, you need to send $1\,\mathrm A$ through the whole thing. In your alternative design, the coils are in parallel connection. To send $1\,\mathrm A$ through each loop, you must send $N\,\mathrm A$ through the coil. For large $N$, supplying the current could become a problem. – celtschk Jul 25 '16 at 9:54
• @celtschk so this is the question: why 2nd case is harder to implement? The movement of electrons is the same then why it is more problematic? – Dims Jul 25 '16 at 11:19

If you have a single tube, the current will flow on it directly without making the $N$ loops. It will result

• a different direction, i.e. different magnetic field,
• its magnetic field will be much weaker.

Having the loops, the magnetic fields created by the induvidual loops is added. Actually, you have "the same current" using $N$ times to produce the field.

If you don't have the loops, you need to multiply the current in the wire, which is in most cases impractical:

• You need to create electronic for higher current (much more costly as to simply looping the wire)
• You need to produce higher voltage
• You need to count with much higher secondary losses, f.e. heat loss.
• See my second picture please I drew. I represent "thick" wire as the same number of turns, but just connected differently – Dims Jul 25 '16 at 9:20
• @Dims Ok! Understood. So the induvidual loops will be parallel connected, which will result that the current will be divided between them. In the normal structure, the loops are connected serially, thus every electron passes all of the loops. In your second picture, the electrons pass only a one of the loops. And the magnetic field strength depends on the current. (Although your structure would have a much lower ohmical resistance.) – peterh - Reinstate Monica Jul 25 '16 at 10:02
• So, the question IS: why do we made (1) but not (2)? What guides technical choice? – Dims Jul 25 '16 at 11:18
• @Dims In the case of (2), the loops are parallel connected, which results that the current will be divided between them. And the magnetic field strength depends on the current. Thus, the sum of the magnetic field won't be added N times. The result will be a similar magnetic field as in (1), but without the N-time multiplication. – peterh - Reinstate Monica Aug 11 '16 at 8:47

CuriousOne's comment has it exactly. Maybe some elaboration is in order, though.

I'm only going to cover the DC case -- as CuriousOne points out, when you move to AC inductive effects enter the picture.

Thinking of the coil as a fused cylinder of fixed shape is an excellent mental move for understanding the problem. No matter how you slice up the cylinder physically, the magnetic field it generates is the sum of the magnetic fields generated by each local part of the cylinder. That is, whether made up of a long thin wire or a short thick wire, the magnetic field produced is determined only by its overall shape and the density of current $\vec J$ everywhere in it.

What does care about how you slice up the cylinder is the current you have to push through the wire -- that's the current density times the cross-sectional area of the wire, and so the current is lower if you use a thinner wire. You can think of it as making the same current work harder for you by having it go around more times.

Note that you don't get something for nothing -- if you make the wire thinner, you also have to make it longer to fill up your cylinder. (The product of the cross-sectional area and the length is the fixed volume of the cylinder.) So what happens is the voltage drop in the wire goes up as the current goes down, and in fact their product, which is the resistive power loss in the coil, is fixed. (You can get that last fact directly -- the power density is $\rho J^2$, where $\rho$ is the resistivity of the metal, and so once you fix the current density you fix the resistive loss.)

This brings us to why we choose a long thin wire instead of a short thick one. You want to deliver current to your coil at a practical current and voltage -- for example, a 9V battery can comfortably deliver 100 mA, which implies a resistance of 90 Ohms. But your wire is probably made up of some good conductor like copper, so it needs to be quite long and thin to have a resistance like that.

I have seen the answers and your edit of question. It is not the serial or parallel connection of the wire that matters, both your coils will generate same magnetic field if the number of turns are same and current flown (in each turn) is same.

Suppose you have 10 turns and the current flown in each coil is 1 ampere than in second case the current flown in straight wire will be 10 amperes and current flown in first case is 1 ampere.

First of all why you get more magnetic field if you have more number of turns. The current flown in a wire will generate magnetic field, the magnetic field at the center of the wire loop will be

$$B=\frac{\mu_0I}{2R}$$

where R is the radius of the loop. Now if you stack the loops with current in same direction then this magnetic field will be added and you will get the $n$ times field. If you add these loops in series you can flow just one ampere and get $n$ times field. It may be noted here that, if by adding the loops you construct the solenoid you will get the magnetic field

$$B=\mu_0NI$$

where N is the number of turns per unit length

Now comes the low current-large number vs high current-low number mode. Both cases are used practically and it depends on your design. If you want to generate high magnetic field by adding larger numbers of turns then you have to make your wire thinner (to the size of magnet). This increases the resistance in the wire and in turn increases the heating of wire. Heating of the wire will result in removal of the insulation over it and then your magnate will fail. Hence there is a rated current over which there is always a probability of the failure of magnet. It may also be noted that the stacking of loops one over the above poses difficulty in heat removal from magnet, which will further reduce the maximum current rating of the magnet.

If you can generate larger currents ($>100$ or $1000$ Amps) it is always preferable to use thicker wires. The wires of section >1 $cm^2$ are often used to generate large magnetic field for large size magnet, because these wires have less resistance and can have center hole for cooling. You can generate much larger magnetic field by flowing chilled low conductivity water from within the wire. It may be noted here that as the size of the magnet increases so is the length of wire per turn, which also increases the resistance i.e. more heating for same current. The question here is not the power consumption but the generation of maximum magnetic field.

As you have noticed, this is a serial vs. parallel problem. We can look at the equation for a magnetic field in such a winding, $B=\frac{NI\mu}{L}$ where B is the magnetic field generated, N is the number of loops, I is the current going through each loop, and L is the length of the circuit through the center of the magnet. We see clearly that N is important, as you noticed. The higher N is, the better. However, we also see that I is important, and I is the current going through each loop.
There's two reasons we don't make them this way. First is that we're very good at making wire, so it's cheap and easy to manufacture this device. The second is resistive losses. Resistive losses in wires is proportional to the square of the current ($P=I^2R$). Now each of your loops has the same current as it does in the normal winding case, so those losses are the same. However, your bus has far more current than anything we see in a normal electromagnet, so the resistive losses in that bus will be great. The only way to solve this is to decrease the resistance of the bus dramatically, and that calls for a huge expensive piece of copper. Copper may not even be viable on the sizes you need, so it may be that you have to switch to more exotic materials.
I'd have to ballpark figures, but I'd expect that, for \$100,000, you could build a device with your design which fares comparably against a \$20 wire-wound electromagnet. Thus, the answer to your question: economics.