# Is voltage important when creating a magnetic field?

Would a DC circuit that has a high current and low voltage have a powerful magnetic field?

I'm trying to create a powerful solenoid. In order to create a powerful magnetic field, I'm focusing more on the current(I). Increasing the number of turns would cause a stronger field, but at what cost?

• The magnetic field strength is proportional to the current and the number of turns. However, the wire won't have 0 resistance, so a certain steady current will require some fixed voltage. – Olin Lathrop Feb 16 '14 at 16:16

Typically the field strength is proportional to the voltage, so to get a higher field strength you need to increase the voltage.

To see why this is you start from the basic formula for the field strength:

$$B = k N I$$

where $B$ is the field strength, $N$ is the number of turns and $I$ is the current in the coil. $k$ is a constant that we'll ignore for now. The current $I$ is given by $I = V/R$ where $R$ is the resistance of the wire, and the resistance of the wire is proportional to the length of the wire, and the length of the wire is proportional to the number of turns. So we end up with:

$$I = k' \frac{V}{N}$$

where $k'$ is another constant that contains things like the resistance per unit length of the wire and the circumference of the coil. If we substitute the second equation into the first we get:

\begin{align} B &= k N k' \frac{V}{N} \\ &= kk' V \end{align}

So the field is just proportional to the voltage. You need an increased voltage to get an increased field strength.

In principle you don't need multiple turns in your coil as just one turn would do. The trouble is that one turn has a very low resistance so there would be a very high current and it would melt the wire. When designing your coil you typically start by deciding what current your wire can safely carry. Then multiply this current by your supply voltage to find out what th resistance should be, and then calculate how many turns of wire it takes to build up the coil resistance to the required value.

Response to comment:

If the resistance per unit length is $R_\ell$, and the circumference of one turn is $l$, then the total length of the wire is $Nl$ and the total resistance is $R_\ell Nl$. So the equation for the current is:

$$I = \frac{V}{R_\ell Nl}$$

The value of $k'$ is then:

$$k' = \frac{1}{R_\ell l}$$

It's the reciprocal of the resistance of a single turn.

• Well that makes things A LOT easier. Now I can evenly split the input power to both V and I so that I could create a powerful field... What is the value of k'? I know that k = relative permeability of free space... – Pupil Dec 18 '13 at 10:54
• @Key: I've edited my answer to respond to your comment – John Rennie Dec 18 '13 at 11:36

In most cases of magnetic driver design, the construction volume $V$ for the winding (including the fill factor) is restricted. At the first shot one does not care about the number of windings. One must get rid of the heat. Therefore, the power $P$ is limited that can be dissipated as heat at the maximal admissible temperature. From this constraint you get the admissible current density $J$ from $P = V\frac{J^2}{\kappa}$ with $J=\sqrt{\frac{P\kappa}V}$ where $\kappa$ is the conductivity of the wire material. The winding number determines the resistance and therefore it defines at which voltage the current density $J$ is reached. This is a constraint from the electrical/electronic driver.

The magnetic field from current carrying wires is given by Ampere's Law, which itself can be derived by considering the magnetic field from isolated moving charges. In the limit of having many charge carriers, you can express $q\vec{v} \rightarrow I\vec{l}$, a current along a wire segment of length l. You should keep in mind that currents are the more fundamental sources of magnetic fields because moving charges create magnetic fields. What you are basically doing then is simply reformulating the fundamental equations to speak of the voltage rather than the current, via Ohms Law.

## protected by Qmechanic♦Nov 1 '16 at 18:42

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).