To find the strength in gauss you must multiply the number of turns of wire in the electromagnet by the amperage.So for a example if you have a electromagnet with 20 turns of wire with 10 amps going through the wire then you multiply 20 by 10 and that will equals 200 so the magnetic feild it's producing is 200 gauss.Hopes this helps:)
This is a typical electromagnet.
Let's refer to your figure and say that $g$ is the width of the gap and $A_c$ the section of the core.
If $g << \sqrt A_c$,the magnetic field $\vec B$ inside the core will be approximately the same as the magnetic field $\vec B_0$ in the air gap (this is because in this case the magnetic field lines will stay approximately parallel to each other in the air gap and the flux $\Phi = B \ A_c$ will be conserved).
If this is the case, you will find using Ampere's law that
$$B = \frac{\mu_0}{g} (N i - H l_c)$$
with $l_c$ mean core length.
If moreover $\vec H = \vec B / \mu$, then
$$B \left( 1 + \frac{l_c}{g \mu_r} \right)=\frac{\mu_0 N i}{g}$$
Notice again that this value is the same in the air gap and inside the core (solenoid included)!
For most ferromagnetic materials, $\mu_r$ is of the order of $10^3-10^5$ (https://en.wikipedia.org/wiki/Permeability_(electromagnetism)), so we can approximate the previous relation neglecting the term $\frac{l_c}{g \mu_r}$:
$$ B \simeq \frac{\mu_0 N i}{g} > \frac{\mu_0 N i}{l_c}$$
since of course $g < l_c$. So the field is indeed stronger than the field inside an empty solenoid of length $l_c$ with the same number of turns $N$.
Best Answer
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.