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I'm working on a science project and need to know what makes a strong electromagnet - more wire turns or more voltage? I understand energy can be lost to heat with too many wire turns.

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What determines the strength of an electromagnet? Turns or voltage, or both? – Bernhard Apr 5 '13 at 5:27
Amps times Turns. – Optionparty Apr 5 '13 at 14:03

The magnetic field in a solenoid of length $L$ around an iron core with $N$ turns is given by: $$ B = \mu \frac{NI}{L}. $$

Assuming some Ohm's law type of resistance in the wire we can replace $I$ with $V/R$ to get $$ B = \mu \frac{NV}{LR}. $$

So the magnetic field strength increases linearly with both the number of turns and the voltage. The resistance of a wire is given by

$$ R = \rho \frac{l}{A}, $$

(where $\rho$ is some material property of the wire, $A$ is the cross sectional area and $l$ is the length) so that $LR = \rho Ll/A$ and

$$ B = \mu \frac{NVA}{\rho Ll}. $$

If $r$ is the radius of the wire ($\pi r^2 = A$) then $2 r N \approx L.$ (Think of stacking the rings of wire on top of each other.) These last substitutions yield

$$ B = \frac{\mu \pi r}{2 \rho }\frac{V}{l}. $$

This seems to suggest that increasing the voltage has the same effect as shrinking the length of wire used.

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Though note that you have to add turns without increasing the length of the coil. – Michael Brown Apr 5 '13 at 2:28
@MichaelBrown I realized that I overlooked that and edited it accordingly. – Pricklebush Tickletush Apr 5 '13 at 2:34
@AlecS It seems like $L$ is used to denote both the length of the solenoid and the length of the wire. This makes the answer confusing. – jkej Apr 5 '13 at 9:31
@jkej Corrected. Thanks. – Pricklebush Tickletush Apr 5 '13 at 10:02
@AlecS Better, but the substitution $2rN=L$ seems to be based on the assumption that the windings are all in a single layer. It is quite common to wind the wire in several layers. Thereby, increasing the number of turns does not necessarily increase $L$. It would seem more obvious to make the substitution $l=Ns$, where $s$ is the length of each turn, since increasing the number of turns will most definitely make the wire longer. – jkej Apr 5 '13 at 18:41

When you're increasing the voltage, you're increasing the current. Recall Ohm's law. Provided a steady current flows through the material at constant temperature, the resistance remains the same. If $I$ is increased, then $V$ has to increase in order for the resistance to remain constant.

The magnetic field along the axis of the solenoid is given by $$B=\mu \frac{NI}{L}$$

The permeability of the medium $\mu$, which can be of the order of some $10^5$ when the coil is wound on a soft iron core, which is really helpful. The current (or voltage) can be increased.

There's a problem with increasing the number of turns. When solving for Ohm's law, the expression when you combine $I=nAev_d$ and $v_d=eE\tau/m$, you get $$V=\frac{mL}{nAe^2\tau}I$$

The large expression in the middle is the resistance $R$. It can be seen that $R\ \alpha\ l/A$. It can be seen that the larger the area, the less resistance there is. So if you're using a long coil, make sure that the area occupied by the coil per turn is more. There's another problem if you're using an AC. The number of turns can also increase the inductance of the coil and hence may lead to impedance, which also adds up with the resistance.

Hence, I suggest the use of soft iron core, increase the current and voltage input, use a wire with low specific resistivity $\rho$ like silver or aluminum. And finally - if you require more number of turns, use thick wires and make sure that the coil occupies a larger area.

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"When you're increasing the voltage, you're decreasing the current"? – Michael Brown Apr 5 '13 at 2:47
@Michael: All are waiting - when this fella makes a typo..! ;-) – Waffle's Crazy Peanut Apr 5 '13 at 2:51
i just dont know how increasing voltage will lead to increase in current.? – idiosincrasia23 Apr 5 '13 at 9:31
@A4KASH: Hi A4kash. I think you should. So, you're saying that increasing current doesn't increase voltage, which leads to a conclusion that a 1 ohm resistor changes its resistance whenever you change the current..? – Waffle's Crazy Peanut Apr 5 '13 at 16:08

The $B$ field does matter, but the force that is exerted from the field is $$F=qVB\ ,$$ where $q$ is charge, $V$ is voltage, and $B$ is magnetic field. Given the equation for $B\ ,$ $$B=\frac{\mu N V }{L R}\ ,$$ we have that $$F= \frac{V^2 N \mu}{L R}\ .$$Thus, in order to optimize the force output you must use voltage more than anything.

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