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In a circular solenoid, the magnetic field in the center of the solenoid, is written with the following equation :

$$ B = \mu_0 Ni/L $$

Where $N$ is the number of spirals, $ i $ is the intensity of the current, and $L$ the length of the solenoid.

But what of the voltage? wire material? wire radius? I am trying to create my own solenoid, and it seems very unsettling for me that I "don't have to worry" about any of those factors..

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    $\begingroup$ Hi fellow! Please be aware your formula is only valid when magnetic field is measured in a classical limit, inside a infinitely-long solenoid with perfectly circular infinitely-many-spirals all crossing same current $i$, having spiral number density $N/L$, in a vacuum environment. Anything other than that, this formula is wrong!! But you are not worrying with anything of this... Why? Why isn't it "unsettling" for you? Well, on ordinary situations, there is no reason to worry, because its a good approximation. Same thing with wire material, radius, etc... So, you shouldn't be "unsettling". $\endgroup$ – Physicist137 Sep 27 '15 at 23:29
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You're really asking about "engineering" issues rather than "physics" issues. The equation you presented gives the B-field in a solenoid. From a physics perspective, that's it. Physicists don't have to "worry" about anything else but engineers - the guys responsible for actually getting things to work in the real world - do. Now as for the wire material, wire radius, and voltage, you're going to have to come up with suitable values for those things so that the solenoid doesn't overheat, cost too much, isn't too heavy, etc.. Copper wire is a good starting point. You can look up tables on the internet which give the recommended maximum current for various wire diameters of copper wire.

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The voltage and current will be determined by the overall wire material, size and length, via the coil's resistance. First determine the electrical parameters you need from the magnetic field you need, choose a wire material (most likely copper magnet wire for availability and cost, could also be aluminum if you need to save some weight; you need to select the wire for its resistance according to the power you intend to flow through the coil), select an appropriate gauge of wire from your electrical characteristics, the length you need for the magnetic field you need, determine the resistance per foot then apply Ohm's law.

$$P_{source} = VI$$ $$I = V/R$$ $$V_{drop} = IR$$

From there you can determine the heat produced in watts from the voltage drop across the coil and current flowing through it:

$$P_{loss} = V_{drop}*I$$

For the temperature increase in the wire, you can get a rough estimate from the heat capacity of the wire material and mass of the coil. For pure copper it is $0.385 J/(g/K)$ @ $25^{\circ}C$, where $1W = 1J/s$.

Then you need to calculate heat dissipation to determine if you can deal with the heat produced and it gets complicated real fast. As a ballpark, for $P_{loss}$ >= $1W$ you need an external heat sink.

Engineering is always a compromise. Write down three desired attributes and you can have two of those.

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