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DESCRIPTION: Hi there, I don't know lot about electricity and I`m having problems with my first solenoid-prototype, I tried to build it watching you tube videos. I have attached the pictures of the wire that I used (diam=1.32mm), and the solenoid that I wired. The core is iron pipe (inner diameter= 1.5cm, outer= 2cm), you could see the size from the picture. There are 500 turns around the core, each turn is about 6.75cm + 50cm additional wire from each side, so the total length of the wire should be 6.75*500+50*2=34.75 meters. I used 1.5v batteries (1st test with 1 battery, 2nd test with 2 of them, so 3v). I also found from the internet that the resistance for this diameter of wire is 12.597 milli-ohms per meter.

PROBLEM: I want it to be a strong electro-magnet. But it doesn't stick any metal at all. Please tell me if the design is wrong, or i wired the circuit wrong, or volts are too little. How much volts do I need to make it work best?

wire diameter = 1.32mm

testing solenoid with 1.5v D-size battery

iron-core solenoid view size

iron-core solenoid up view

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    \$\begingroup\$ Did you strip the enamel off the ends where you're connecting the battery? If not, the wires are insulated and no current will flow. \$\endgroup\$
    – John U
    Mar 25, 2014 at 19:38
  • \$\begingroup\$ John, I have checked, the current flow. Both sides are stripped with sand paper \$\endgroup\$
    – Hasan A.
    Mar 25, 2014 at 19:48
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    \$\begingroup\$ How much current is flowing through your solenoid? For a perfect 1.5V source I calculate the max current you can get is ~3.43A, but batteries don't like high current draws so your actual current could be significantly lower. \$\endgroup\$
    – helloworld922
    Mar 25, 2014 at 19:52
  • \$\begingroup\$ Are those scissors actually ferrous? Many scissors are stainless-steel or aluminium, neither of which are magnetic. (Before someone else says it, yes some stainless is magnetic, but sod's law dictates it won't be when you need it) \$\endgroup\$
    – John U
    Mar 25, 2014 at 20:03
  • \$\begingroup\$ John, the scissors and other stuff that I tried to stick are all ferrous, they stick to the ones that are on the fridge :D \$\endgroup\$
    – Hasan A.
    Mar 25, 2014 at 20:11

3 Answers 3

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That wire is capable of carrying 3A or so, depending on cooling.

Given the length, and your resistance per meter, the total coil is probably around 0.4 ohms.

Plugging that into Ohm's law, I=V/R, which is V=IR, you find the voltage required is V=3*0.4=1.2.

You need a power supply capable of 1.2V at 3A to get as much power from that solenoid as you can reasonably expect.

The battery you have is a Panasonic Hyper Manganese D size cell. Sadly they don't provide detailed information on discharge rates, but looking at a similar competitor's datasheet you'll find the battery isn't rated above 500mA of discharge current.

However, a careful check of the datasheet will show that the battery has an internal resistance of 150 to 300milliohms. This is pretty close to your coil's resistance, meaning that you're wasting a third your energy inside the battery. If the battery were capable of delivering 2-3A you might be able to get a decently strong magnetic field from the solenoid.

But it's not. You'll need to consider using a power supply rated for the current and voltage you need, or several D cells in parallel to supply the current you need.

If you put 7 D cells in parallel, each would still be discharging at a fast rate, and wouldn't last long, but you will see a much more noticeable magnetic field from your solenoid.

Lastly, note that the magnetic field will appear to emanate from the ends, not the side, of the pipe, and is strongest closest to the coil. So the open end of the pipe closest to the coil will appear to have a stronger magnetic force than the other end, or the sides of the pipe.

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  • \$\begingroup\$ Thanks for these tips, I`ll try them tomorrow, and if successful post it here \$\endgroup\$
    – Hasan A.
    Mar 25, 2014 at 20:22
  • \$\begingroup\$ The battery is capable of high current discharge, it's life just suffers. A fresh D drained at 1 Amp would bring the battery down to 1.2V in 1 hour. You can get 1+ amp from regular Alkaline 1.5v cells no problem. It's maintaining that draw for a useful amount of time that's the problem. \$\endgroup\$
    – Passerby
    Mar 26, 2014 at 2:28
  • \$\begingroup\$ @Passerby go ahead and measure the voltage on a D cell under a one amp draw. Due to internal resistance you'll find the voltage is much lower than 1.5v while the current draw is that high. Then do the same with several cells in parallel. You should find that a single D cell will not supply 1.2V at 3A into a 0.400ohm load for any useful length of time. \$\endgroup\$
    – Adam Davis
    Mar 26, 2014 at 3:22
  • \$\begingroup\$ Speaking of alkaline cells, anyway. Nicd, nimh, and other cell chemistries have much lower internal resistance and can supply much more power at higher current draws than average alkaline cells. \$\endgroup\$
    – Adam Davis
    Mar 26, 2014 at 3:25
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There are a few things I can see that might be the problem:

The magnetic field is strongest "inside" the solenoid, and it quickly fall off outside the solenoid. hyperphysics image of solenoid magnetic field: enter image description here

However, this image ignores the fact that your steel core extends beyond the wire turns. Good conductors tend to prevent the "diffusion" of magnetic fields into them, thus the theoretical strongest magnetic field locations outside the solenoid is at the two ends of the steel cylinder.

The second issue is with how weak this electromagnet is.

Using this solenoid calculator, I found that for ~3.43A through your solenoid (theoretical current for a perfect 1.5V source and your given dimensions) the magnetic field strength was 35.9 mT. If I took into account the internal resistance of the battery (~0.15 ohms), the current reduces to 2.55A, with a corresponding magnetic field of 26.7 mT. This is only a few times stronger than a fridge magnet. Note that this is the magnetic field inside the solenoid. The magnetic field outside will be weaker.

Another caveat is I calculated these magnetic fields assuming the steel core was the same size as the portion wrapped with wires. It's not. It isn't immediately clear to me which length to use for calculating the magnetic field strength, So I simply used the length of the wrapped portion. The true length will probably be somewhere in between, so the magnetic field strength will be lower still.

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  • \$\begingroup\$ I'm glad you addressed this from the perspective of the magnetic field this device can generate. Thanks! \$\endgroup\$
    – Adam Davis
    Mar 25, 2014 at 20:22
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    \$\begingroup\$ You can't really use that calculator determine the solenoid magnetic field because that calculator assumes a core that is all the same material. In this case, the core is a hollow tube, so the core is made of iron and air. Most of the flux is going to go through the iron pipe, not the air inside the pipe. The magnetic field inside the iron pipe is going to be close to 0. I did a quick simulation in FEMM and if I use 2.55A, I get a max mag field of ~.4T and a mag field of ~.25T in the area where the scissors is located in the picture. But the actual current will be closer to .5 A as Adam said. \$\endgroup\$
    – Eric
    Mar 25, 2014 at 21:46
  • \$\begingroup\$ @helloworld922 Failed projects it’s an excellent opportunity to test knowledge and theory. To summarize: 1_Personally, never calculate the magnetic field outside of a solenoid. Usually examine the plunger movement inside the solenoid and the relevant force. 2_Almost never extend the core of an electromagnet out of coil former. 3_The only electromagnet with hollow core I have construct, is one to magnetize tools, but not to lift iron pieces. Solid core electromagnets design is a strait foreword process. It is a good example to post it at “featured questions” with a genereus points. \$\endgroup\$
    – GR Tech
    Mar 26, 2014 at 21:46
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The magnetic attraction force from an electromagnet is calculated as: -

Force = \$(N\cdot I)^2\cdot\dfrac{\mu_0\cdot A}{2\cdot g^2}\$ where

  • N is number of turns
  • I is current
  • \$\mu_0\$ is permeability of free-space = 4\$\pi\times 10^{-7}\$
  • A is cross sectional area
  • g is gap from end of solenoid to piece you want to attract with force

With 2A, 500 turns, a 1" diameter solenoid and a 0.5" gap, the force is 0.08 newtons.

See this calculator

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