How to build a stonger Electro-Magnet

[DHS Science]

A friend of mine and I have been working on a science fair project but have run into a small problem. The Electro-Magnet that we have built is melting the wire that we have and we cannot buy any stronger wire because we do not have a place where we could buy any stonger wire. We were wondering if anyone had an idea about how we could get about how we could get the magnet under control or if we should use a small battery. If we were to use a weaker battery tho we need to have the magnetc field to have at least the same strength. Thanks in advance to everyone that posts a responce.

 

[hover]

I personally used a power supply with my electromagnet with resistors. Batteries will often die to easily. As for increasing the strength of your electromagnet, just either add more current (amps) or add more turns of your wire. Make sure you place an iron solenoid. A good way to greatly increase the strength of your magnet is to get a horseshoe shaped solenoid.

Hope this helps

 

[ranger]

DHS Science, what are you using as the core? You should use a material that is ferromagnetic, for example an iron core. The existing magnetic field (caused by the current in the solenoid) will align the magnetic domains of the iron core and thus the core will act like magnet increasing the strength of your overall magnetic field. Adding such a core could increase your magnetic field strength by hundreds or even thousands.

 

[wxrocks]

This link talks about the magnets used in the MINOS experiment...

http://www-numi.fnal.gov/Minos/info/tdr/mintdr_4.pdf

Maybe that will give you some ideas or at least contacts that might help you with your supply problem.

 

[silver-rose]

http://en.wikipedia.org/wiki/Mumetal <--- use Mu-metal as the core instead of iron? wouldn't that increase B field?

 

[DHS Science]

We are using a steel rod as a core and a 12 volt battery as our power supply at the moment but we would galdly take any suggestions on what we should use instead. If you could please try and explain what a "ferromagnet" is?

Thank for your help.

 

[hover]

you mean ferromagnetism? simple. All it is is an object that can let magnetic fields pass through it easily.

If you really want to know how strong your electromagnet is use http://sci-toys.com/scitoys/scitoys/magnets/calculating/calculating.html" to calculate the strength of an electromagnet. I use it sometimes. If you don't know how to calculate the strength of an electromagnet from this page then give me the specs of your magnet.
example-
amount of amps used
number of wire turns
length of electromagnet
and so on...

 

[DHS Science]

I will get those specs to you within the next couple of days because I will not be able to get into the school. But I have another question for you if that would be alright. I was wondering if what exactly you meant when you said that you used a "power supply." Did you mean that you used a power supply from a computer or not. Sorry about all of the questions.

 

[marcusl]

http://en.wikipedia.org/wiki/Mumetal <--- use Mu-metal as the core instead of iron? wouldn't that increase B field?

No, Mumetal saturates at lower applied fields ("H") than iron, and it is hideously expensive. Stick to plain iron for your magnet core, mild steel (i.e., not alloyed or heat treated, the type used in framing nails) is fine too.

 

[marcusl]

I will get those specs to you within the next couple of days because I will not be able to get into the school. But I have another question for you if that would be alright. I was wondering if what exactly you meant when you said that you used a "power supply." Did you mean that you used a power supply from a computer or not. Sorry about all of the questions.

Your school might have a variable power supply that they could loan you. You could then adjust the voltage down from 12V to where your coil doesn't overheat (this would reduce the field proportionally, however). This web page shows some typical power supplies so you know what they look like:
http://www.testextra.com/unisource_dc_power_supplies_selection_guide.htm"
You'll need a supply that can source enough current and voltage, which means computer supplies won't cut it. One of your science classrooms might have a "multimeter" that you can use to measure the resistance R of your coil. You can then figure out the current I using Ohm's law
I = V / R
where V is the battery or supply voltage.

 

[hover]

I will get those specs to you within the next couple of days because I will not be able to get into the school. But I have another question for you if that would be alright. I was wondering if what exactly you meant when you said that you used a "power supply." Did you mean that you used a power supply from a computer or not. Sorry about all of the questions.

What i ment was i used a power supply(cord) to an old laptop i had. It was rated around 19 volts and about 3 amps. I perfer this over batteries because short circuiting a battery will easily drain it. With a power supply you have constant power, which means your electromagnet never loses energy bacause the battery is weakening. You can probably get power cords where ever they sell computer supplies. Make sure you try stay low with voltage though.

I want to make myself clear MAKE SURE YOU BUY RESISTORS http://www.radioshack.com/product/i...2032058&s=A-StorePrice-RSK&parentPage=search". With out resistors you will short out the power supply which is not good. Radioshack has a good section for resistors. I recommend you stick with the 100 or 50 ohm resistor. In practice the resistors will get hot, this is normal. This depends on how much current and amps you put through. For the resistors on that page they can at max take 10 watts of power, any more or they will overheat. For example say that i am using the 50 ohm resistor and with 19volts. So to get current i use
V=IR
V= voltage
I= current(amps)
R= resistance
So for my example i put in 19volts=I*50ohms. Doing algebra I get .38 amps. After that i want the equation for watts. Which is
VI=W
V= voltage
I= current(amps)
W= watts
So 19volts*.38amps=7.22watts. In this example the resistor will get hot but it should be fine since the max for these resistors is 10 watts.

hope this helps

 

[DHS Science]

Will that resistor be able to run for 24 hours straight tho or will that be to long for it?

Thanks for all of the help and sorry again about all of the questions.

 

[ranger]

Will that resistor be able to run for 24 hours straight tho or will that be to long for it?

Thanks for all of the help and sorry again about all of the questions.

Yes it can. But just make sure that you purchase a resistor that meets the power requirements of your circuit. Use the formula that hover suggested to find power (voltage times current). For example, if you get a total power of 5W and you have resistor with a higher power rating, you should be good.

 

[hover]

Thanks for all of the help and sorry again about all of the questions.

Thats the point of physics forums, to ask questions . Without questions this site would be useless.

 

[DHS Science]

The only problem with using a resistor is that we cannot find one powerful enough for the application that we are using the magnet for. I have thought about using a breaker like what is in an electrical box but I am not for sure how much power the magnet will be pulling from the wall and I cannot afford to buy more supplies if the magnet does not work the first time.

 

[ranger]

The only problem with using a resistor is that we cannot find one powerful enough for the application that we are using the magnet for. I have thought about using a breaker like what is in an electrical box but I am not for sure how much power the magnet will be pulling from the wall and I cannot afford to buy more supplies if the magnet does not work the first time.

Wow! How much power would you want to dissipate across the resistor? Mayb you could use 2 resistors with each one having a resistance of \frac {R_{TOTAL}}{2}, in that way each resistor would have half the total voltage across it and would dissipate half the power as one resistor. You would now have the same resistance that you are aiming for and the same current.

Mayb you could provide use details has to what resistor value you plan on using and the voltage of the power supply. I'd advice not to take power directly from the wall outlet. Use a [step down] transformer or a power supply. Using a wall outlet directly will call for more power dissipation within the circuit

 

[hover]

Wow! How much power would you want to dissipate across the resistor? Mayb you could use 2 resistors with each one having a resistance of \frac {R_{TOTAL}}{2}, in that way each resistor would have half the total voltage across it and would dissipate half the power as one resistor. You would now have the same resistance that you are aiming for and the same current.

Mayb you could provide use details has to what resistor value you plan on using and the voltage of the power supply. I'd advice not to take power directly from the wall outlet. Use a [step down] transformer or a power supply. Using a wall outlet directly will call for more power dissipation within the circuit

This is certainly true. If you have 2 of the same resistors parralell in a circuit, it would decrease the amount of resistance by 2. Or if you really wanted you could add 3 or more in paralell, but remember that the amount of ohms of 1 resistor gets divided by how many resistors you have in parrallel. Example-

100 ohms/ 2 resistors in parallel = 50 ohms
100 ohms/4 resistors in parallel = 25 ohms

Keep in mind thought that the number of ohms change when finding how many watts you are putting in a resistor. Example-

50 ohms/ 3 resistors in parellel = 16.66 ohms... so if you used 19 volts/16.66 ohms, you will get 1.14amps...so 19 volts * 1.14 amps= 21.66 watts. For the resistors i listed before this is double the amount of watts they are supose to take and will greatly overheat!

 

[DHS Science]

The resistor that I am using is a 1-megaohm resistor that is 1/2 watt and has 5% tolerance. I have 5 of them in all, so would it be better to use like 2 or more of them or should I try and find one that can handle more watts and less ohms.

So if i were to use more than one of the resistors will that take the number of ohms down or will that just take the number of ohms and separate them evenly between the two resistors.

The school that I go to does not have a power supply or a step down transformer.

Also the power coming out of the wall has 15 amps so if that makes any difference.

 

[ranger]

The resistor that I am using is a 1-megaohm resistor that is 1/2 watt and has 5% tolerance. I have 5 of them in all, so would it be better to use like 2 or more of them or should I try and find one that can handle more watts and less ohms.

So if i were to use more than one of the resistors will that take the number of ohms down or will that just take the number of ohms and separate them evenly between the two resistors.

The school that I go to does not have a power supply or a step down transformer.

Also the power coming out of the wall has 15 amps so if that makes any difference.

How do you know how much current is coming out? The current you draw would depending on the resistance of the circuit. And 15A is alot! Unil you have decided what resistances you will have, only then can we know the power dissipation needed. These are values you should have before you purchase your components.

So since you are using the wall outlet, you already know your voltage. I'm assuming that you have a numerical quantity of the magnetic field which you wish to achieve?

It depends on how you connect the resistors, in series, the resistances would add. In parallel they would decrease (hover's example). Since in both of these cases, either the voltage drop or current of the resistor changes, the power dissipation could be "spread out" across the resistors.

If you use 2 of your 1 mega ohm resistors is series, you would have a total resistance of 2 mega ohms.
Knowing that the wall voltage is 120V, we get the total current in the circuit:
I = 120/2000000 = 60 micro amps

Each resistor would have 60V across it:
P = IV
P = 60 micro amps * 60 = 3.6 milli watts

Your resistors will handle the power dissipation, but your current is small!

 

[hover]

The resistor that I am using is a 1-megaohm resistor that is 1/2 watt and has 5% tolerance. I have 5 of them in all, so would it be better to use like 2 or more of them or should I try and find one that can handle more watts and less ohms.

So if i were to use more than one of the resistors will that take the number of ohms down or will that just take the number of ohms and separate them evenly between the two resistors.

The school that I go to does not have a power supply or a step down transformer.

Also the power coming out of the wall has 15 amps so if that makes any difference.

I have to agree with ranger. Even though those 1-megohm resistors are well able to handle the outlet, they barely give you any current and current is "the thing" that makes an electromagnet an electromagnet. You shouldn't use an outlet anyway because the voltage is so high. Even if you got lower ohm resistors it would be a pain finding something that could handle the watts. Search around for a power cord for a laptop or something like i did. Something else to look out for are resistors that have a high wattage rating.

 

[DHS Science]

I am using a power cord from a regular computer. But that does not allow the whole 120V thru it correct?

If I were to buy different resistors then I would want some that allow less ohms to run thru them?

I don't mean to ask stupid questions but I am really confused about some of this stuff because I have never really worked with this technical of stuff on this subject.

 

[ranger]

I am using a power cord from a regular computer. But that does not allow the whole 120V thru it correct?

If I were to buy different resistors then I would want some that allow less ohms to run thru them?

I don't mean to ask stupid questions but I am really confused about some of this stuff because I have never really worked with this technical of stuff on this subject.

If you are using a power supply from a computer, the output voltage could be 12V(?). if you plan on using 1 mega ohm resistors, do you get an idea of the quantity of current in your circuit?

Also I must ask, when you first built your circuit, what was your current through the circuit that caused the wires to melt?

 

[hover]

i have 2 power cords, one 19volts and the other 16volts. As for resistors get something like http://www.radioshack.com/product/i...2032058&s=A-StorePrice-RSK&parentPage=search". Resistors like that from radioshack are good for power supplies and come two in a pack. They are also less than $2, you could easily buy it.

 

[hover]

DHS Science i have a question for you, how strong are you trying to make your electromagnet? Are you just trying to run some current through a wire to create a simple electromagnet or are you trying to create the strongest electromagnet you can make??

 

[Averagesupernova]

All I can say is WOW. This should be moved to electrical engineering. There are some things in here that seriously don't make any sense. DHS, you haven't told us what gauge wire you are using in your electromagnet. You are using 1 megohm resistors with a 12 volt battery? There is NO WAY a one megohm series resistor can allow enough current through from a 12 volt battery to melt your wire. You'd have to have 10000 of them in parallel to get down to 100 ohms. There is no way that could even allow enough current through from a 12 volt battery to melt your wire. As for the 15 amps getting through what you guys are calling the 'power cord', well, that IS a power supply. It converts the 120 VAC down to a lower DC voltage. 19 and 16 volts was mentioned. If the thing has a rating of 3 amps, then that is all the current that you will get out of it. If the resistance of the coil calls for more than 3 amps at the power supply's rated voltage, then the power supply will not be able to handle it and its output voltage will drop.
-
Hover, you refer to 3 50 ohms resistors in parellel = 16.66 ohms with a total power dissipated in the resistors of 21.66 watts. This is NOT the power in each resistor. In this case each resistor will take one third of the total power of 21.66 watts. You are also assuming zero resistance of the electromagnet coil. We don't know what this resistance is since DHS hasn't told us nor has he told us what kind of wire is being used.
-
DHS, I'm not sure what kind of wire you are using, but try installing some car headlights in series with your coil. Older sealed beam bulbs are cheap and the ones that have 2 filaments in them can be wired so that both filaments are in series, both in parallel, or just one at a time to get close to the desired current. Obviously a car headlight is able to take the full 12 volts so you can't hurt them. They are simply there to limit the current in your coil so you don't destroy it.

 

[hover]

Hover, you refer to 3 50 ohms resistors in parellel = 16.66 ohms with a total power dissipated in the resistors of 21.66 watts. This is NOT the power in each resistor. In this case each resistor will take one third of the total power of 21.66 watts. You are also assuming zero resistance of the electromagnet coil. We don't know what this resistance is since DHS hasn't told us nor has he told us what kind of wire is being used.
-.

I am only trying to help. Plus i DON'T know what wire gauge he is using so i just avoided it using the resistance in the wire. The only reason i am even helping is because i did a electromagnet project by myself once and i thought i could give some advice.

Averagesupernova i quote you "Hover, you refer to 3 50 ohms resistors in parellel = 16.66 ohms with a total power dissipated in the resistors of 21.66 watts. This is NOT the power in each resistor.". Are you saying that in order to get the amount of watts in each resistor i do 21.66/3??

 

[Averagesupernova]

I am saying that the power in a resistor is equal to the voltage across it squared and then divided by its resistance.

 

[hover]

That way or my way works out exactly although your way is a shorter version. You seemed to miss the point of what i just asked. I said " Are you saying that in order to get the amount of watts in each resistor i do 21.66/3??". I reread your previous post and got this "In this case each resistor will take one third of the total power of 21.66 watts." so i guess that anwsers my question.

 

[Averagesupernova]

I said "in this case" because it only pertains to this case. The safest most sure way of getting the correct answer is E^2/R.

 

[hover]

Just to be clear, say you had 5 resistors i parrallel. Say the amount of power(watts) in the resistors is "X". Would i divide "X" by 5 to get the amount of power in each resistor??

 

[Averagesupernova]

You would only if the 5 resistors are of the same ohm value.

 

[hover]

ok i see now.

Thx

 

[Averagesupernova]

Let me guess. You did some more math?

 

[hover]

How did you know ? I never put three 50 ohm resistors in series on my electromagnet before because i thought they would all take 21.66 watts of power and overheat.

 

[DHS Science]

Hover:
"DHS Science i have a question for you, how strong are you trying to make your electromagnet? Are you just trying to run some current through a wire to create a simple electromagnet or are you trying to create the strongest electromagnet you can make??"

I am trying to make the stongest that i can with the supplies that i have.

Averagesupernova:
DHS, you haven't told us what gauge wire you are using in your electromagnet. You are using 1 megohm resistors with a 12 volt battery? There is NO WAY a one megohm series resistor can allow enough current through from a 12 volt battery to melt your wire.

Sorry about not saying the size of wire, it is 6 gauge. And the reason that the batter melt thru the wire last year was because i never used a resistor then. I just ran a wire straight off of the battery thru the coil. That is the main reason why i have so many questions. Sorry.

Last night i went out and bought a pack of 2 resistors that allow 100 ohms and 10W. So if hook both of them up parallel in the line then i should be fine? Because if i do then the power will be divided evenly between the two resistors and will still only allow 10W thru in all, correct?

 

[NoTime]

Last night i went out and bought a pack of 2 resistors that allow 100 ohms and 10W. So if hook both of them up parallel in the line then i should be fine? Because if i do then the power will be divided evenly between the two resistors and will still only allow 10W thru in all, correct?

With six gauge wire the resistance of the coil will be close to 0.
Power in resistor is E squared times R.
In this case 12^2 /100 = 1.4 watts.

The resistor does not limit watts.
It only provices resistance.

#6 wire should be able to handle about 30A to 40A without heating up too much.

Current in the coil is E/R.
Or in this case 12/50 = 0.24A

You may want to use smaller resistance value.
For 10W resistors 30 ohms is about as low as you want to go with a 12v source.
This provides a safety factor since the 10W rating is only for ideal conditions.

 

[Averagesupernova]

Go here: http://www.solenoidcity.com/

Lots of stuff to learn about winding coils here.

 

[DHS Science]

If we were to use a power cord from a computer then how much power (amps, volts, watts) would that allow to enter the magnet if we were getting our power straight out of the wall?

Sorry if I am asking stupid questions but I would really like it if we would not burn the school down.

 

[hover]

If we were to use a power cord from a computer then how much power (amps, volts, watts) would that allow to enter the magnet if we were getting our power straight out of the wall?

Sorry if I am asking stupid questions but I would really like it if we would not burn the school down.

Are you saying how much power would you get out of the wall if you plugged in a power cord and used that?? That all depends on the rating of voltage from the power cord. It should be listed right on it.

 

[DHS Science]

The power cord does not give an answer directly if just gives (what appears to me at least) a random configuration of numbers and letters.

 

[Averagesupernova]

Why the heck is this still in this section of the forums? Someone move this to electrical engineering.

 

[hover]

The power cord does not give an answer directly if just gives (what appears to me at least) a random configuration of numbers and letters.

It should be right on it somewhere. If you can't find it get a volt meter and measure it that way.

 

[ranger]

Why the heck is this still in this section of the forums? Someone move this to electrical engineering.

You know there is a feature called "Report Post" on the lower left section of a post (exclamation mark). Just report the thread and request that it be moved.

 

[DHS Science]

We decided to use a car battery so it is pushing out 12 volts and the wire that we are using is still getting really hot. In order for the coil to maintain the same strength we hooked a battery charger up to it, which did the job but should we put a breaker (like from a fuse box) in it to help maintain the amount of amps?

 

[ranger]

We decided to use a car battery so it is pushing out 12 volts and the wire that we are using is still getting really hot. In order for the coil to maintain the same strength we hooked a battery charger up to it, which did the job but should we put a breaker (like from a fuse box) in it to help maintain the amount of amps?

But to use a fuse, you would need to know how much current the wire can handle before it becomes a danger. The amount of amps depend on the wire gauge. There is no use in getting a fuse that's rated for 7 amps when your wire can handle more.

 

[DHS Science]

well we have 6 gauge wire and i was going to put a 15 amp fuse in the circut but i am not sure if that is still to many amps

 

[ranger]

There are many tables on the internet that will tell you how many amps your wire can handle. Look it up and then you will be able to choose the appropriate fuse rating.

 

[Averagesupernova]

There is significantly more to it than just sizing the fuse to the amount of current that the wire will handle. If you wind several hundred turns of #6 onto a large rod and hook it up to a 12 volt battery with a fuse that is sized at the maximum safe current allowed you will most likely blow the fuse right away. The fuse does not 'regulate' the amount of current allowed to flow. It will allow any current to pass up to a given point and then open up preventing the current from flowing. You need to wind enough wire to get enough resistance so that the natural resistance in the wire will not allow more than X current to pass at a given voltage. Naturally this causes problems because to get enough wire wound to get to this resistance we may have a coil with a 5 foot diameter. The link I gave earlier in this thread should explain it well.
-
http://www.solenoidcity.com/electromagnet/E-28-150p1.htm
This link goes to one of their products which shows wire size related to reisistance of the windings related to the duty cycle at various voltages as well as a few other things.

 

[kvmadan]

i went through the discussion...i'm making an electromagnet with a cylindrical iron core(2.5 cm dia, 8 cm height)... i used a 12v 1.3ah battery and it got drained in no time ! 20 seconds i guess... I'm thinking of buying a 12v 10a battery ... is it necessary? i want the electromagnet to work for 30 mins. what resistor should i use ... will adding bridge rectifiers help ??

 

[ChummyJigger]

Hello fellow hobbyist,
I was wondering if using an actual magnet (cylindrically shaped ) as the core itself in an electromagnet would make a stronger magnet, make no difference, or make it weaker...thanks - I look forward to hearing from you guys...