How to build a stonger Electro-Magnet

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In summary, two students are working on a science fair project involving an electromagnet, but they are facing issues with the wire melting and not being able to find stronger wire. They are seeking advice on how to control the magnet or if they should use a different power source. Suggestions are given to use a ferromagnetic core and a power supply rather than batteries. The students are advised to use resistors to prevent short circuits.
  • #1
DHS Science
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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.
 
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  • #2
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
 
  • #3
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.
 
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  • #6
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.
 
  • #7
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" [Broken]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...
 
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  • #8
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.
 
  • #9
silver-rose said:
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.
 
  • #10
DHS Science said:
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" [Broken]
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.
 
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  • #11
DHS Science said:
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:biggrin:
 
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  • #12
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.
 
  • #13
DHS Science said:
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.
 
  • #14
DHS Science said:
Thanks for all of the help and sorry again about all of the questions.

Thats the point of physics forums, to ask questions:rolleyes: . Without questions this site would be useless.
 
  • #15
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.
 
  • #16
DHS Science said:
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 [itex]\frac {R_{TOTAL}}{2}[/itex], 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
 
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  • #17
ranger said:
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 [itex]\frac {R_{TOTAL}}{2}[/itex], 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!
 
  • #18
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.
 
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  • #19
DHS Science said:
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!
 
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  • #20
DHS Science said:
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.
 
  • #21
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.
 
  • #22
DHS Science said:
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?
 
  • #24
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??
 
  • #25
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.
 
  • #26
Averagesupernova said:
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??
 
  • #27
I am saying that the power in a resistor is equal to the voltage across it squared and then divided by its resistance.
 
  • #28
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.
 
  • #29
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.
 
  • #30
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??
 
  • #31
You would only if the 5 resistors are of the same ohm value.
 
  • #32
ok i see now.

Thx:smile:
 
  • #33
Let me guess. You did some more math?
 
  • #34
How did you know:tongue2: ? 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.
 
  • #35
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?
 
<h2>1. How does an electro-magnet work?</h2><p>An electro-magnet works by using an electric current to create a magnetic field. When an electric current flows through a wire, it creates a circular magnetic field around the wire. By coiling the wire and increasing the electric current, the magnetic field becomes stronger, creating a stronger electro-magnet.</p><h2>2. What materials are needed to build a strong electro-magnet?</h2><p>The main materials needed to build a strong electro-magnet are a power source, a coil of wire, and a ferromagnetic core. The power source can be a battery or a power supply, the coil of wire should be made of a conductive material such as copper, and the ferromagnetic core can be made of iron, steel, or other magnetic materials.</p><h2>3. How can I increase the strength of an electro-magnet?</h2><p>To increase the strength of an electro-magnet, you can increase the number of coils in the wire, increase the electric current flowing through the wire, or use a stronger ferromagnetic core. Additionally, using a larger diameter wire and tightly wrapping the coils can also help increase the strength of the electro-magnet.</p><h2>4. What is the difference between a permanent magnet and an electro-magnet?</h2><p>A permanent magnet is a magnet that retains its magnetic properties without the need for an external electric current. On the other hand, an electro-magnet only has magnetic properties when an electric current is flowing through it. This allows for greater control and adjustability of the strength of an electro-magnet compared to a permanent magnet.</p><h2>5. Can I control the strength of an electro-magnet?</h2><p>Yes, the strength of an electro-magnet can be controlled by adjusting the electric current flowing through the wire. By using a variable power supply or a rheostat, you can increase or decrease the strength of the electro-magnet as needed. This makes electro-magnets useful in a variety of applications where adjustable magnetic strength is required.</p>

1. How does an electro-magnet work?

An electro-magnet works by using an electric current to create a magnetic field. When an electric current flows through a wire, it creates a circular magnetic field around the wire. By coiling the wire and increasing the electric current, the magnetic field becomes stronger, creating a stronger electro-magnet.

2. What materials are needed to build a strong electro-magnet?

The main materials needed to build a strong electro-magnet are a power source, a coil of wire, and a ferromagnetic core. The power source can be a battery or a power supply, the coil of wire should be made of a conductive material such as copper, and the ferromagnetic core can be made of iron, steel, or other magnetic materials.

3. How can I increase the strength of an electro-magnet?

To increase the strength of an electro-magnet, you can increase the number of coils in the wire, increase the electric current flowing through the wire, or use a stronger ferromagnetic core. Additionally, using a larger diameter wire and tightly wrapping the coils can also help increase the strength of the electro-magnet.

4. What is the difference between a permanent magnet and an electro-magnet?

A permanent magnet is a magnet that retains its magnetic properties without the need for an external electric current. On the other hand, an electro-magnet only has magnetic properties when an electric current is flowing through it. This allows for greater control and adjustability of the strength of an electro-magnet compared to a permanent magnet.

5. Can I control the strength of an electro-magnet?

Yes, the strength of an electro-magnet can be controlled by adjusting the electric current flowing through the wire. By using a variable power supply or a rheostat, you can increase or decrease the strength of the electro-magnet as needed. This makes electro-magnets useful in a variety of applications where adjustable magnetic strength is required.

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