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Is it possible to create a powerful electromagnet at home? With use of a ferromagnet it seems so... Using the following formula: $B(Tesla)= k\mu_0nI$. I understand some ferromagnet's like iron could have permeability above 10,000? That would easily boost the field above a 1 Telsa?

Relative permeability of raw iron is: 200,000! There must be something wrong here, see this wikipedia table.

UPDATE: How much input power might be needed?

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    $\begingroup$ Magnetic fields well above 1T are possible. See magnet.fsu.edu/mediacenter/factsheets/records.html for some examples. $\endgroup$ – John Rennie Dec 18 '13 at 12:06
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    $\begingroup$ I mean a "home made" one. $\endgroup$ – Pupil Dec 18 '13 at 12:08
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    $\begingroup$ see physicsforums.com/showthread.php?t=297170 - it can be done though it's hard. $\endgroup$ – John Rennie Dec 18 '13 at 12:17
  • $\begingroup$ How much power is needed for such a concept... $\endgroup$ – Pupil Dec 19 '13 at 0:44
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    $\begingroup$ In addition to being hard it poses certain risks. There is a lot of energy in the field and very high forces on the materials involved. Make a mistake engineering this beast and it can fail explosively. Don't take it lightly. $\endgroup$ – dmckee Dec 19 '13 at 4:26
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I oppose the comments which say that it is pretty risky and/or dangerous as it involves high energy fields simply because it does not !

To make a powerful solenoid :
Let's see : { $ B =k \mu_0 n I$} Increasing magnetic field strength :
1. $n$ is no of turns per unit length (m in SI), take sufficiently thin wire and you can have this above 1000 !
2. $k$ take a good core and you can have this above 1000 ! as well.

Let's assume both $k$ and $n$ are above 1000 this would give us :

$ B = 0.4 \pi I $
$ B = 1.25 I $

Lets say you took awg 38 wire, it has resistance per unit length as $2.19 \Omega/m$ so net resistance of your device will be {$ R = 2\pi r n 2.19$} here $r$ is radius of loops you are making for solenoid.

Assuming you have a nail of material with $k > 1000$ and of cross-sectional dimater $5mm$ and you carefully wind awg 38 wire on it, your loop's radius will be $ 2.5 × 10^{-3} m$ : 1. Rssistance of device : $ R = 34.4 \Omega$.
2. Required current : $ I = B/1.25A $.
3. Required voltage : $ V = IR = 27.52 B$.
4. Required power : $ P = 22 B^2 J$

All of the above are easily achievable in the comfort of your home, BUT this field would be obtained only at the heart of your solenoid, although their will be appreciable field near the end points, the field would reduce by {$10^{-3} $} outside. Also if you operate it using either AC/DC the wires would be extremely hot and untouchable, if you try to increase the field by increasing applied voltage it may even burn down.

PS : check $Ahr$ of battery it will give you an estimate of how long your device will run before draining your battery of everything, I suppose it would be short.

This was also made In home : enter image description here

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  • $\begingroup$ Thank you for that input, really interesting. But you stated the field would be in the heart of the solenoid, could you elaborate more about that point? I assume in order to use such a field the gap should be near 0. $\endgroup$ – Pupil Jan 26 '14 at 21:07
  • $\begingroup$ The formula of resistance n =? $\endgroup$ – Pupil Jan 27 '14 at 0:41
  • $\begingroup$ In the formula for resistance "n" is the number of turns of wire you make. I mean that the used formula gives magnetic field right at the centre of solenoid, as we move away from the centre there is significant change in the field strength and when you move out, as medium changes field becomes negligible ! $\endgroup$ – Rijul Gupta Jan 27 '14 at 14:28
  • $\begingroup$ Ah, but this seems simple for a very small electromagnet, what I was planning for is a large one. Assuming the same kind of wire is used, with a 1000 turns an the cross sectional area = 0.196 m^2 since r = 0.25m. Using the same formula above Voltage would probably be above 1500V. The power needed for a smaller electromagnet is certainly low. $\endgroup$ – Pupil Jan 27 '14 at 19:23
  • $\begingroup$ You cant use the same wire for big magnets as you are planning as it will burn down with that current ! You will need thicker wire which will reduce $n$ as well as $R$ then you must calculate requirements, as for the voltage you can use a step up transformer step up an alternating current to desired voltage and then convert to direct current, it can still be achieved easily at home but it is certainly more risky, mainly because of high voltage. $\endgroup$ – Rijul Gupta Jan 27 '14 at 22:43
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The number you have found for relative permeability is fine if you embed all the magnet into a piece of iron, then you will have a quite high field but you won't be able to "feel" it as it just stays inside the metal.

What you are probably thinking to do is an iron core solenoid, where the magnetic field lines close into air. Then you need an effective magnetic permeability which depends also on the geometry of the core, and in general it's hard to go above one or two hundreds.


Take a look at this picture:

enter image description here

That's fine: you add an iron core and you get a multiplication factor $k$ into your formula. But $k \ll \mu_r\approx 1\cdot 10^6$ (the constant you red in wikipedia) because if you follow the path of a field line, you don't stay always in the iron, but also go into air.

To get the formula: $$B=\mu_0 \mu_r n I$$ and easily get a huge field you need to put iron even outside the coil in such a way that the whole path of the field line is in metal.

Another possibility is to shape the iron core like a C. Then the field will stay much longer in the iron and will have to travel just a small air gap, thus your $k$ will be enhanced.

This link also contains some calculations: http://sci-toys.com/scitoys/scitoys/magnets/calculating/calculating.html


It's not hard to compute the power requirement:

$$P = R I^2 = R \left(\frac{B}{k\mu_0 n}\right)^2$$

where $R$ is the resistance of the wire and can be estimated with: $$R=\frac{\rho L}{A}$$ where $\rho$ is the resistivity of the coils' material, L is the total length of the wire and A is its cross section.

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  • $\begingroup$ Could you explain the second paragraph a bit more please? So using that same iron above wont work? I could not find any Neodymium magnet on the market with a field of 1 Tesla most of them are below it. $\endgroup$ – Pupil Dec 18 '13 at 12:57
  • $\begingroup$ @Key The point is that you have part of the field in the iron, but part is in air (outside the solenoid), so you cannot consider only the iron permeability. That would work only if the solenoid is fully immersed into iron. I've found for sale online a cylindrical magnet: 4" dia. x 1" thick, NdFeB, Grade N52 which on the edges seems to achieve a field of 10663 gauss. $\endgroup$ – DarioP Dec 18 '13 at 13:34
  • $\begingroup$ I read your comment and post multiple times and still can't figure out what you mean... An electromagnet is based off the idea of a ferromagnetic core inside a solenoid. I'm adding permeability of air and the ferromagnet int the calculations ... An solenoid(without a core) generates a weak magnetic field... By adding a high permeability ferromagnet with a high saturation it could increase the total magnetic field by 100 or 1000 times. Have I understood you correctly? All of this is based on:B = ku(0)nI $\endgroup$ – Pupil Dec 19 '13 at 1:58
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I find this all very intriguing..

Making an electromagnet requires you understand the formula shown here hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html

You need to use thick wires and provide high amperage. Thicker the wire, lower the resistance and higher the current carrying capacity. You cannot keep on increasing the number of turns using small wires and give a low input. It would not work. It is not necessary to give high voltage to get a powerful electromagnet but low voltage and high amperage would do. The rods or core of the solenoid also carries the current and so when you do these experiments put a paper on the entire floor of the room and then wood and then place on the the solenoid in the horizontal position and then give current. use a fuze to control the input. I have built a number of electromagnets to study this in my office and if you stand on insulated material and the solenoid is also on Insulated material there is no risk. No water any where near the experimenting place. You must wear rubber shoes. Nothing more.

The iron will roar like a tiger if it is goes in to the saturation levels and I have reached many times very powerful and roaring electromagnets that are discernible about 7 feet from electromagnet core. You take a permanent magnet in your hand and it will oscillate and from that you can find out how long the field is effective.

Be careful in doing these things but it can be done at home. If you exceed a critical value the iron is supposed to reverse the hysterisis curve or demagnetise. About 50 volts and 16 amps input is more than sufficient to make a very powerful electromagnet.

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Let:

  • $R$: The radius of the coil, $h$ the height of the coil, $n$: spiral density, ie, the number of spirals per height.

  • $r$: The radius of the wire, $A$: The area of the cross section of the wire.

  • $L$: The total size of the wiring, $N$: The amount of spirals in the coil.

  • $\bar R$: The overall resistance of the coil, $\rho$: resistivity of the material of the wire.

  • $B$: The magnetic field (ofc), $k$: The relative magnetic permeability of the core.

  • $V$: The voltage difference across the ends of the coil, $I$: The current passing thru the coil

  • $P$: Dissipated power by the coil.

With that in mind, some formulas relate our quantities: $$ B = k\mu_0nI, \quad n = \frac{1}{2r}, \quad L = 2\pi Rnh, \quad A = \pi r^2, \quad P = VI, \quad V = \bar RI, \quad \bar R = \frac{\rho L}{A} $$

These formulas comes from physical laws, or simple geometry. Also, we can relate the amount of spirals $N$ by $N = nh$. Don't forget that we want to minimize dissipated power, and maximize the amount of magnetic field generated. With that in mind, we shall fix $B$, and find $P$. $$ P = VI = \bar RI^2 = \frac{\rho L}{A}\left(\frac{B}{k\mu_0 n}\right)^2 = \frac{\rho\cdot 2\pi Rnh}{\pi r^2}\left(\frac{B}{k\mu_0 n}\right)^2 = \frac{2\rho RhB^2}{nr^2k^2\mu_0^2} = \frac{4\rho RhB^2}{r k^2\mu_0^2} = \frac{8\rho RB^2}{k^2\mu_0^2}N $$

So, your goal is to maximize the most you can all variables in the denominator, and minimize the numerator, for a fixed $B$, in order to minimize the power you require to operate such a thing. I actually find quite interesting that $r$ should be maximized, in order to minimize $P$ for a fixed $B$. Intuition would say otherwise isn't it? You can do the same for voltage: $$ V = \bar R I = \frac{\rho L}{A}\frac{B}{k\mu_0 n} = \frac{\rho\cdot 2\pi Rnh}{\pi r^2}\frac{B}{k\mu_0 n} = \frac{2\rho RhB}{k\mu_0r^2} $$

And for last, to find the necessary current, its a direct relationship with the magnetic field: $$ I = \frac{B}{k\mu_0n} = \frac{2rB}{k\mu_0} $$

Hereby, our final results: $$ I = \frac{2rB}{k\mu_0},\quad\quad V = \frac{2\rho RhB}{k\mu_0r^2},\quad\quad P = \frac{4\rho RhB^2}{r k^2\mu_0^2} = \frac{8\rho RB^2}{k^2\mu_0^2}N $$

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