Talk:Electromagnet

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I changed the part about the left-hand rule giving the direction of the magnetic field inside a solenoid - definitely not the case! Magnetic flux inside a solenoid follows the right-hand rule. (I'm sure one or two people got one or two answers wrong in their test!) —The preceding

Try using other sources. --Smack (talk) 22:26, 12 May 2005 (UTC)[reply]

This attitude is the very reason why Wikipedia is sooo useless. That is why people don't donate: we can't use it.
Here, it doesn't show how to calculate the force excerted on a piece of iron by the magnetic force at 'd' distance from the coil. All academic bull, no practical use. — Preceding unsigned comment added by 115.70.177.64 (talk) 13:53, 16 July 2013 (UTC)[reply]

If someone could come up with some drawings and images, I can do the math. Something simple, like a U-bolt from a hardware store would be nice. Or simpler yet, a square core with one end cut off. Please don't include the standard nail example -- it's way too hard to analyze in closed form. Here are a couple of example from Edmunds Scientific. Madhu 00:13, 31 October 2005 (UTC)[reply]

If you're talking about a closed magnetic circuit, such as an electromagnet of wire wound on a U-shaped bolt picking up a steel plate bridged across its poles, eq. (1) and (2) in the article can be used. You can calculate the magnetic field B from the number of turns and the current using eq. (1) if you measure the length of the looping magnetic path L through the bolt and the plate, provided you can find the magnetic permeability μ of the steel. Then eq. (2) will give you the force exerted by each pole face on the steel plate, from the area A of the pole face. The main problem will be finding the permeability of the steel. I'd just go on the Internet and use a value for mild steel. It won't be very accurate, but you will be within a factor of two. --Chetvorno^TALK 14:32, 16 July 2013 (UTC)[reply]

I think an intereseting application of electro-magnets is in its use as a weapon, namely rail guns.

Yeah, the incredibly high magnetic field needed to make a heavy object shoot out of the electromagnet will leave tumours in your brain, apart from killing another person with the weapon, but why use springs or air pressure or maybe even gunpowder in your gun for a better, cheaper, safer result, not to mention the exclusion of the necessity to plug your "weapon" into a wall socket. How about junkyards? They seem to lift pretty heavy objects with those electromagnetic cranes. Slartibartfast1992 23:51, 13 March 2007 (UTC)[reply]

Another place of use in the ag industry are in electric clutches so pumps or shafts can be turned on and off immediately with the flick of a switch. — Preceding unsigned comment added by Kenton007 (talk • contribs) 13:49, 5 May 2013 (UTC)[reply]

who made this page

Lots of people. See the history page. --Smack (talk) 05:11, 6 October 2005 (UTC)[reply]

The force per unit area is given in lbft/in², when all the quantities that go into the calculation are in SI. If I do the calculation in SI units, the F/A comes out to 398 kN/m² for the 1 Tesla case.

Does anyone else agree that we should have consistent units all the way through those calculations? --Syrthiss 02:58, 28 October 2005 (UTC)[reply]

Sure, include the SI numbers. The only reason I used mixed units is that most people don't have a feel for what a kN/m^2 (kPa) is. All I ask is that you keep the mixed unit numbers since it's easy to get your head around PSI. Think of it this way: I need to lift 50 pounds, how big of a magnet do I need? 1 Tesla is a reasonable number for flux density, so a square inch is in the ballpark. In SI, 50 pounds (force) is about 200 N, so you would need some very small area in m^2. Square centimeters would be easier, a centimeter is easy to guestimate. I like to run numbers in my head or the back of an envelope... Madhu 00:05, 31 October 2005 (UTC)[reply]

I think we should insist on SI. Add American units in parentheses if you want, but it's counterproductive to perpetuate them. --Smack (talk) 22:43, 11 November 2005 (UTC)[reply]

WHERE is the calculation of force exerted on ferromagnetic magetic material across an air gap???? I've looked everywhere and I can't find it.

Thecoolsundar (talk) 07:28, 15 August 2010 (UTC)[reply]

Another question

Come to think of it, what is this section doing in an article on electromagnetism? Shouldn't it be somewhere else, like ferromagnetism? --Smack (talk) 04:59, 16 November 2005 (UTC)[reply]

Well, I think he was using it as a basis for how to calculate the turnings and such an electromagnet would need to develop a field of XX strength. So, yes it could be used in Ferromagnetism but also seems applicable here. That reminds me though, I need to put in the SI calculations. Then we can still consider whether to move it or leave it here. --Syrthiss 05:32, 16 November 2005 (UTC)[reply]

Done. I just looove Mathcad. It has to be one of the greatest software applications ever developed. Madhu 01:48, 17 November 2005 (UTC)[reply]

Thanks :) --Syrthiss 04:44, 17 November 2005 (UTC)[reply]

could you tell me more about magnetic calculations?thanking you,[email protected]

Sure, it's simple. The fundamental definition of energy is force times distance. The energy in a magnetic field can be derived from first principles, you can find that in any electromagnetics text. Fields and Waves in Communication Electronics by Ramo, Whinnery, and Van Duzer is an excellent book. If you have a magnetic field of a known field strength, you can calculate the energy of that field. If anything is added to that field that changes the field strength, the energy changes. To change the energy requires a force, which is energy divided by change in distance (or dE/dx to be more accurate). I assumed the field was in air, but it could be in any medium. The derivation is in Magnetic_field Madhu 17:25, 18 November 2005 (UTC)[reply]

I need to create a electrical magnet that can preform the operations od a linear actuator EG propel the inner rod with reasonable force but not exceeding the stroke. I do not waht the rod to fly out im trying to achieve a popper like motion when a current is applied can anyone help?

The article should include something about remanent magnetism.

What happened to the cool image of an industrial lifting electromagnet? Why was that removed? Madhu 20:50, 11 August 2006 (UTC)[reply]

You say there is no point in using more than 787 A-turns per metre (not sure exactly where this figure comes from), because of saturation of the electromagnet core. However this only happens when the air gap of the magnet is closed. If you want to project a large field into an air gap, rather than just hold onto a chunk of iron, then I think you can usefully use a far larger mmf per length of core because the dominant reluctance of the air gap lowers the H field.

Simon80.41.41.166 23:37, 31 August 2006 (UTC)[reply]

That value is rougly the saturation field intensity for steel. You are correct about projecting large fields into an air gap. Feel free to adjust the wording as appropriate. Madhu 03:08, 1 September 2006 (UTC)[reply]

Done. --Chetvorno^TALK 19:30, 27 September 2008 (UTC)[reply]

Somebody Please include the description on force calculation between the electromagnets ( attractive or repulsive ).

vijay anand

WP:SOFIXIT. --Slartibartfast1992 19:26, 11 February 2008 (UTC)[reply

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Does anyone know how one could block the field in one direction and/or amplify the field in the other direction? 71.185.202.190 (talk) 22:19, 1 May 2008 (UTC)[reply]

The Introduction section says Much stronger magnetic fields can be produced if a "core" of paramagnetic or ferromagnetic material (commonly soft iron) is placed inside the coil. Paramagnetic? I've never heard of a paramagnetic material being used for the core of an electromagnet, at least not to increase it's permeability. Yes, a paramagnetic material would produce a slightly greater H than no core, but the increase is of the order 10^-5, totally negligible. Even if paramagnetic cores have been used for some specialized research purposes, I think it is misleading to list it next to and on an equal basis with 'ferromagnetic'. All ordinary electromagnets use ferromagnetic cores. Unless there's a good reason for leaving it in, I'm going to delete it. --Chetvorno^TALK 05:52, 9 August 2008 (UTC)[reply]

Done --Chetvorno^TALK 23:33, 12 August 2008 (UTC)[reply]

I am doing a year 13 (grade 12) physics paper, and must cite where i get any outside equations. Rather than cite wikipedia, does anyone have specific links to where these equations are derived or stated? Especialy the ones for a closed magnetic cercuit and force between electromagnets. - Josh, UK. —Preceding unsigned comment added by 90.219.63.103 (talk) 17:59, 3 November 2008 (UTC)[reply]

Where? Air gap between turns? Air gap between winding and core? Air gap between sections of core? The latter makes sense from the discussion but isnt at first obvious. All terms should be clarified with a diagram or description, as one does with an equation. Small ambiguities can lead to hours of futility for a practical desighn engineer where accurate information has to be applied to create working systems. —Preceding unsigned comment added by Q5101997 (talk • contribs) 02:48, 17 December 2008 (UTC)[reply]

Thanks, you're right, that's unclear. Its the air gap between sections of the core. I was going to include a diagram of an electromagnet illustrating air gaps, fringing fields, and leakage flux, but didn't get around to it. --Chetvorno^TALK 03:31, 17 December 2008 (UTC)[reply]

Added diagram clarifying the above terms. --Chetvorno^TALK 15:01, 6 October 2010 (UTC)[reply]

Your article was very informative and accurate. The use of diagrams and explanations of terms is imperitive. Contininuue to elaborate on this and it will be a fine and usefull article. 124.177.127.47 (talk) 10:56, 22 December 2008 (UTC)[reply]

hi —Preceding unsigned comment added by 99.251.238.177 (talk) 14:24, 6 February 2010 (UTC)[reply]

A 15:59, May 14 edit by 94.237.95.24 removed the discussion of minimum heat dissipation in the "Ohmic heating" section with this comment: "It was forgotten that when N increases -> also R increases due to increased length. Further analysis based on this misconception all removed." I reverted this removal; the original content was correct. Below is a derivation of the disputed point: that the ohmic heat losses in an electromagnet can be reduced by increasing the number of turns and proportionately decreasing the current, but that there is a minimum imposed by the area available for the windings. This is shown in many texts on electric machinery; I'll try to find a source.

The total resistance R of a DC electromagnet's windings is given by:

${\displaystyle R={\frac {\rho L}{a}}={\frac {\rho Nl}{a}}}$

where ${\displaystyle \rho }$ is the resistivity of the wire, ${\displaystyle L}$ is the total length of the winding, ${\displaystyle l}$ is the average length of one turn, ${\displaystyle N}$ is the number of turns, and ${\displaystyle a}$ is the cross-sectional area of the wire. So the total power dissipation ${\displaystyle P}$ due to ohmic heating is:

${\displaystyle P=I^{2}R=I^{2}{\frac {\rho Nl}{a}}\qquad \qquad (1)}$

As discussed in the article, the magnetic field produced by the magnet is a function of the magnetomotive force ${\displaystyle {\mathfrak {F}}=NI}$ so ${\displaystyle N}$ and ${\displaystyle I}$ can be varied to minimize the power dissipation as long as their product is constant. Substituting for ${\displaystyle I}$ into (1):

${\displaystyle P=\left({\frac {\mathfrak {F}}{N}}\right)^{2}{\frac {\rho Nl}{a}}={\frac {{\mathfrak {F}}^{2}\rho l}{Na}}\qquad \qquad (2)}$

So the power dissipation is inversely proportional to ${\displaystyle N}$. Thus the heat losses can be minimized by increasing the number of turns, while decreasing the current proportionally to keep ${\displaystyle {\mathfrak {F}}}$ constant. It would seem that the dissipation could by reduced indefinitely this way. However the product ${\displaystyle Na}$ in the denominator is equal to the cross-sectional area of the winding (ignoring insulation thickness and wasted space). It increases with ${\displaystyle N}$. There is some limit to the area available between the arms of the core for the winding, call it ${\displaystyle A}$. Therefore ${\displaystyle Na\leq A\,}$. Substituting (2) into this:

${\displaystyle P\geq {\frac {{\mathfrak {F}}^{2}\rho l}{A}}}$

This is the minimum ohmic heat dissipation of a resistive electromagnet. To show that the minimum dissipation increases with ${\displaystyle B^{2}}$ the Faraday law expression for a closed core electromagnet (eq. (1) from the article) can be used:

${\displaystyle NI={\mathfrak {F}}={\frac {B}{\mu }}}$

So:

${\displaystyle P\geq {\frac {B^{2}\rho l}{\mu ^{2}A}}}$

--Chetvorno^TALK 05:15, 16 May 2010 (UTC)[reply]

I can't seem to correct this error in the first paragraph. Anyone with more Wiki mojo want to try? Pammalamma (talk) 03:59, 20 November 2010 (UTC)[reply]