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Wikipedia states that -

The only power consumed in a DC electromagnet is due to the resistance of the windings, and is dissipated as heat.

Is this true? So powerful magnets like junkyard magnets don't consume electricity apart from resistance?

How much does electricity does a junkyard or any other powerful super magnet consume?

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Yes it is true. The only DC magnets that use "no" power are superconducting magnets (like in MRI systems). Of course for those, there is significant power needed to keep the windings at superconducting temperature... and the cooling system will typically use several kW.

"How much power does a junkyard magnet use" is not an easy thing to answer: but according to a post on this page, 50 A at 250 V DC powers a pretty big magnet capable of lifting 1000's of pounds. That corresponds to about 10 kW.

Note that the "force" of a magnet depends greatly on the geometry: if you can get really close to the poles over a large area, they work extremely well; but if you are picking up an oddly shaped object, it will be much harder (saturation effects will rapidly come into play).

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Yes, in electrical circuits of only passive elements (Resistors, Capacitors, and Inductors) only the Resistors dissipate power (as heat). Active elements like transistors can also dissipate power, and if the currents in the circuit are changing with time, then power can be radiated away in electromagnetic waves. Therefore, the electrical resistance of the magnet coil dissipates power.

Additionally, power may be dissipated in the iron core of the electromagnet or in iron the magnet is picking up. As the DC current is turned on, a voltage is induced in the magnet's iron core by the changing magnetic flux. This causes power to be dissipated in the resistance of the iron. Likewise, as the magnet is brought near a piece of iron being picked up, the magnetic flux in the piece changes, a voltage is generated, and power is dissipated by the piece's resistance.

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