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| Calculating electromagnet – magnetism vs heat |
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| radiolistener:
--- Quote from: perik on August 22, 2019, 11:30:02 am ---Would you agree on this calculation? --- End quote --- Here is my draft calc for your case: Wire material = Copper Wire diameter = 0.5 [mm] Wire turns = 500 Solenoid diameter = 0.01 [m] Wire length = 15.707 [m] Wire resistance at DC = 1.3423 [Ohm] Voltage = 5 [V] Current = 5 / 1.3423 = 3.725 [A] Solenoid length = 0.25 [m] Solenoid H = 7450 [A/m] Solenoid B = 0.009354466 [T] (= 93.54466 [Gauss]) Solenoid L = 97.0397 [uH] Heating = 5^2 / 1.3423 = 18.6 [W] Magnetic flux depends on coil winding density and current. Heating depends on resistance, which depends on wire thickness. If you will use fat wire (for less heating), you will get smaller winding density and smaller magnetic flux. And vice versa. For comparison, Neodymium magnet N52 has about B = 1.43 - 1.48 [T] ( = 14300 - 14800 [Gauss]) https://www.aliexpress.com/item/32978000774.html And it don't consume electricity and no heating at all :) In order to get the same magnetic flux from the coil calculated above, you will need to apply 590 Amps / 792 Volts. |
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