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Engineering
Electrical Engineering
How many Amp turns around the Earth for this flux density?
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[QUOTE="Tom.G, post: 5983742, member: 581973"] Let's see how to build a coil to cancel that field. Since I don't expect an implementation. I'll be rounding off values here. Choose AWG #0 Copper wire with a diameter over insulation of 0.5inch. Dia. 0.5in. = 24 Turns/Ft. = 126 720 Turns/mile Res Ohm/1000Ft = 0.1 Ohms = [COLOR=blu][S]5.28[/S][/COLOR] [COLOR=red]0.528[/COLOR] Ohms/Mile Weight = 3.13ft/pound = 5280/3.13 = 1687lb/mi Current rating 150A 5E8 Ampere-Turns needed Earth Circumference = 24000 Miles 5E8AT/150A = 3.3E6 Turns * 24000 = 8E10 miles of wire 3.3E6/126720 Turns per mile = one-layer winding 26.3 Miles wide 3.3E6 Turns * 24000 = 8E10 miles of wire 8E10 * 16871lb/mi = 1.35E15 pounds of wire 8E10 miles * [COLOR=blu][S]5.28[/S][/COLOR] [COLOR=red]0.528[/COLOR] Ohms per mile = 4.224E[COLOR=blu][S]11[/S][/COLOR] [COLOR=red]10[/COLOR] Ohms 150A * 4.224E[COLOR=blu][S]11[/S][/COLOR] [COLOR=red]10[/COLOR] = 6.3E[s]13[/s][color=red]12[/color] Volts to drive the coil With 1.35E15 pounds of Copper wire needed and Global Copper production of 5E10 pounds per year, you would need 27000 years to mine the Copper. It looks like Earth-sized Helmholtz coils won't be in our lifetime. :wink: Cheers, Tom [/QUOTE]
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How many Amp turns around the Earth for this flux density?
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