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lily25
Joined: 07 Jul 2012 Posts: 4 Location: Toronto

Posted: Sat Jul 07, 2012 8:48 pm Post subject: Magnetic field of earth, how much current needed to cancel i 


It can be shown the magnetic field in a long narrow tube that is 5 meters around which 1000 turns of wire are wrapped. its uniform and given by B(Tesla )=1.26E6 N I(amp)?
where N is number of wire loops/meter. Assuming the solenoid is aligned along the Earth's magnetic field,how much current is needed to cancel the Earth's field?
earth's magnetic field is 10^4 Tesla
Last edited by lily25 on Sat Jul 07, 2012 9:16 pm; edited 1 time in total 

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hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Sat Jul 07, 2012 9:12 pm Post subject: 


Are you sure this is all the information? Don't you have the number of turns in the tube? 

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lily25
Joined: 07 Jul 2012 Posts: 4 Location: Toronto

Posted: Sat Jul 07, 2012 9:16 pm Post subject: 


yes sorry there are 1000 turns 

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hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Sun Jul 08, 2012 1:52 pm Post subject: 


In this case,
N per meter = 1000 turns /5 metres
N = 200 turns per metre.
B=1.26E6 x 200 x I
and you know the Earth's B is 10^4 T
so
10^4 T = 1.26E6 x 200 x I
Therefore, I = 10^4/(1.26E6 x 200) = 0.39 Amps 

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hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Sun Jul 08, 2012 2:13 pm Post subject: 


In this case,
N per meter = 1000 turns /5 metres
N = 200 turns per metre.
B=1.26E6 x 200 x I
and you know the Earth's B is 10^4 T
so
10^4 T = 1.26E6 x 200 x I
Therefore, I = 10^4/(1.26E6 x 200) = 0.39 Amps 

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lily25
Joined: 07 Jul 2012 Posts: 4 Location: Toronto

Posted: Sun Jul 08, 2012 6:03 pm Post subject: 


Thanks so much i got 0.40Amps just wanted to make sure I was on the right track 

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