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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 )=1.26E-6 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 TeslaLast edited by lily25 on Sat Jul 07, 2012 9:16 pm; edited 1 time in total
hepcj

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?
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
hepcj

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=1.26E-6 x 200 x I and you know the Earth's B is 10^4 T so 10^4 T = 1.26E-6 x 200 x I Therefore, I = 10^-4/(1.26E-6 x 200) = 0.39 Amps
hepcj

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=1.26E-6 x 200 x I and you know the Earth's B is 10^4 T so 10^4 T = 1.26E-6 x 200 x I Therefore, I = 10^-4/(1.26E-6 x 200) = 0.39 Amps
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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