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What current is needed in the solenoid’s wires?

  1. Jul 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A researcher would like to perform an experiment in zero magnetic field, which means that the field of the earth must be cancelled. Suppose the experiment is done inside a solenoid of diameter 1.0 m, length 4.0 m, with a total of 5000 turns of wire. The solenoid is oriented to produce a field that opposes and exactly cancels the field of the earth. What current is needed in the solenoid’s wires?


    2. Relevant equations
    ∑B ∆length = (µ0)I


    3. The attempt at a solution
    B = 5 * 10^-5 T
    length = 4 m
    radius = 0.5 m


    i = [4*(5*10^-5)]/(5000*2*pie*0.5)
    i = 1.27*10^-8

    have i done this correctly at all?!?!
     
  2. jcsd
  3. Jul 9, 2009 #2
    using Ampere's law the magnetic field ina solenoid will be given by

    B= magnetic permeability x current i x number of turns N / length L

    and for cancelling earth's magnetic field, the current needed is

    i = B x L / N x magnetic permeability

    = 0.4x10^-4 x 4 / 5000x 4 pi x 10^-7 = 2.55 x 10^-2 Amp

    = 25.5 milli amp.

    Or is this correct??
     
  4. Jul 9, 2009 #3

    rock.freak667

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    That would look correct but I don't know the strength of Earth's magnetic field off-hand
     
  5. Jul 9, 2009 #4
    my book says that the earth's magnetic field strenght is 5 *10^-5...so would i use that instead of the .4 x 10^-4??
     
  6. Jul 9, 2009 #5

    rock.freak667

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    Yes.
     
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