What current is needed in the solenoid’s wires?

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Homework Statement


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?


Homework Equations


∑B ∆length = (µ0)I


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??
 
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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??
 
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??