Answer:
We need about 8769 meters of wire to produce a 2.6 kilogauss magnetic field.
Explanation:
Recall the formula for the magnetic field produced by a solenoid of length L. N turns, and running a current I:
[tex]B=\mu_0\,\frac{N}{L} \,I[/tex]
So, in our case, where B = 2.6 KG = 0.26 Tesla; I is 3 amperes, and L = 0.57 m, we can find what is the number of turns needed;
[tex]B=\mu_0\,\frac{N}{L} \,I\\0.26=4\,\pi\,10^{-7}\frac{N}{0.57} \,3\\N=\frac{0.26*0.57\,10^7}{12\,\pi} \\N=39311.27[/tex]
Therefore we need about 39312 turns of wire. Considering that each turn must have a length of [tex]\pi\,D[/tex], where D is the diameter of the plastic cylindrical tube, then the total length of the wire must be:
[tex]Length=39312\,(\pi\,D)=39312\,(\pi\,0.071)\approx 8768.66\,\,m[/tex]
We can round it to about 8769 meters.
