Question

To construct a solenoid, you wrap insulated wire uniformly around a plastic tube 7.1 cm in diameter and 57 cm in length. You would like a 3.0 A current to produce a 2.6 kG magnetic field inside your solenoid. What is the total length of wire you will need to meet these specifications?

Answer 1

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:

$B=\mu_0\,\frac{N}{L} \,I$

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;

$B=\mu_0\,\frac{N}{L} \,I\\0.26=4\,\pi\,10^{-7}\frac{N}{0.57} \,3\\N=\frac{0.26*0.57\,10^7}{12\,\pi} \\N=39311.27$

Therefore we need about 39312 turns of wire. Considering that each turn must have a length of $\pi\,D$, where D is the diameter of the plastic cylindrical tube, then the total length of the wire must be:

$Length=39312\,(\pi\,D)=39312\,(\pi\,0.071)\approx 8768.66\,\,m$

We can round it to about 8769 meters.