Question

Your employer asks you to build a 23-cm-long solenoid with an interior field of 6.0 mT . The specifications call for a single layer of wire, wound with the coils as close together as possible. You have two spools of wire available. Wire with a #18 gauge has a diameter of 1.02 mm and has a maximum current rating of 6 A. Wire with a #26 gauge is 0.41 mm in diameter and can carry up to 1 A.

A) Which wire should you use?
B) What current will you need?

Answer 1

We are going to use the magnetic field equation for a solenoid:

mu is the permeability of vacum

n is the turn density (the number of turns per length)

N is the number of turns, and L is the length of the solenoid

I is the current flowing thru the solenoid

So is we get n:

for the #18 gauge wire at his maximum current rating of 6A:

The maximum number of turns possible for #18 gauge wire is:

For the #26 gauge wire at his maximum current rating of 1A:

the maximum number of turns for the #26 gauge:

We notice tha n1 is less than the maximum possible number of turns for #18 gauge wire, and that n2 is greaterthan the maximum possible number of turns for #26 gauge wire. So we can produce a 6.0mT field using a single layer of #26 gauge wire.

So the wire we should use is #18 gauge.

ok for figuring out how much current we would need we replace n in the equation by the maximum possible number of turns of the #18 gauge wire and we get I to the other side of the equation.