Your employer asks you to build a 26-cm-long solenoid with an interior field of 4.1 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 #26 gauge is 0.41 mm in diameter and can carry up to 1 A. Which wire should you use, and what current will you need?

This wire is not suitable to make the solenoid.

The suitable wire is #18 gauge wire.

Explanation:

The magnetic field in a solenoid is

         B = μ₀ N / L I

Where N is the number of turns, L the solenoid length and I the current

         N = B L /  μ₀  I

Let's calculate

       N = 4.1 × 10⁻³ × 0.26 / 4 π 10⁻⁷ × 1

       N = 848.3 laps

Let's find the solenoid length

   For this we use a rule of proportions

               L_solenoid = Turns * wire diameter

               L_ solenoid = 848.3 * 0.41 × 10⁻³

               L_solenoid = 0.3478 m