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

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees. 

their crystal patterns have been disrupted by stress