Question

Your employer asks you to build a 24-cm long solenoid with an interior field of 4.6 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

Step-by-step explanation:

Formula for magnetic field in a solenoid

B = μ₀ N/ L x I

N is total number of turns , L is length of solenoid , I is current , B is magnetic field , μ₀ is constant equal to 4π x 10⁻⁷

Putting the values for 18 gauge wire

N = length of wire / diameter of wire

N = 24 / .102

= 235.3

B = μ₀ N/ L x I

4.6 x 10⁻³ = 4π x 10⁻⁷ x (235.3 / .24) x I

I = 3.735 A

Putting the values for 26 gauge wire

N = length of wire / diameter of wire

N = 24 / .041

= 585.36

B = μ₀ N/ L x I

4.6 x 10⁻³ = 4π x 10⁻⁷ x (585.36 / .24) x I

I = 1.5 A

a ) So 18 gauge wire will be used

b ) as current required in this case is

3.735 A

In the second case current required is 1.5 A which is beyond its tolerence limit.