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Your employer asks you to build a 18-cmcm-long solenoid with an interior field of 5.8 mTmT . 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?

Respuesta :

Answer:

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 = 5.8 10⁻³ 0.18 / 4 π 10⁻⁷ 1

        N = 8.3 102 laps

        N = 831 laps

Let's find the solenoid length

    For this we use a rule of proportions

                L_solenoid = Turns * wire diameter

                L_ solenoid = 831 * 0.41 10--3

                L_solenoid = 0.3407 m

We see that two turns are needed in the wire to have a length of 0.18 m

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