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A solenoid 10.0 cm in diameter and 85.1 cm long is made from copper wire of diameter 0.100 cm, with very thin insulation. The wire is wound onto a cardboard tube in a single layer, with adjacent turns touching each other. What power must be delivered to the solenoid if it is to produce a field of 8.90 mT at its center

Sagot :

Answer:

P = 29.3 W

Explanation:

The magnetic field in a solenoid is

          B = μ₀  n i

          i = B /μ₀ n

where n is the density of turns

           

We can use a direct rule of proportions or rule of three to find the number of turns, 1 a turn has a diameter of 0.100 cm = 10⁻³ m, in the length of

L= 85.1 cm = 0.851 m how many turns there are

         #_threads = 0.851 / 10⁻³

         #_threads = 8.50 10³ turns

the density of turns is

          n = # _threads / L

          n = 8.51 103 / 0.851

          n = 104 turn / m

the current that must pass through the solenoid is

          i = 8.90 10-3 / 4pi 10-7 104

          i = 0.70823 A

now let's find the resistance of the copper wire

         R = ρ L / A

the resistivity of copper is ρ = 1.72 10⁻⁸ Ω m

wire area

         A = π r²

         A = π (5 10⁻⁴)

         A = 7,854 10⁻⁷ m²

let's find the length of wire to build the coil, the length of a turn is

         Lo = 2π r = ππ d

         Lo = π 0.100

         Lo = 0.314159 m / turn

With a direct proportion rule we find the length of the wire to construct the 8.5 103 turns

          L = Lo #_threads

          L = 0.314159 8.50 10³

          L = 2.67 10³ m

resistance is

         R = 1.72 10⁻⁸ 2.67 10₃ / 7.854 10⁻⁷

         R = 5,847 10¹

         R = 58.47 ohm

The power to be supplied to the coil is

          P = VI = R i²

          P = 58.47 0.70823²

          P = 29.3 W

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