Question

Your employer asks you to build a 20 cm-long solenoid with an interior field of 5.0 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.


Which wire should you use, and what current will you need?

Answer 1

Answer: wire with #18 and at 4.07 A

Explanation:

Given

Length of solenoid, l = 20 cm

Magnetic field strength, B = 5.0 mT

B = µ ° N / L * I, where

B = field strength

µ = permeability of free space

N = number of turns

L = length of the solenoid

I = current of the solenoid

µ = 4 π 10^-7

N = 180 mm / 1.02 = 176 turns

B = 4π*10^-7 * 176 / 0.18 * I

5.0*10^-3 = 4π*10^-7 * 977.78 * I

I = 5.0*10^-3 / 4 π 10^-7 * 977.77 = 5.0*10^-3 / 12287*10^-7 = 4.07 A (which is less than 6A)

Wire with a #18 gauge has a diameter of 1.02mm --- 176 turns ( 17.952 cm)

I = 4.07 A---- B = 0.005 T = 5.0 mT

Answer 2

The wire with 18 gauge should be used and it carries a current of 4.06A

Magnetic field and current inside a solenoid:

Length of solenoid, L = 20 cm = 200mm = 0.2m

Magnetic field strength, B = 5.0 mT

The magnetic field inside a solenoid is given by:

B = µ₀NI / L,

where

B = magnetic field strength

N = number of turns

L = length of the solenoid

I = current in the solenoid

For 18 gauge wire:

µ₀ = 4π × 10⁻⁷

N = 200 mm / 1.02mm = 196 turns

B = µ₀NI / L

$5.0\times10^{-3} = 4\pi\times10^{-7} \times 196 \times I/0.2\\\\I = 0.2\times5.0\times10^{-3} / (4\pi \times10^{-7} \times 196 )$

I = 4.06 A  (which is less than 6A)

For 26 gauge wire:

N/L = 200mm/0.41mm = 488 turns

B = µ₀NI / L

$5.0\times10^{-3} = 4\pi\times10^{-7} \times 488 \times I/0.2\\\\I = 0.2\times5.0\times10^{-3} / (4\pi \times10^{-7} \times 488 )$

I = 1.6 A (which is greater than 1A)

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