Your employer asks you to build a 30-cm-long solenoid with an interior field of

Question

Your employer asks you to build a 30-cm-long solenoid with an interior field of 5.5 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. Wife with a #26 gauge is 0.41 mm in diameter and can carry up to 1 A. Which wire should you use? # 26 #18 What current will you need? Express your answer to two significant figures and include the appropriate units. I = _____________ ____________

Explanation / Answer

part A:

B = µ ° N / L * I
µ° = 4 10^-7
N = 300 mm / 1.02 = 294 turns
B = 4 10^-7 * 294 / 0.30 * I
5.5* 10^-3 = 4 10^-7 * 980.39 * I
I = 5.5 10^-3 / 4 10^-7 * 980.39 = 5.5 10^-3 / 12313.7 *10^-7 = 4.06 A ( 6A)

Wire with a #26 gauge has a diameter of 0.41mm

B = µ ° N / L * I
µ° = 4 10^-7
N = 300 mm / 0.41 = 732 turns
B = 4 10^-7 * 732 / 0.30 * I
5.5* 10^-3 = 4 10^-7 * 2439 * I
I = 5.5 10^-3 / 4 10^-7 * 2439 = 5.5 10^-3 / 30634 *10^-7 = 1.8 A ( greater than 1A)

so that gauge 18 wire is useful so that answer is 18

part B:

current I = 4.06 A

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