Your employer asks you to build a 32-cm-long solenoid with an interior field of

Question

Your employer asks you to build a 32-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. Which wire should you use? What current will you need? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

(A) We know that the magnetic field inside the solenoid is
B = uonI
where B is magnetic field , n is the no. of turns per unit length and I is the current
so for first wire
4.6*10-3 = (4Pi*10-7)*n*6
And for second wire
4.6*10-3 = (4Pi*10-7)*n*1
therefore for second wire the value of n will be more so we need large length for that.
So we will use the first wire of 18 gauge.
(B) B = µ ° N / L * I
µ° = 4 10^-7
N = 320 mm / 1.02 =313.72 turns
B = 4 10^-7 * 313.72 / 0.32 * I
4.6 10^-3 = 4 10^-7 * 980.375 * I
I = 4.6 10^-3 / 4 10^-7 * 980.375 = 3.7338 ( 6A)

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.