A 45.C mA current is carried by a uniformly wound air-core solenoid with 460 tur

Question

A 45.C mA current is carried by a uniformly wound air-core solenoid with 460 turns, a 11.5 mm diameter, and 15.0 cm length. (a) Compute the magnetic field inside the solenoid. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. mu T (b) Compute the magnetic flux through each turn. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. T middot m^2 (c) Compute the inductance of the solenoid. mH (d) Which of these quantities depends on the current? (Select all that apply.) magnetic field inside the solenoid magnetic flux through each turn inductance of the solenoid

Explanation / Answer

Here ,

current , I = 45 mA = 0.045 A

N = 460

L = 0.15 m

a) magnetic field inside the solenoid = u0 * N * I/L

magnetic field inside the solenoid = 4pi *10^-7 * 460 * 0.045/(0.15)

magnetic field inside the solenoid = 1.734 *10^-4 T

b)

flux through each turn = area * magnetic flux

flux through each turn = 1.734 *10^-4 * pi * (0.0115/2)^2

flux through each turn = 1.801 *10^-8 Wb

c)

indutance of the solenoid = flux * N/current

indutance of the solenoid = 1.8101 *10^-8 * 460/.045

indutance of the solenoid = 0.000185 = 1.85 mH

d)

the correct answer is

magnetic field inside the solenoid

magnetic flu through each turn

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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