A 250-turn solenoid carries a current of 9.0 A.The radius of the solenoid is 0.075 m;and its length is 0.14 m.Determine the magnetic flux through the circular cross-sectional area at the center of the solenoid. A)1.8 × 10^-⁵ Wb B)9.9 × 10^-⁵ Wb C)3.6 × 10^-⁴ Wb D)7.0 × 10^-⁴ Wb E)2.2 × 10^-³ Wb

Question

Quizplus · Accepted Answer

The magnetic flux through the circular cross-sectional area at the center of the solenoid can be calculated using the formula:

ϕ = μ₀n²πr²I

where μ₀ is the permeability of free space (4π x 10^-7 T m/A), n is the number of turns per unit length (n = N/L), N is the total number of turns (N = 250), L is the length of the solenoid (L = 0.14 m), r is the radius of the solenoid (r = 0.075 m), and I is the current (I = 9.0 A).

Substituting the values:

n = N/L = 250/0.14 = 1785.71 turns/m
ϕ = (4π x 10^-7 T m/A) x (1785.71 turns/m)² x π x (0.075 m)² x 9.0 A
ϕ = 3.6 x 10^-4 Wb
ϕ = 3.6 x 10^4 μWb

Therefore, the magnetic flux through the circular cross-sectional area at the center of the solenoid is 3.6 x 10^-4 Wb or 3.6 x 10^4 μWb, which is closest to the option C (3.6 × 10⁴ μWb).

A 250-Turn Solenoid Carries a Current of 9