A cylindrical insulated wire of diameter 5.0 mm is tightly wound 200 times around a cylindrical core to form a solenoid with adjacent coils touching each other. When a 0.10 A current is sent through the wire, what is the magnitude of the magnetic field on the axis of the solenoid near its center? (μ₀ = 4π × 10^-7 T ∙ m/A) A) 6.6 × 10^-5 T B) 2.5 × 10^-5 T C) 1.3 × 10^-5 T D) 3.6 × 10^-5 T E) 9.8 × 10^-5 T

Question

Quizplus · Accepted Answer

The magnetic field at the center of a solenoid is given by:
B = μ(n/L)I
where μ is the permeability of free space, n is the number of turns per unit length (n = N/L, where N is the total number of turns and L is the length of the solenoid), and I is the current.
Given: diameter = 5.0 mm, so radius = 2.5 mm = 0.0025 m.
Number of turns per unit length, n = N/L = 200/L.
Length of solenoid, L = πdN = π(0.005)(200) = 1 m.
Current, I = 0.10 A.
Substituting these values into the formula, we get:
B = (4π×10-7 T∙m/A) (200/L) (0.10 A)
B = (8π×10-8/L) T
B at the center of the solenoid is given by the expression above when L = 0.5 m (half the length of the solenoid, since the center is halfway between the ends of the solenoid).
Therefore, B = (8π×10-8/0.5) T = 1.6×10-7 T = 2.5 × 10-5 T (in scientific notation) or 0.000025 T (in decimal notation).
The best choice is B, 2.5 × 10-5 T, which matches the answer in scientific notation.

A Cylindrical Insulated Wire of Diameter 5