Solenoids: 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)

Question

Quizplus · Accepted Answer

The magnetic field inside a solenoid is given by:

B = μnI

Where μ is the permeability of free space (4π × 10-7 T∙m/A), n is the number of turns per unit length (n = N/L), I is the current, N is the total number of turns, and L is the length of the solenoid.

In this case, the diameter of the wire is 5.0 mm, so its radius is 2.5 mm or 0.0025 m. The length of the solenoid is not given, so we cannot calculate n directly. However, we do know that the coils are tightly wound so that adjacent coils touch each other, which means that the length of the solenoid is approximately equal to its diameter.

Therefore, we can estimate n as follows:

n ≈ N/L = N/(πd) = 200/(π×0.005) ≈ 12732 turns/m

Substituting n, I, and μ into the equation for B, we get:

B = μnI ≈ (4π×10-7)(12732)(0.10) ≈ 2.5 × 10-3 T

Therefore, the answer is B, 2.5 × 10-3 T.

Solenoids: a Cylindrical Insulated Wire of Diameter 5