A single-layer winding (turn to turn) of wire with a diameter of d = 0.2 mm is laid on a long cardboard

A single-layer winding (turn to turn) of wire with a diameter of d = 0.2 mm is laid on a long cardboard frame with a diameter of D = 5 cm. Determine the magnetic flux Ф created by such a solenoid at a current strength I = 0.5 A.

D = 5 cm = 0.05 m.

d = 0.2 mm = 0.2 * 10 ^ -3 m.

I = 0.5 A.

μ0 = 1.25 * 10 ^ -6 m * kg / s2 * A2.

F -?

The magnetic flux of the solenoid Ф is determined by the formula: Ф = I * L / N, where I is the current in the coil, L is the inductance of the coil, N is the number of turns in the winding.

N = 1 / d.

The inductance of the solenoid L is expressed by the formula: L = μ0 * n2 * V, n is the number of turns per unit length of the solenoid, V is the volume of the solenoid.

n = 1 / d, V = 1 * S = P * D ^ 2/4.

L = μ0 * P * D ^ 2 / d * 4.

Ф = I * μ0 * П * D ^ 2/4 * d.

Ф = 0.5 A * 1.25 * 10 ^ -6 m * kg / s2 * A ^ 2 * 3.14 * (0.05 m) ^ 2/4 * 0.2 * 10 ^ -3 m = 6 * 10 ^ -6 Wb.

Answer: Ф = 6 * 10 ^ -6 Wb.