A direct current I = 5A passes through a coil with a radius of R = 2 cm, containing N = 500 turns.

A direct current I = 5A passes through a coil with a radius of R = 2 cm, containing N = 500 turns. Determine the inductance of the coil if the magnetic field strength in its center is H = 10kA / m.

Let’s use the formula:
B = m * u * (I / 2pi * R);
Where:
m * u – absolute magnetic permeability;
I is the current strength;
R is the radius of the coil;
Let’s transform the formula:
B = (m * u * N * I) / L;
Let us express:
L = (m * u * N2 * S) / l;
Let’s transform:
L = m * u * N2 * pi * R2 / (m * u * N * I / B) = pi * N * R2 * B / I;
Substitute the values:
L = (3.14 * 500 * 0.0004 * 1.26 * 10-2) / 5;
L = 0.0015 H;
L = 15 * 10-4 G.

Answer: L = 15 * 10-4 G.