When the current strength in the solenoid was changed from 6 A to 10 A

When the current strength in the solenoid was changed from 6 A to 10 A, the magnetic field energy changed to | delta W | = 6.4 J. The inductance of the solenoid is?

Initial data: I1 (initial current in the solenoid) = 6 A; I2 (end current) = 10 A; ΔW (change in magnetic field energy) = 6.4 J.

We express the solenoid inductance from the formula: ΔW = L * Ik ^ 2/2 – L * In ^ 2/2 = L / 2 * (Ik ^ 2 – In ^ 2), whence L = 2 * ΔW / (Ik ^ 2 – In ^ 2).

Let’s calculate: L = 2 * 6.4 / (10 ^ 2 – 6 ^ 2) = 2 * 6.4 / 64 = 0.2 H.

Answer: The inductance of this solenoid is 0.2 G.