Find the inductance of a conductor in which, with a uniform change in current strength by 2A

Find the inductance of a conductor in which, with a uniform change in current strength by 2A for 0.25 s, an EMF of self-induction of 20 mV is excited.

Initial data: Esi (EMF of self-induction, which is excited in the conductor) = 20 mV = 20 * 10-3 V; Δt (time interval) = 0.25 s; ΔI (change in current strength in the conductor) = 2 A.

We determine the inductance of the conductor from the formula for calculating the EMF of self-induction: Esi = -L * ΔI / Δt and L = | Esi * Δt / ΔI |.

Let’s calculate: L = 20 * 10-3 * 0.25 / 2 = 2.5 * 10-3 H = 2.5 mH.

Answer: The inductance of the conductor is 2.5 mH.