An air-core inductor has 1 000 turns/m and an internal volume of 3.0 cm³.What is its inductance? A)3.8 mH B)38 mH C)0.38 H D)Insufficient information is given.

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The inductance of an air-core inductor is given by:$$L = \frac{\mu_0 N^2 A}{l}$$where:* L is the inductance in henrys (H)* $\mu_0$ is the permeability of free space (4π × 10^-7 H/m)* N is the number of turns* A is the cross-sectional area of the inductor in square meters (m^2)* l is the length of the inductor in meters (m)In this problem, we are given:* N = 1000 turns/m* A = 3.0 cm^3 = 3.0 × 10^-6 m^3* l = 1 mSubstituting these values into the equation, we get:$$L = \frac{(4π × 10^-7 H/m) (1000 turns/m)^2 (3.0 × 10^-6 m^3)}{1 m} = 3.8 mH$$Therefore, the inductance of the inductor is 3.8 mH.

An Air-Core Inductor Has 1 000 Turns/m and an Internal