An air-core inductor has 920 turns/m and an internal volume of 3.0 cm³.What is its inductance? A)3.2 μH B)2.8 mH C)0.35 H D)32 μH E)Insufficient information is given.

Question

Quizplus · Accepted Answer

The inductance of an air-core inductor can be calculated using the formula $L = \mu_0 N^2 A / l$, where $L$ is the inductance, $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$), $N$ is the total number of turns, $A$ is the cross-sectional area, and $l$ is the length of the inductor. Given 920 turns/m and a volume of 3.0 cm³ (or $3.0 \times 10^{-6}$ m³), we can find $N$ and $A$ given $l$. However, the volume and turns per meter alone are not enough to directly calculate $A$ and $l$ without additional information about the shape or dimensions of the inductor. Despite this, the question seems to expect a calculation based on the provided information, assuming a direct relationship between turns, volume, and inductance without requiring the dimensions. This might be an oversight in the question's design, as typically, either the length or the cross-sectional area, along with the number of turns, is needed to calculate inductance accurately. The correct answer, based on the options provided and assuming a direct calculation is possible, is marked as A) 3.2 μH, but this is under the assumption that there's an implicit method or additional information not provided in the question that allows for this calculation.

An Air-Core Inductor Has 920 Turns/m and an Internal Volume