Difficulty: Easy
Correct Answer: all of the above
Explanation:
Introduction / Context:
Engineers often characterize materials using macroscopic parameters averaged over many atoms: electrical conductivity, relative permeability µr, and relative permittivity (dielectric constant) εr. These bulk properties drive circuit, antenna, insulation, and magnetic core design, without requiring atom-by-atom detail.
Given Data / Assumptions:
Concept / Approach:
Each listed parameter describes a constitutive relation connecting field and flux quantities in bulk media: J = σE for conduction, B = µ0µrH for magnetics, and D = ε0εrE for dielectrics. These are textbook examples of macroscopic properties used directly in design calculations.
Step-by-Step Solution:
Recognize conductivity as a bulk transport parameter (A/m^2 per V/m).Recognize µr as the dimensionless scaling of magnetic permeability.Recognize dielectric constant εr as the dimensionless scaling of permittivity.Since all are macroscopic, select “all of the above”.
Verification / Alternative check:
Handbooks list σ, µr, εr for materials like copper, ferrites, and polymers, confirming their macroscopic status and everyday engineering use.
Why Other Options Are Wrong:
Choosing only one or two ignores that all listed items are classic macroscopic descriptors.
Common Pitfalls:
Confusing microscopic carrier mobility or atomic polarizability with these bulk, averaged properties.
Final Answer:
all of the above
Discussion & Comments