Question E6D06:
What core material property determines the inductance of an inductor?  

Choices:  
A: Permittivity  
B: Resistance  
C: Reactivity  
D: Permeability    

Correct Answer is  D: Permeability  

The correct answer is D: Permeability.

Why D is correct:
- Inductance depends on how easily the core supports magnetic flux. This is set by the core’s magnetic permeability, $\mu$.
- For a simple inductor (e.g., a solenoid), the inductance is approximately:
  $$L = \mu \frac{N^2 A}{\ell}$$
  where:
  - $N$ = number of turns
  - $A$ = core cross-sectional area
  - $\ell$ = magnetic path length
  - $\mu = \mu_0 \mu_r$ (absolute permeability), with $\mu_r$ being the core’s relative permeability (the material property).
- A higher $\mu_r$ (e.g., ferrite) increases $L$ compared to an air core ($\mu_r \approx 1$).

Why the other choices are incorrect:
- A: Permittivity ($\epsilon$) relates to electric fields and capacitance, not inductance. For capacitors: $$C = \epsilon \frac{A}{d}$$
- B: Resistance is a loss parameter; it affects the Q and heating but does not set the inductance value.
- C: Reactivity (reactance) is a frequency-dependent circuit quantity, not a material property. For inductors: $$X_L = \omega L = 2\pi f L$$ It results from $L$, not the material directly.

Useful formulas and rules of thumb:
- Inductance scaling with core:
  $$L_{\text{core}} \approx \mu_r \, L_{\text{air}}$$
- Core constant method (common in ferrite/powdered iron design):
  $$L = A_L N^2$$
  where $A_L$ is provided by the core manufacturer and depends on permeability and geometry.
- Typical $\mu_r$: air ≈ 1, powdered iron ≈ 10–100, ferrite ≈ 100–5000.