Question

The self inductance of an air-core inductor (solenoid) is \(0.03\) mH. By introducing an iron core into inductor the self-inductance increases to \(30\) mH. The relative permeability of the core used is

• A. \(10^{-3}\)
• B. \(10^{-2}\)
• C. \(10^2\)
• D. \(10^3\)

Hint:

Inductance with a core is mu_r times the air-core value.

Answer 1

The Correct Option is D

Step 1: Ratio of inductances:
$\frac{L_2}{L_1} = \frac{30}{0.03}$. Dividing: $30 \div 0.03 = 1000$.

Step 2: Link to permeability:
$L \propto \mu$, so $\frac{\mu}{\mu_0} = 1000$.

Step 3: Answer:
$\mu_r = 10^3$.

Final Answer:
mu r is 10 cubed, option (D). \[ \boxed{10^3} \]