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
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
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Inductance with a core is mu_r times the air-core value.
Updated On: Oct 1, 2026
- \(10^{-3}\)
- \(10^{-2}\)
- \(10^2\)
- \(10^3\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
The self inductance of a solenoid is \(L = \mu n^2Al\). Replacing air (\(\mu_0\)) by an iron core (\(\mu = \mu_r\mu_0\)) multiplies \(L\) by \(\mu_r\), when the geometry stays the same.
Step 2: Compute:
\[ \mu_r = \frac{L_{\text{iron}}}{L_{\text{air}}} = \frac{30\text{ mH}}{0.03\text{ mH}} = 1000 = 10^3 \]
Step 3: Why the other options are wrong.
\(10^{-3}\) and \(10^{-2}\) would make the inductance decrease. \(10^2\) would give only 3 mH.
Final Answer:
The relative permeability is \(10^3\), option (D). \[ \boxed{10^3} \]
The self inductance of a solenoid is \(L = \mu n^2Al\). Replacing air (\(\mu_0\)) by an iron core (\(\mu = \mu_r\mu_0\)) multiplies \(L\) by \(\mu_r\), when the geometry stays the same.
Step 2: Compute:
\[ \mu_r = \frac{L_{\text{iron}}}{L_{\text{air}}} = \frac{30\text{ mH}}{0.03\text{ mH}} = 1000 = 10^3 \]
Step 3: Why the other options are wrong.
\(10^{-3}\) and \(10^{-2}\) would make the inductance decrease. \(10^2\) would give only 3 mH.
Final Answer:
The relative permeability is \(10^3\), option (D). \[ \boxed{10^3} \]
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