Question:
If an air core solenoid with self-inductance of $0.5\,mH$ is filled with soft iron of relative permeability of $1500$, its self-inductance becomes
If an air core solenoid with self-inductance of $0.5\,mH$ is filled with soft iron of relative permeability of $1500$, its self-inductance becomes
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Inductance increases directly with relative permeability.
Updated On: Apr 24, 2026
- $0.5\,H$
- $1.5\,H$
- $0.25\,H$
- $1.25\,H$
- $0.75\,H$
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The Correct Option is
Solution and Explanation
Concept:
• Self inductance: \[ L \propto \mu_r \]
Step 1: Given
\[ L_0 = 0.5\,mH = 0.5 \times 10^{-3}H \]
Step 2: New inductance
\[ L = \mu_r L_0 = 1500 \times 0.5 \times 10^{-3} \] \[ = 0.75H \] Final Conclusion:
Option (E)
• Self inductance: \[ L \propto \mu_r \]
Step 1: Given
\[ L_0 = 0.5\,mH = 0.5 \times 10^{-3}H \]
Step 2: New inductance
\[ L = \mu_r L_0 = 1500 \times 0.5 \times 10^{-3} \] \[ = 0.75H \] Final Conclusion:
Option (E)
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