A coil is wound having a diameter'd' its inductance is 10 mh. If it is rewound on the same core with a wire having diameter '0.5d' and total weight of tue wire is the same as before. The inductance will be?

Question

Answer

Inductance of the Coil
When rewinding a coil with a different wire diameter while maintaining the same total wire weight, the inductance will change based on several factors.
Initial Coil Parameters
- Diameter of the Wire: 'd'
- Inductance: 10 mH
- Length of Wire: Determined by the original coil dimensions and wire diameter.
New Coil Parameters
- New Wire Diameter: '0.5d'
- Total Weight of Wire: Remains constant
Impact of Wire Diameter on Inductance
1. Cross-Sectional Area: The cross-sectional area of the wire decreases with the new diameter. The area is proportional to the square of the diameter. Thus:
- Original Area = $ \frac{\pi d^2}{4} $
- New Area = $ \frac{\pi (0.5d)^2}{4} = \frac{\pi d^2}{16} $
2. Number of Turns: With the same weight of wire, the total length of wire will remain constant, but due to the smaller diameter, more turns can be accommodated in the same space.
3. Inductance Formula: The inductance $ L $ of a coil is given by:
$$
L \propto N^2 \cdot A \cdot \mu
$$
Where $ N $ is the number of turns, $ A $ is the cross-sectional area, and $ \mu $ is the permeability of the core material.
Conclusion
- Increased Turns: The number of turns $ N $ increases as the wire diameter decreases, leading to a higher inductance.
- Reduced Area: The cross-sectional area $ A $ decreases, which tends to lower inductance.
Considering these factors, the overall change in inductance depends on the balance between increased turns and reduced area. In this scenario, the increase in turns will outweigh the reduction in area, resulting in an inductance greater than 10 mH.

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