Question

A wire of fixed length is wound on a solenoid of length l and radius r. its self inductance is found l . now if same wire is wound on a solenoid of length l/2 and radius r/2, then the self inductance will be

Answer 1

Answer:

The self-inductance of the coil will get doubled, i.e., 2L.

Explanation:

The self-inductance of a coil is defined as the property of the coil by virtue of which it opposes the change in the flow of current through itself by inducing an emf.  

The self inductance of a solenoid:

$\rm L = \dfrac{\mu_o N^2\pi r^2}{l}.$

where,

• L = self inductance of the solenoid.

• $\mu_o$ = magnetic permeability of free space.

• N = number of turns of the solenoid.

• r = radius of cross section of solenoid.

• l = length of the solenoid.

The length of the wire of the coil is equal to the product of the circumference of the coil and the number of turns of the coil.

Since the same wire is taken in both the cases, therefore, the product of the circumference of the coil and the number of turns of the coil should be same.

$\rm Length\ of\ the\ wire,\ L_o = N \times 2\pi r\\r = \dfrac{L_o}{2\pi N}.\\\therefore\ L = \dfrac{\mu_o N^2 \pi \left (\dfrac{L_o}{2\pi N} \right ) ^2}{l}=\dfrac{\mu_o L_o^2}{4\pi l}.$

When the length of the wire is l and the radius is r, then,  

$\rm L=\dfrac{\mu_o L_o^2}{4\pi l}.$

When the length of the wire is l/2 and the radius is r/2, then,

$\rm L'=\dfrac{\mu_o L_o^2}{4\pi \dfrac l2}=2 \dfrac{\mu_o L_o^2}{4\pi l}=2L.$

Thus, the self-inductance of the coil will get doubled.

Answer 2

Answer:

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