# An ideal air-core coil has an inductance of 2 mH. The number of turns of the coil is doubled and its length is halved.

An ideal air-core coil has an inductance of 2 mH. The number of turns of the coil is doubled and its length is halved. Assuming that the inner cross-sectional area of the core remains constant the new inductance of this altered air-core coil is

### Right Answer is: 16 mH

#### SOLUTION

Concept:

The inductance of a coil is given by,

$L=\frac{{{\mu }_{0}}{{\mu }_{r}}{{N}^{2}}A}{l}$

Where

μ0 is the magnetic permeability of free space

μr is relative permeability

N is number if turns

A is cross-sectional area

l is length

Calculation:

Self-inductance of the solenoid is given by

${L_1} \propto \dfrac{{{N^2}}}{l} = 2mH$………(1)

Now the number of turns in the coil is doubled and the length is halved

N = 2N2

l = l/2

${L_2} = \dfrac{{{{(2N)}^2}}}{{l/2}}$……..(2)

Dividing equations 1 and 2

$\dfrac{{{L_2}}}{{{L_1}}} = \dfrac{{8{N^2}}}{l} \times \frac{l}{{{N^2}}}$

L2 = 8 × 2 = 16 mH

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