# 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

μ_{0} is the magnetic permeability of free space

μ_{r} 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 = 2N^{2}

l = l/2

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

Dividing equations 1 and 2

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

**L _{2} = 8 × 2 = 16 mH**