When the length of the solenoid is doubled without any change in the number of turns and the area of the coil.  Then its self-inductance will 

CONCEPT:

Self-Induction

• Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
• As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
• This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
• Self-inductance of a solenoid is given by –

\(⇒ L=\frac{{{\mu }_{o}}{{N}^{2}}A}{l}\)

Where μo = Absolute permeability, N = Number of turns, l = length of the solenoid, and A = Area of the solenoid.

EXPLANATION:

Given - l2 = 2l1

• Self-inductance of a solenoid is given by:

\(⇒ L=\frac{{{\mu }_{o}}{{N}^{2}}A}{l}\)

• According to the question, the length of the solenoid is doubled without any change in the number of turns and the area of the coil

\(⇒ L\propto \frac{1}{l}\)

⇒ L₁l₁ = L2l2

\(⇒ \frac{{{L}_{1}}}{{{L}_{2}}}=\frac{l_2}{l_1}=\frac{2l_1}{l_1}=2\)

\(⇒L_2=\frac{L_1}{2}\)​