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The self-inductance of a solenoid of N-turns is proportional to

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\frac{1}{N}

N

N^{2}

\frac{1}{N^{2}}

answer is C.

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Detailed Solution

The self-inductance of a solenoid of N-turns is proportional to

N^{2}

.
According to 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-inductance, 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{μ_{0} N^{2} A}{l}

Where

μ_{0} =

Absolute permeability

N =

Number of turns

l =

Length of the solenoid

A =

Area of the solenoid

\Rightarrow L α N^{2}

From the equation of self inductance of solenoid, we can say that the self-inductance of solenoid is proportional directly to the square of the number of turns of the solenoid.

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