Does inductance depend on frequency ?

Hint:Here, we are asked if the inductance of an inductor depends on the frequency or not. By frequency we mean that the inductance depends on the frequency of the applied voltage source. In order to answer this question, what you need to do is just find the inductance of an inductor and see whether it depends on the frequency or not.

Complete step by step answer:
You know that whenever a current is passed through a closed conducting loop, it is observed that the current carrying loop produces a magnetic field depending on the current flowing in it and the geometry of the loop. So, if a magnetic field is produced, you will also have a magnetic flux that will pass through the area bounded by the loop. Now, if the current changes with time, the magnetic field will also change and therefore the magnetic flux will also change which will induce an emf in the loop. This is called self-inductance.

The magnetic field at any point is found out to be proportional to the current and therefore the magnetic field will also be proportional to the current in the loop. So, mathematically we can write magnetic flux as $\Phi = Li$ where \[L\] is a constant and is called self-inductance of the loop or simply inductance. The induced emf is given by Faraday’s law of electromagnetism as $\xi = - \dfrac{{d\Phi }}{{dt}}$. Let us substitute the value of flux in the emf expression, we get,
$\xi = - \dfrac{{d\Phi }}{{dt}} \\
\Rightarrow\xi = - \dfrac{{d\left( {Li} \right)}}{{dt}} \\
\Rightarrow\xi = - L\dfrac{{di}}{{dt}} \\ $
Now, for a solenoid, we have
\[\xi = - N\dfrac{d}{{dt}}\left( {\int {\overrightarrow B .d\overrightarrow S } } \right) \\
\Rightarrow\xi = - N\dfrac{d}{{dt}}\left( {{\mu _0}ni} \right)\left( {\pi {r^2}} \right) \\ \]
Since the magnetic field produced inside a solenoid is \[B = {\mu _0}ni\], where $n = \dfrac{N}{l}$ and $r$ is the radius of the loop.
So, we have,
\[\xi = - N\dfrac{d}{{dt}}\left( {{\mu _0}ni} \right)\left( {\pi {r^2}} \right) \\
\Rightarrow\xi = - N{\mu _0}n\pi {r^2}\dfrac{{di}}{{dt}} \\
\therefore\xi = - {\mu _0}{n^2}\pi {r^2}l\dfrac{{di}}{{dt}} \\ \]
If you compare it with the equation which we previously derived, we will get \[L = {\mu _0}{n^2}\pi {r^2}l\]. As you can see that the inductance is independent of the frequency.

Therefore, inductance does not depend on frequency.

Note:We have discussed in detail about the inductance and came to a conclusion that the inductance does not depend on the frequency. Also, keep in mind that the inductance is a constant which depends only on the geometry of the loop or solenoid. Also, you should remember the Faraday’s law of electromagnetism.