Question

The length of a thin wire require to manufacture a solenoid of length $l=100\,cm$ and inductance $L=1\,mH$, if the solenoid’s cross-sectional diameter is considerably less than its length is:

Answer 1

Here before solving the problem we will know about the self-inductance of solenoid and use the formula of self-induction of solenoid to solve the given problem.The property of a current-carrying coil that resists or opposes the shift in current flowing through it is known as self-inductance. This is primarily due to the self-induced emf produced by the coil. Self-inductance is a phenomenon that occurs when a voltage is inducted in a current-carrying wire.

Formula used:
Self-inductance of solenoid:
$L=\dfrac{({{\mu }_{0}}{{N}^{2}}A)}{l}$
$L$ is the inductance of the solenoid, ${{\mu }_{0}}$ is the permeability of free space, $N$ is the number of turns of coil, $A$ is the area of the wire and $l$ is the length of the wire.

Complete step by step answer:
A solenoid is an electric current-carrying coil of wire that functions as a magnet. It's a helix made up of a cylindrical coil twisted into a densely packed helix. In Greek, solen means "pipe, tube," and oid means "form, shape." The word solenoid was coined by Andre-Marie Ampere, a French physicist.

The solenoid's core material is ferromagnetic, which means that magnetic lines of flux are concentrated, increasing the coil's inductance. At the end of the core material, the magnetic flux can be seen outside the coil, but the majority of the flux is contained inside the core material. Self-inductance of solenoid can be given as:
$L=\dfrac{({{\mu }_{0}}{{N}^{2}}A)}{l}$
Here given parameters are:
$L={{10}^{-3}}H$
$\Rightarrow l=100\,cm={{10}^{2}}\times {{10}^{-2}}\,m=1\,m$
Let ${{l}_{1}}$​ be the length of wire,
${{l}_{1}}=2\pi rN$
By rearranging,
$Nr=\dfrac{{{l}_{1}}}{2\pi }$
By squaring both sides, we get
${{N}^{2}}{{r}^{2}}=\dfrac{{{l}_{1}}^{2}}{4{{\pi }^{2}}}$
Now by the formula mentioned above,
$L=\dfrac{({{\mu }_{0}}{{N}^{2}}\pi {{r}^{2}})}{l}$
Putting value of ${{N}^{2}}{{r}^{2}}$, we get
$\Rightarrow L=\dfrac{{{\mu }_{0}}\pi }{l}\dfrac{{{l}_{1}}^{2}}{4{{\pi }^{2}}}$
$\Rightarrow {{l}_{1}}=\sqrt{\dfrac{Ll4\pi }{{{\mu }_{0}}}} \\\ \Rightarrow {{l}_{1}} =\sqrt{\dfrac{{{10}^{-3}}\times 1\times 4\pi }{4\pi \times {{10}^{-7}}}} \\\ \Rightarrow {{l}_{1}} ={{10}^{2}}\,m \\\ \therefore {{l}_{1}} =0.10\,km$

Therefore the length of the solenoid is 0.10 km.

Note: The coil's inductance is only present when the current is changing, such as alternating current, and not when the current is constant. Self-inductance, which is calculated in Henrys, is often in opposition to the changing current (SI unit). If the current in the circuit is increasing or decreasing, induced current often opposes the change in current. Self-inductance is one of the types of electromagnetic inductance.