Question

You are given a length (l) of wire that has radius (a)and are told to wind it into an inductor in the shape of a helix that has a circular cross section of radius (r). The windings are to be as close together as possible without overlapping. Show that the self-inductance of this inductor is L = 1/4nrl/a.

Answer 1

To solve this problem it is necessary to apply the concepts related to inductance on an inductor, which is mathematically described as:

$L = \mu_0 n^2 A l$

Where,

$\mu_0$= Permeability constant (Described as 'n' at the problem equation)

l = Length

A = Cross-sectional Area

n = No of turn per unit length

The number of turns N is given by

$N = \frac{l}{2a}$

The number of turns per unit length n is

$n = \frac{N}{l} = \frac{1}{2a}$

The relationship of the cable lengths starts from assuming that the length 'a' is less than the length 'r', and therefore the length of the wire d would be related by:

$d = N(2\pi r)$

$d = \frac{l}{2a}2\pi r$

$d = \frac{\pi r}{a}l$

Solving to obtain l,

$l = \frac{ad}{\pi r}$

Substituting at the first equation,

$L = \mu_0 n^2 A l$

$L = \mu_0 (\frac{1}{2a})^2(\pi r^2)(\frac{ad}{\pi r})$

$L = \mu_0 (\frac{rd}{4a})$