Question

A wooden toroidal core with a square cross section has an inner radius of $10$ cm and an outer radius of $12$ cm. It is wound with one layer of wire (of diameter $1.0$ mm and resistance per meter $0.020\Omega {m^{ - 1}}$). What are
(a) the inductance and
(b) the inductive time constant of the resulting toroid? Ignore the thickness of the insulation on the wire.

Answer 1

We can find the inductance of the toroidal coil by using the concept of Magnetic induction and flux through the toroidal coil, then we calculate the resistance of the coil with a square cross section and the inductive time constant of the toroid. The inductive time constant is inductance per unit resistance of the toroid.

Formulae used:
The magnetic induction of the toroid is given by
$B = \dfrac{{{\mu _0}iN}}{{2\pi r}}$
And self-inductance of the toroid is given by
$N{\Phi _B} = Li$
Where, $i$ - current flowing through the toroid, $r$ - distance between the point and center of toroid, $N$ - number of turns of the toroid, ${\Phi _B}$ - flux through the toroid and $L$ - self-inductance of the toroid.

Complete step by step answer:
Let us denote the terms in the given problem,
(a) $h = b - a = 0.12 - 0.10 = 0.02m$
Where, $h$ - height of the toroid, $a,b$ - inner and outer radius of the toroid, respectively.
Using the two formulae, we can derive the equation for inductance as
$L = \dfrac{{{\mu _0}{N^2}h}}{{2\pi }}\ln \left( {\dfrac{b}{a}} \right) - - - - - - - - - (1)$
The circumference of the inner toroid is $2\pi a = 2(3.14)(10) = 62.8$ cm
Thus, the number of turns of the toroid is $N = \dfrac{{62.8cm}}{{1.0mm}} = 628$ .
From equation $(1)$ , we get
$L = \dfrac{{\left( {4\pi \times {{10}^{ - 7}}} \right){{\left( {628} \right)}^2}\left( {0.02} \right)}}{{2\pi }}\ln \left( {\dfrac{{0.12}}{{0.10}}} \right)$
Solving, we get
$\therefore L = 2.9 \times {10^{ - 4}}\,H$

(b) We know that the perimeter of the square is four times its size. So, the total length of the wire is given by
$l = 628 \times 4 \times 0.2cm = 50m$
The resistance of the wire is given by
$R = \dfrac{{50m}}{{0.02\Omega {m^{ - 1}}}} = 1.0\Omega$
Now, the inductive time constant of the toroid is given by
$\tau = \dfrac{L}{R} = \dfrac{{2.9 \times {{10}^{ - 4}}}}{{1.0}} \\\ \therefore \tau= 2.9 \times {10^{ - 4}}$ s
Thus, the inductance of the toroid is $2.9 \times {10^{ - 4}}H$ and the inductive time constant is $2.9 \times {10^{ - 4}}$ s.

Note: When an alternating electric current flows through the wire of a toroid, it will create a changing magnetic field within the toroid. This changing magnetic field creates a changing magnetic flux that will induce an emf inside the loops of wire (known as inductance). The inductive time constant is the transient time of any inductive circuit determined by the relationship between inductance and resistance.