Question

A wooden toroidal core with a square cross section has an inner radius of10 cm and an outer radius of 12 cm. It is wound with one layer of wire (of diameter1.0 mmand resistance per meter $\mathbf{0}\mathbf{.}\mathbf{020}\mathbf{}\mathit{\Omega }\mathbf{/}\mathit{m}$). (a) What is the inductance? (b) What is the inductive time constant of the resulting toroid? Ignore the thickness of the insulation on the wire.

Short answer

1. Inductance,$L=2.9\times {10}^{-4}\mathrm{H}$
2. Inductive time constant, $\tau =2.9\times {10}^{-4}sec$

Step-by-step solution

Given

1. Inner radius is $a=0.10\mathrm{m}$
2. Outer radius is$b=0.12\mathrm{m}$
3. Resistance per unit length is$0.020\mathrm{ohm}/\mathrm{m}$
4. Diameter of wire, $d=1.0\mathrm{mm}=1.0\times {10}^{-3}\mathrm{m}$

Understanding the concept

We have to use the formula for inductance in terms of number of turns, magnetic flux, and current, and time constant can be found in terms of inductance and resistance.

Formula:

$$\begin{gathered} L=\frac{N{\varphi }_B}{i} \\ B=\frac{{\mu }_{0}Ni}{2\mathrm{\pi r}} \\ \tau =\frac{L}{R} \end{gathered}$$

(a) Calculate Inductance of toroid

The inner circumference of toroid is as follows:

$$\begin{gathered} l=2\pi a \\ l=2\pi \times 0.1 \\ l=0.628\mathrm{m} \end{gathered}$$

The number of turns of toroid is asfollows:

$$\begin{gathered} N=\frac{l}{d}=\frac{0.628}{1.0\times {10}^{-3}} \\ N=628 \end{gathered}$$

The magnetic flux linked with toroid is

${\varphi }_B={\int }_a^{b}B.dA$

Here$B=\frac{{\mu }_{0}Ni}{2\pi r}$

The cross sectional area of toroid is $dA=\left(b-a\right)dr$

role="math" localid="1661854001705" $$\begin{gathered} {\varphi }_B={\int }_a^b\frac{{\mu }_{0}Ni}{2\mathrm{\pi r}}.\left(b-a\right)dr \\ {\varphi }_B=\frac{{\mu }_{0}Ni}{2\pi }.\left(b-a\right){\int }_a^b\frac{dr}{r} \\ {\varphi }_B=\frac{{\mu }_{0}Ni}{2\pi }.\left(b-a\right)\mathrm{In}\left(\frac{b}{a}\right) \end{gathered}$$

The inductance of toroid is

$$\begin{gathered} L=\frac{N{\varphi }_B}{i} \\ L=\frac{N}{i}\times \frac{{\mu }_{0}Ni}{2\pi }.\left(b-a\right)\mathrm{In}\left(\frac{b}{a}\right) \\ L=\frac{{\mu }_0{N}^2}{2\pi }.\left(b-a\right)\mathrm{In}\left(\frac{b}{a}\right) \\ L=\frac{4\mathrm{\pi }\times {10}^{-7}\times {628}^2}{2\mathrm{\pi }}.\left(0.12-0.1\right)\mathrm{In}\left(\frac{0.12}{0.1}\right) \\ L=2.9\times {10}^{-4}\mathrm{H} \end{gathered}$$

(b) Calculate inductive time constant of toroid.

The toroid square side is$12-10=2\mathrm{cm}=0.02\mathrm{m}$

The perimeter is$4\left(0.02\right)=0.08\mathrm{m}$

Therefore, the total length ofthewire is

$$\begin{gathered} l\text{'}=628\times 0.08 \\ l\text{'}=50.24\mathrm{m} \end{gathered}$$

The resistance ofthewire is

$$\begin{gathered} R=50.24\mathrm{m}\times 0.02\Omega /\mathrm{m} \\ R=1.0048\mathrm{ohm} \end{gathered}$$

The inductive time constant is

$$\begin{gathered} \tau =\frac{L}{R} \\ \tau =\frac{2.9\times {10}^{-4}}{1.0048} \\ \tau =2.9\times {10}^{-4}\mathrm{s} \end{gathered}$$