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.

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, $d=1.0\mathrm{mm}=1.0\times {10}^{-3}\mathrm{m}$

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:

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

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

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

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

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