91Ó°ÊÓ

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

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:

$$\begin{gathered} L=\frac{N{\mathrm{Ï•}}_B}{i} \\ B=\frac{{\mathrm{μ}}_{0}Ni}{2\mathrm{Ï€°ù}} \\ \mathrm{Ï„}=\frac{L}{R} \end{gathered}$$

The inner circumference of toroid is as follows:

$$\begin{gathered} l=2\mathrm{π}a \\ l=2\mathrm{π}×0.1 \\ l=0.628\mathrm{m} \end{gathered}$$

The number of turns of toroid is asfollows:

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

The magnetic flux linked with toroid is

${\mathrm{ϕ}}_B={∫}_a^{b}B.dA$

Here$B=\frac{{\mathrm{μ}}_{0}Ni}{2\mathrm{π}r}$

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

role="math" localid="1661854001705" $$\begin{gathered} {\mathrm{Ï•}}_B={∫}_a^b\frac{{\mathrm{μ}}_{0}Ni}{2\mathrm{Ï€°ù}}.\left(b-a\right)dr \\ {\mathrm{Ï•}}_B=\frac{{\mathrm{μ}}_{0}Ni}{2\mathrm{Ï€}}.\left(b-a\right){∫}_a^b\frac{dr}{r} \\ {\mathrm{Ï•}}_B=\frac{{\mathrm{μ}}_{0}Ni}{2\mathrm{Ï€}}.\left(b-a\right)\mathrm{In}\left(\frac{b}{a}\right) \end{gathered}$$

The inductance of toroid is

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

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

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

The resistance ofthewire is

$$\begin{gathered} R=50.24\mathrm{m}×0.02\mathrm{Î\copyright }/\mathrm{m} \\ R=1.0048\mathrm{ohm} \end{gathered}$$

The inductive time constant is

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