Question

A wooden toroidal core with a square cross section has an inner radius of 14 cm and an outer radius of 16 cm. It is wound with one layer of wire (of diameter 1.0 mm and resistance per meter 0.020 Ω/m).

(a) What is the inductance of the toroid? (Ignore the thickness of the insulation on the wire.)
I got 3.82e-4 H.

(b) What is the inductive time constant of the toroid?

Answer 1

Answer with Explanation:

We are given that

Inner radius of wooden toroidal core=
`r_1=14`cm

Outer radius of wooden toroidal core=
`r_2=16` cm

Diameter of wire=1.0 mm

Resistance per meter=0.020 ohm/m

a.We have to find the inductance of the toroid.

Inner circumference of toroid=
`2\pi r_1=2* (22)/(7)* 14=88` cm=880 mm

`\pi=(22)/(7)`

1 cm=10 mm

Number of turns is roughly ,N=
`(880)/(1)\approx 880`

h=
`r_2-r_1=16-14=2 cm=0.02 m`

1 m=100 cm

Inductance of the toroid=
`(\mu_0N^2h)/(2\pi)ln(r_2)/(r_1)`

Substitute the values then, we get

`L=(4\pi* 10^(-7)* (880)^2* 0.02)/(2\pi)* ln(16)/(14)`

`\mu_0=4\pi* 10^(-7)`

`L=4.1* 10^(-4) H`

Hence, the inductance of the toroid=
`4.1* 10^(-7)` H

b.We have to find the inductive time constant of toroid.

Total length of wire=
`4(880)* (2)/(100)=70.4 m`

Because, total number of turns=880

Perimeter of square = 4 times the side of the square

Side of square shaped loop=2 cm

Resistance of wire=
`70.4* 0.02=1.408` ohm

Inductive time constant
`=(L)/(R)`

Inductive time constant=
`(4.1* 10^(-4))/(1.408)=2.91* 10^(-4) s`

Hence, the inductive time constant of toroid=
`2.91* 10^(-4) s`