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3 years ago

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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 o

ne 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?

1 answer:

allochka39001 [22]3 years ago

6 0

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= $2\pi r_1=2\times \frac{22}{7}\times 14=88$ cm=880 mm

$\pi=\frac{22}{7}$

1 cm=10 mm

Number of turns is roughly ,N= $\frac{880}{1}\approx 880$

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

1 m=100 cm

Inductance of the toroid= $\frac{\mu_0N^2h}{2\pi}ln\frac{r_2}{r_1}$

Substitute the values then, we get

$L=\frac{4\pi\times 10^{-7}\times (880)^2\times 0.02}{2\pi}\times ln\frac{16}{14}$

$\mu_0=4\pi\times 10^{-7}$

$L=4.1\times 10^{-4} H$

Hence, the inductance of the toroid= $4.1\times 10^{-7}$ H

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

Total length of wire= $4(880)\times \frac{2}{100}=70.4 m$

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= $70.4\times 0.02=1.408$ ohm

Inductive time constant $=\frac{L}{R}$

Inductive time constant= $\frac{4.1\times 10^{-4}}{1.408}=2.91\times 10^{-4} s$

Hence, the inductive time constant of toroid= $2.91\times 10^{-4} s$

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