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USPshnik [31]
3 years ago
9

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?
Physics
1 answer:
allochka39001 [22]3 years ago
6 0

Answer with Explanation:

We are given that

Inner radius of wooden toroidal core=r_1=14cm

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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