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A child (approximately 4 years old) takes her metal “slinky Toy” (a flexible coiled metal spring) and does various tests to determine that the Slinky has an inductance 130 µH, when it has been stretched to a length of 3 m. The permeability of free space is 4π × 10−7 N/A 2 . If a slinky has a radius of 4 cm, what is the total number of turns in the Slinky?

asked
User DeniSHow
by
7.5k points

1 Answer

4 votes

Answer:

248

Step-by-step explanation:

L = Inductance of the slinky = 130 μH = 130 x 10⁻⁶ H


l = length of the slinky = 3 m

N = number of turns in the slinky

r = radius of slinky = 4 cm = 0.04 m

Area of slinky is given as

A = πr²

A = (3.14) (0.04)²

A = 0.005024 m²

Inductance is given as


L = (\mu _(o)N^(2)A)/(l)


130* 10^(-6) = ((12.56* 10^(-7))N^(2)(0.005024))/(3)

N = 248

answered
User Hansemann
by
7.7k points
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