Answer:
[tex]l=0.068 m[/tex]
Explanation:
given,
inductance of solenoid = 2.07 μ H
winding of the wire = 1.19 m
Using formula of inductance
[tex]L = \dfrac{\mu_0N^2A}{l}[/tex]
L is the inductance
N is number of turns of the coil
μ₀ is permeability of free space
L is length of winding
N (2π r) = 1.19
squaring both side
4π(N²(πr²))=1.19²
N² A = 0.113
now
[tex]2.07 \times 10^{-6}= \dfrac{4\pi\times 10^{-7}\times 0.113}{l}[/tex]
[tex]l= \dfrac{4\pi\times 10^{-7}\times 0.113}{2.07 \times 10^{-6}}[/tex]
[tex]l=0.068 m[/tex]
This question involves the concepts of the inductance of an inductor and the length of the winding.
The tube should be "6.84 cm" long.
The inductance of an inductor can be given by the following formula:
[tex]L=\frac{N^2\mu_oA}{l}[/tex]
where,
L = inductance = 2.07 μH = 2.07 x 10⁻⁶ H
N = No. of turns in coil = [tex]\frac{length\ of\ wire}{circumference\ of\ tube} = \frac{1..19\ m}{2\pi r}[/tex]
μ₀ = permeability of free space = 4π x 10⁻⁷ Wb/A.m
A = Area of tube = πr²
l = length of tube = ?
Therefore,
[tex]2.07\ x\ 10^{-6}\ H = \frac{(\frac{1.19\ m}{2\pi r})^2(\pi r^2)(4\pi\ x\ 10^{-7}\ Wb/A.m)}{l}\\\\l = \frac{1.4161\ x\ 10^{-7}}{2.07\ x\ 10^{-6}}\\\\[/tex]
l = 0.0684 m = 6.84 cm
Learn more about inductance here:
https://brainly.com/question/10254645?referrer=searchResults
The attached picture shows the inductance formula.
