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3 years ago

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You wind a small paper tube uniformly with 153 turns of thin wire to form a solenoid. The tube\'s diameter is 5.11 mm and its le

ngth is 2.47 cm. What is the inductance, in microhenries, of your solenoid

1 answer:

lina2011 [118]3 years ago

3 0

The concept required to solve this problem is linked to inductance. This can be defined as the product between the permeability in free space by the number of turns squared by the area over the length. Recall that Inductance is defined as the opposition of a conductive element to changes in the current flowing through it. Mathematically it can be described as

$L = \frac{\mu N^2 A}{l}$

Here,

$\mu$ = Permeability at free space

N = Number of loops

A = Cross-sectional Area

l = Length

Replacing with our values we have,

$L = \frac{(4\pi *10^{-7})(153)^2(\pi (\frac{5.11*10^{-3}}{2})^2)}{2.47*10^{-2}}$

$L = 0.00002442H$

$L = 24.42\mu H$

Therefore the Inductance is $24.42\mu H$

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The equation for the variable $\rho$ is

$\rho =\frac{2\gamma \Phi+\Psi}{rg}$

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$\rho rg = 2\gamma \Phi + \Psi$

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$[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]$

2)

The term $2\gamma \Phi$ must have the same units of $\Psi$ in order to be added to it. Therefore,

$[\gamma \Phi] = [\frac{mol}{m\cdot s^2}]$

We also know that the units of $\gamma$ are $[\frac{J}{kg}]$ , therefore

$[\frac{J}{kg}][\Phi]= [\frac{mol}{m\cdot s^2}]$

And so, the fundamental units of $\Phi$ are

$[\Phi]= [\frac{mol\cdot kg}{J\cdot m\cdot s^2}]$

However, the Joules can be written as

$[J]=[kg][\frac{m^2}{s^2}]$

Therefore

$[\Phi]= [\frac{mol\cdot kg}{(kg \frac{m^2}{s^2})\cdot m\cdot s^2}]=[\Phi]= [\frac{mol}{m^3}]$

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