Why does the resistivity of superconductors become zero?

Deanna Gregory
2022-10-09
Answered

Why does the resistivity of superconductors become zero?

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Lamar Esparza

Answered 2022-10-10
Author has **8** answers

Step 1

given statement - " why the resistivity of superconductors become zero" ?

Step 2

The superconductors are mostly normal conductors which are cooled to a temperature below their critical temperature, hence they start to act as superconductors.

The electrical energy is transported with zero or almost no energy losses in a superconductor because the resistance offered to the flow of charges is almost zero.

The resistivity of superconductors is zero because at temperature lower than the critical temperature, the inter electronic collisions and scattering of charge carriers become negligible and hence the charge carriers flow without any must hindrance in their path and ultimately the resistivity of super conductors is said to be zero.

given statement - " why the resistivity of superconductors become zero" ?

Step 2

The superconductors are mostly normal conductors which are cooled to a temperature below their critical temperature, hence they start to act as superconductors.

The electrical energy is transported with zero or almost no energy losses in a superconductor because the resistance offered to the flow of charges is almost zero.

The resistivity of superconductors is zero because at temperature lower than the critical temperature, the inter electronic collisions and scattering of charge carriers become negligible and hence the charge carriers flow without any must hindrance in their path and ultimately the resistivity of super conductors is said to be zero.

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That is the general equation for damped harmonic motion. What is the term or name that describes gamma ?

Is it called the damping constant ? I know its the ration between the resistive coefficient (b) and mass of the system (m) but what do we actually call it ?

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The equations are as follows:

${\rho}_{a}(s)={s}^{2}{\int}_{0}^{\mathrm{\infty}}T(\lambda ){J}_{1}(\lambda s)ds\phantom{\rule{2em}{0ex}}(4)$

Where ${\rho}_{a}$ is the apparent resistivity and $T(\lambda )$ is a resistivity transform function. Using a Hankel Transform, we get:

$T(\lambda )={\int}_{0}^{\mathrm{\infty}}{\rho}_{a}(s){J}_{1}(\lambda s)\left[\frac{1}{s}\right]ds\phantom{\rule{2em}{0ex}}(7)$

New variables are then defined by

$x=ln(s)\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}y=ln\left(\frac{1}{\lambda}\right)\phantom{\rule{2em}{0ex}}(8)$

Substituting (8) in (7), we get

$T(y)={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{\rho}_{a}(x){J}_{1}\left(\frac{1}{{e}^{y-x}}\right)dx\phantom{\rule{2em}{0ex}}(9)$

Please note that any missing equations were for another resistivity equation.

${\rho}_{a}(s)={s}^{2}{\int}_{0}^{\mathrm{\infty}}T(\lambda ){J}_{1}(\lambda s)ds\phantom{\rule{2em}{0ex}}(4)$

Where ${\rho}_{a}$ is the apparent resistivity and $T(\lambda )$ is a resistivity transform function. Using a Hankel Transform, we get:

$T(\lambda )={\int}_{0}^{\mathrm{\infty}}{\rho}_{a}(s){J}_{1}(\lambda s)\left[\frac{1}{s}\right]ds\phantom{\rule{2em}{0ex}}(7)$

New variables are then defined by

$x=ln(s)\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}y=ln\left(\frac{1}{\lambda}\right)\phantom{\rule{2em}{0ex}}(8)$

Substituting (8) in (7), we get

$T(y)={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{\rho}_{a}(x){J}_{1}\left(\frac{1}{{e}^{y-x}}\right)dx\phantom{\rule{2em}{0ex}}(9)$

Please note that any missing equations were for another resistivity equation.

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