
The ratio of current density and an electric field is called
A. Resistivity
B. Conductivity
C. Drift velocity
D. Mobility
Answer
483.3k+ views
Hint: In this question, we need to comment on the ratio of the current density and the electric field. For this, we will use the definition of the terms involved in the problem along with the properties of the same.
Complete step by step answer:
The current density of the material is the product of the number of the electrons, charge on an electron and the drift velocity of the electron. Mathematically, $J = ne{v_d}$ where ‘J’ is the current density, ‘n’ is the number of the electrons, ‘e’ is the charge on an electron and ${v_d}$ is the drift velocity of the electrons.
Moreover, the drift velocity is also given as the ratio of the product of the charge on an electron, the magnitude of the electric field applied and the average time elapsed between the successive collisions to the mass of the electron. Mathematically, ${v_d} = \dfrac{{eET}}{m}$.
Substituting the value the drift velocity as ${v_d} = \dfrac{{eET}}{m}$ in the equation $J = ne{v_d}$, we get
$
J = ne{v_d} \\
= ne\left( {\dfrac{{eET}}{m}} \right) \\
= \dfrac{{n{e^2}ET}}{m} - - - - (i) \\
$
Now, we know that the current density is the product of the conductivity of the material and the magnitude of the electric field. So, $J = \sigma E$ where, $\sigma $ is the conductivity.
Also, the conductivity of a material is given as $\sigma = \dfrac{{n{e^2}T}}{m}$.
So, substituting the value of conductivity as $\sigma = \dfrac{{n{e^2}T}}{m}$ in the equation, $J = \sigma E$ we get
$
J = \sigma E \\
= \dfrac{{n{e^2}TE}}{m} - - - - (ii) \\
$
Now, by the equations (i) and (ii), we can say that the ratio of the current density and the electric field is the electrical conductivity of the material.
So, the correct answer is “Option B”.
Note:
Resistivity and conductivity are the properties of the material mean that every material has its unique values of resistivity and conductivity whereas resistance and conductance of a material are dependent on other variables and can be altered. The resistivity tells you how strong E needs to be to produce a given J.
Complete step by step answer:
The current density of the material is the product of the number of the electrons, charge on an electron and the drift velocity of the electron. Mathematically, $J = ne{v_d}$ where ‘J’ is the current density, ‘n’ is the number of the electrons, ‘e’ is the charge on an electron and ${v_d}$ is the drift velocity of the electrons.
Moreover, the drift velocity is also given as the ratio of the product of the charge on an electron, the magnitude of the electric field applied and the average time elapsed between the successive collisions to the mass of the electron. Mathematically, ${v_d} = \dfrac{{eET}}{m}$.
Substituting the value the drift velocity as ${v_d} = \dfrac{{eET}}{m}$ in the equation $J = ne{v_d}$, we get
$
J = ne{v_d} \\
= ne\left( {\dfrac{{eET}}{m}} \right) \\
= \dfrac{{n{e^2}ET}}{m} - - - - (i) \\
$
Now, we know that the current density is the product of the conductivity of the material and the magnitude of the electric field. So, $J = \sigma E$ where, $\sigma $ is the conductivity.
Also, the conductivity of a material is given as $\sigma = \dfrac{{n{e^2}T}}{m}$.
So, substituting the value of conductivity as $\sigma = \dfrac{{n{e^2}T}}{m}$ in the equation, $J = \sigma E$ we get
$
J = \sigma E \\
= \dfrac{{n{e^2}TE}}{m} - - - - (ii) \\
$
Now, by the equations (i) and (ii), we can say that the ratio of the current density and the electric field is the electrical conductivity of the material.
So, the correct answer is “Option B”.
Note:
Resistivity and conductivity are the properties of the material mean that every material has its unique values of resistivity and conductivity whereas resistance and conductance of a material are dependent on other variables and can be altered. The resistivity tells you how strong E needs to be to produce a given J.
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