Define the term ‘conductivity’ of a metallic wire. Write its SI unit.

Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E.

Q: 1. Define the term ‘conductivity’ of a metallic wire. Write its SI unit. 2. Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E.

1. Conductivity of a metallic wire is defined as the conductance (reciprocal of resistance) of a metallic wire of unit side and unit cross sectional area perpendicular to the current flow. Alternate Answer Conductivity is defined as the reciprocal of resistvity of a metallic wire. S.I. unit $\Omega^{-1}\text{m}^{-1}\ \text{or mho (m}^{-1}).$ 2. $\text{R}=\frac{\rho\text{l}}{\text{A}}$ $\rho=\frac{\text{RA}}{\text{l}}$ but $\text{R}=\frac{\text{ml}}{\text{ne}^2\text{A}\tau}$ So, $\rho=\Big(\frac{\text{ml}}{\text{ne}^2\text{A}\tau}\Big)\frac{\text{A}}{\text{l}}$ $\rho=\frac{\text{m}}{\text{ne}^2\tau}\ \text{or}\ \sigma=\frac{\text{ne}^2\tau}{\text{m}}\ ....(1)$ where $\sigma=\frac{1}{\rho}=$ conductivity $\tau=$ relaxation time. n = number density of electron. e = charge on one electron. m = mass of one elecrtron. Now, current density is given by: $\text{J}=\text{neV}_\text{d}$ $\text{J}=\text{ne}\Big(\frac{\text{eE}\tau}{\text{m}}\Big)$ $\Big[\because\ \text{V}_{\text{d}}=\frac{\text{eE}\tau}{\text{m}}\Big]$ $\text{J}=\frac{\text{ne}^2\tau}{\text{m}}\text{E}$ From equation (1) $\text{J}=\sigma\text{E}\ \text{or }\ \vec{\text{J}}=\sigma\vec{\text{E}}$

Question

Define the term ‘conductivity’ of a metallic wire. Write its SI unit.
Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E.

CBSE DELHI - OUTSIDE DELHI - FOREIGN SET 2 2018

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Solution

Conductivity of a metallic wire is defined as the conductance (reciprocal of resistance) of a metallic wire of unit side and unit cross sectional area perpendicular to the current flow.

Alternate Answer
Conductivity is defined as the reciprocal of resistvity of a metallic wire.
S.I. unit $\Omega^{-1}\text{m}^{-1}\ \text{or mho (m}^{-1}).$

$\text{R}=\frac{\rho\text{l}}{\text{A}}$

$\rho=\frac{\text{RA}}{\text{l}}$
but $\text{R}=\frac{\text{ml}}{\text{ne}^2\text{A}\tau}$
So, $\rho=\Big(\frac{\text{ml}}{\text{ne}^2\text{A}\tau}\Big)\frac{\text{A}}{\text{l}}$
$\rho=\frac{\text{m}}{\text{ne}^2\tau}\ \text{or}\ \sigma=\frac{\text{ne}^2\tau}{\text{m}}\ ....(1)$
where $\sigma=\frac{1}{\rho}=$ conductivity
$\tau=$ relaxation time.
n = number density of electron.
e = charge on one electron.
m = mass of one elecrtron.
Now, current density is given by:
$\text{J}=\text{neV}_\text{d}$
$\text{J}=\text{ne}\Big(\frac{\text{eE}\tau}{\text{m}}\Big)$ $\Big[\because\ \text{V}_{\text{d}}=\frac{\text{eE}\tau}{\text{m}}\Big]$
$\text{J}=\frac{\text{ne}^2\tau}{\text{m}}\text{E}$
From equation (1)
$\text{J}=\sigma\text{E}\ \text{or }\ \vec{\text{J}}=\sigma\vec{\text{E}}$

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