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Current Electricity — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsCurrent Electricity3 Marks
Question
  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.
✓

Answer

  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 $\Omega^{-1}\text{m}^{-1}\ \text{or mho (m}^{-1}).$
  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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