
What is drift velocity? Obtain the equation of Ohm’s law (\[\overset{\to }{\mathop{J}}\,=\sigma \overset{\to }{\mathop{E}}\,\]) on the basis of drift velocity where parameters are in usual meaning.
Answer
507.6k+ views
Hint: In the above given question to deduce the Ohm’s law with the help of drift velocity first we will discuss what is drift velocity then we will further use the formula of specific resistance, current density and specific conductivity so obtain the given equation.
Complete step-by-step solution:
Let us first discuss what is drift velocity?
The relaxation time of free electrons drifting in a conductor is defined as the average time elapsed between two successive collisions.
Now let us Deduce the Ohm's Law: Consider a conductor of length L and cross-sectional area A. When a potential difference of V is applied across the end of the conductor, the current produced is I. If n is the number of electrons per unit volume in the conductor and \[{{V}_{d}}\] the drift velocity electrons, then the relation between current and drift velocity is
\[I=-neAVd\] let this be equation (1)
In the above equation -e is the charge on the electron
Electric field produced at each point on the wire will be equal to
\[E=\dfrac{V}{l}\]Let this be equation (2)
Where E is the electric field produced.
Let \[\Gamma \]be the relaxation time and we know that E is the electric field so, drift velocity will be
\[{{V}_{d}}\]=\[\dfrac{erE}{m}\] let this be equation (3)
\[{{V}_{d}}\]=\[\dfrac{er}{m}\dfrac{V}{l}\] on substituting the value of equation (2)
Also, we know that
\[{{V}_{d}}=\dfrac{I}{Ane}\]
So, we can say that
\[\dfrac{erE}{m}\]=\[\dfrac{I}{Ane}\]
V= \[\dfrac{ml}{e\Gamma Ane}I\]
V=\[\dfrac{m}{n{{e}^{2}}\Gamma }\dfrac{l}{A}I\]
\[\dfrac{m}{n{{e}^{2}}\Gamma }\] is the specific resistance of the substance denoted by \[\rho \]
If the physical quantities do not change and the temperature remains constant then L and A will also be constant
So, we can say that
\[\rho \dfrac{l}{A}\]is also constant which is equal to R which is the resistance of the conductor.
Now, we can say that
\[V=RI\]
So, \[V\alpha I\]
“the potential difference developed across a conductor is directly proportional to current flowing through the conductor provided the physical conditions of the conductor remain unchanged.” This is Ohm’s law.
V=\[\dfrac{m}{n{{e}^{2}}\Gamma }\dfrac{l}{A}I\]
\[\overset{\to }{\mathop{E}}\,=\dfrac{m}{n{{e}^{2}}\Gamma }\overset{\to }{\mathop{J}}\,\]
As \[\overset{\to }{\mathop{J}}\,=\dfrac{I}{A}\]which is the current density
\[\dfrac{m}{n{{e}^{2}}\Gamma }\] is the specific resistance of the substance denoted by \[\rho \]
So, we can say that
\[\overset{\to }{\mathop{E}}\,=\rho \overset{\to }{\mathop{J}}\,\]
\[\dfrac{\overset{\to }{\mathop{E}}\,}{\rho }=\overset{\to }{\mathop{J}}\,\]
\[\overset{\to }{\mathop{J}}\,=\sigma \overset{\to }{\mathop{E}}\,\]
As \[\dfrac{1}{\rho }=\sigma \] which is the specific conductivity.
Note: To solve the above question and deduce the equation we used the fundamental relationship between the drift velocity and current. Further we reduced the equations to using specific resistance and also the current density. We also used the specific conductivity to deduce the above equation.
Complete step-by-step solution:
Let us first discuss what is drift velocity?
The relaxation time of free electrons drifting in a conductor is defined as the average time elapsed between two successive collisions.
Now let us Deduce the Ohm's Law: Consider a conductor of length L and cross-sectional area A. When a potential difference of V is applied across the end of the conductor, the current produced is I. If n is the number of electrons per unit volume in the conductor and \[{{V}_{d}}\] the drift velocity electrons, then the relation between current and drift velocity is
\[I=-neAVd\] let this be equation (1)
In the above equation -e is the charge on the electron
Electric field produced at each point on the wire will be equal to
\[E=\dfrac{V}{l}\]Let this be equation (2)
Where E is the electric field produced.
Let \[\Gamma \]be the relaxation time and we know that E is the electric field so, drift velocity will be
\[{{V}_{d}}\]=\[\dfrac{erE}{m}\] let this be equation (3)
\[{{V}_{d}}\]=\[\dfrac{er}{m}\dfrac{V}{l}\] on substituting the value of equation (2)
Also, we know that
\[{{V}_{d}}=\dfrac{I}{Ane}\]
So, we can say that
\[\dfrac{erE}{m}\]=\[\dfrac{I}{Ane}\]
V= \[\dfrac{ml}{e\Gamma Ane}I\]
V=\[\dfrac{m}{n{{e}^{2}}\Gamma }\dfrac{l}{A}I\]
\[\dfrac{m}{n{{e}^{2}}\Gamma }\] is the specific resistance of the substance denoted by \[\rho \]
If the physical quantities do not change and the temperature remains constant then L and A will also be constant
So, we can say that
\[\rho \dfrac{l}{A}\]is also constant which is equal to R which is the resistance of the conductor.
Now, we can say that
\[V=RI\]
So, \[V\alpha I\]
“the potential difference developed across a conductor is directly proportional to current flowing through the conductor provided the physical conditions of the conductor remain unchanged.” This is Ohm’s law.
V=\[\dfrac{m}{n{{e}^{2}}\Gamma }\dfrac{l}{A}I\]
\[\overset{\to }{\mathop{E}}\,=\dfrac{m}{n{{e}^{2}}\Gamma }\overset{\to }{\mathop{J}}\,\]
As \[\overset{\to }{\mathop{J}}\,=\dfrac{I}{A}\]which is the current density
\[\dfrac{m}{n{{e}^{2}}\Gamma }\] is the specific resistance of the substance denoted by \[\rho \]
So, we can say that
\[\overset{\to }{\mathop{E}}\,=\rho \overset{\to }{\mathop{J}}\,\]
\[\dfrac{\overset{\to }{\mathop{E}}\,}{\rho }=\overset{\to }{\mathop{J}}\,\]
\[\overset{\to }{\mathop{J}}\,=\sigma \overset{\to }{\mathop{E}}\,\]
As \[\dfrac{1}{\rho }=\sigma \] which is the specific conductivity.
Note: To solve the above question and deduce the equation we used the fundamental relationship between the drift velocity and current. Further we reduced the equations to using specific resistance and also the current density. We also used the specific conductivity to deduce the above equation.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

