Relaxation time refers to the time gap between two successive collisions of electrons in a conductor.
The relationship between the relaxation time and drift velocity is as given below.
\(\begin{array}{l}v_{d} = \left ( e\frac{E}{m} \right )T\end{array}\)
Where
- vd = drift velocity
- e = charge of electron
- E = field
- m = mass of an electron
- T = Relaxation time
So the expression for relaxation time (T) is
\(\begin{array}{l}T = \left ( v_{d}\frac{m}{e} \right )E\end{array}\)
Let
- L be the length of the conductor
- A be the area of the conductor
- n is the current density
Then, the current flowing through the conductor is \(\begin{array}{l}I = -neAv_{d}\end{array}\)
\(\begin{array}{l}I = neA\left ( e\frac{E}{m} \right )T\end{array}\)
\(\begin{array}{l}I = \frac{ne^{2}EA}{m}T\end{array}\)
The field E can be therefore expressed as –
E = V/L
Then the current flowing through the conductor becomes \(\begin{array}{l}I = \frac{ne^{2}VA}{mL}T\end{array}\)
\(\begin{array}{l}\frac{V}{I} = \frac{mL}{ne^{2}TA}\end{array}\)
From Ohm’s law
V = IR
R = V/I
\(\begin{array}{l}R = \left ( \frac{m}{ne^{2}T} \right )\frac{L}{A}\end{array}\)
\(\begin{array}{l}R = \rho \frac{L}{A}\end{array}\)
Related Questions
- What is the relation between drift velocity and electric field?
- Give two examples of drift velocity.
- Two conducting wires X and Y of the same diameter but different materials are joined in series across a battery. If the number density of electrons in X is twice that in Y, find the ratio of the drift velocity of electrons in the two wires.
- Explain the relation between current and drift velocity.
- It is known that the drift velocity of electrons is only a few mm/s for a current of a few amperes. How is it possible that a current is established almost instantaneously when a circuit is closed? For example, a bulb glows as soon as the connection is switched on. Explain.
Read More:






Comments