Define relaxation time of the free electrons drifting in a conductor? How is it related to the drift velocity of free electrons? Use this relation to deduce the expression for the electrical resistivity of the material.

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

  1. What is the relation between drift velocity and electric field?
  2. Give two examples of drift velocity.
  3. 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.
  4. Explain the relation between current and drift velocity.
  5. 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.

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CBSE CLASS XII Related Questions

  • 1.

    The number of free electrons per unit volume ($\text{m}^3$) in a wire made of copper (resistivity $1.72 \times 10^{-8}\,\Omega\,\text{m}$) is $8.5 \times 10^{28}$. Calculate the average number of collisions made by an electron in one second when an electric field is maintained in the wire.


      • 2.

        A magnet held vertically, with its north pole down, is dropped along the axis of a closed solenoid placed vertically on a table. If the observer looks down from the top,

          • the induced current will flow in the anticlockwise direction.
          • the induced current will flow in the clockwise direction.
          • no induced current will flow in the solenoid.
          • the magnet will fall with a constant velocity.

        • 3.
          A wire of length \(L\) is bent round into (i) a square coil having \(N\) turns and (ii) a circular coil having \(N\) turns. The coil in both cases is free to turn about a vertical axis coinciding with the plane of the coil, in a uniform, horizontal magnetic field and carry the same currents. Find the ratio of the maximum value of the torque acting on the square coil to that on the circular coil.


            • 4.

              A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

                • 5 V
                • 3.5 V
                • 2.5 V
                • Zero V

              • 5.

                Find the ratio $\dfrac{\lambda_\alpha}{\lambda_p}$ of the de Broglie wavelength $\lambda_\alpha$ associated with an alpha particle to the de Broglie wavelength $\lambda_p$ associated with a proton if both are moving with the same velocity.


                  • 6.
                    In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

                      • \(10\ \text{N/C}\) along \(+x\)
                      • \(10\ \text{N/C}\) along \(-x\)
                      • \(50\ \text{N/C}\) along \(+x\)
                      • \(50\ \text{N/C}\) along \(-x\)
                    CBSE CLASS XII Previous Year Papers

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