A potential difference V is applied across a conductor of length ‘l’. How is the drift velocity affected when V is doubled and l is halved?

The formula for drift velocity is given as,

|vd| = eτ/mxE

However, we know that E = V/l

where V refers to the potential difference

l is the length of conductor.

Substituting this value of E in the drift velocity formula, we have:

|vd| = eτ/m(Vl)

  • Therefore, the drift velocity is proportional to the potential difference applied across the conductor.
  • The length of the conducting wire is inversely proportional to drift velocity. V becomes 2V when the potential difference applied across the conductor is doubled.
  • When the conductor’s length is half, l becomes l/2.

Putting these values in the drift velocity equation, we have:

|vd′|=eτ/m(4V/l)

On simplifying the equation further, we get:

|vd′|=4eτ/m(V/l)

We know that eτm(V/l)=|vd|

Substituting this in the above equation we get

|vd′|=4|vd|

Therefore, we can observe that when the potential difference of a conductor is doubled and the length of the conductor is half, the drift velocity increases by four times.


Related Questions

  1. What is the relation between drift velocity and electric field?
  2. Define mobility of electron.
  3. Give two examples of drift velocity.
  4. Explain the relation between current and drift velocity.
  5. Define the drift velocity of electrons in a conductor?
  6. Write the mathematical relation between Mobility And Drift Velocity of charge carriers in a conductor.
  7. 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.
  8. 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.
  9. 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:

CBSE CLASS XII Related Questions

  • 1.
    Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.


      • 2.
        Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.


          • 3.
            What is displacement current (\( i_d \))? Considering the case of charging of a capacitor, show that \( i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \). What is the value of \( i_d \) for a conductor across which a constant voltage is applied?


              • 4.
                If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.


                  • 5.
                    Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.


                      • 6.
                        The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

                          • \( \frac{q}{4 \pi \epsilon_0 l^2} \) pointing along MA
                          • \( \frac{q}{\pi \epsilon_0 l^2} \) pointing along AM
                          • \( \frac{q}{2 \pi \epsilon_0 l^2} \) pointing along AM
                          • Zero
                        CBSE CLASS XII Previous Year Papers

                        Comments


                        No Comments To Show