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Drift velocity refers to the average velocity acquired by moving electrons under the influence of an applied electric field. In a conductor, electrons move freely and randomly unless they are not subjected to an external force. When an electric field is applied, it forces the electrons to move towards the higher potential. The net velocity of these electrons is termed as drift velocity. It is given by;
Vd = I / neA
where; Vd = Drift velocity of electrons
I = Current
n = free electron density
e = charge of an electron
A = area of cross section
The SI unit of drift velocity is m/s. It is directly proportional to the current and the applied electric field. The current generated inside the conductor through which the electrons flow is called drift current.
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Important Questions on Drift Velocity
Ques 1. When electrons drift in a metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction? (2 Marks)
Ans. When an electric field is applied all the electrons start moving towards the end with higher potential but in the absence of electric field, electrons move randomly without any specific direction with random velocities.
Ques 2. Define the term mobility of charge carriers in a conductor. Write its SI unit. (2 Marks)
Ans. Mobility of charge carriers is the magnitude of drift velocity per unit electric field. It is expressed as;
μ = | Vd / E | = e\(\tau\) / m
The SI unit of mobility is m2/V.s
Ques 3. Explain how the mobility of electrons in a conductor changes when the potential difference applied is doubled, keeping the temperature of the conductor constant. (2 Marks)
Ans. Mobility, μ = vd/E = vd/(V/l)
So, when potential is doubled, drift velocity also gets doubled, therefore, there will be no change in mobility.
Ques 4. Derive an expression for the resistivity of a good conductor in terms of the relaxation time of electrons. (3 Marks)
Ans. Drift velocity of an electron under the influence of Electric field is given by;
Vd = (eE/m) \(\tau\)
Vd = (eV/ml) \(\tau\)
We Know, I = neA (eV/ml) \(\tau\)
R = V/I = ml/ne2\(\tau\)A
R = ρ (l/A)
ml/ne2\(\tau\)A = ρ (l/A)
Ques 5. Obtain the expression for the current through a conductor in terms of ‘drift velocity’. (3 Marks)
Ans. We know, electric field is given by;
E = V/l
Let n be the number of free electrons per unit volume of the conductor.
Total number of free electrons in the conductor = n volume of the conductor
Hence, Q = (nAI)e
Time taken by the charge to cross the conductor of length l is given by;
t = 1/Vd (Vd = drift velocity of the electrons)
I = Q/t
I = nAle / (1/Vd)
I = neAVd
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| Related Topics | ||
|---|---|---|
| Average Velocity Formula | Derivation of Drift Velocity | Potentiometer |
| Carbon Resistor | Circuit Diagram | Electric Charge |
| Kirchhoff’s Laws | Current Density | Unit of Voltage |
Ques 6. Write a relation between current and drift velocity of electrons in a conductor. Use this relation to explain how the resistance of a conductor changes with the rise in temperature. (3 Marks)
Ans. Relation between current and drift velocity is given by;
I = A neVd
When temperature is increased, resistance of the conductor increases due to increase in frequent collisions of electrons with each other. This leads to a decrease in the drift velocity.
Ques 7. Estimate the average drift speed of conduction of electrons in a copper wire of cross-sectional area 1 x 10-7 m2 carrying a current of 1.5 A. Assume the density of conduction electrons to be 9 x 1028 m-3. (3 Marks)
Ans. Given, A = 1 x 10-7 m2 , I = 1.5 A, n = 9 x 1028 m-3, Vd = ?
I = neAVd
Vd = I/neAVd
Vd = 1.5 / (9 x 1028) x (1.6 x 10-19)(1 x 10-7)
Vd = 10.4 x 10-4 m/s
Ques 8. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 2.5 x 10-7 m2 carrying a current of 1.8 A. Assume the density of conduction electrons to be 9 x 1028 m-3. (3 Marks)
Ans. A = 2.5 x 10-7 m2, I = 1.8 A, m = 9 x 1028 m-3, Vd = ?
I = neAVd
Vd = I/neA
Vd = 1.8 / (9 x 1028) (1.6 x 10-19)(2.5 x 10-27)
Vd = 0.05 x 10-2
Vd = 0.5 x 10-3m
Vd = 0.5 mm
Ques 9. Using the concept of drift velocity of charge carriers in a conductor, deduce the relationship between current density and resistivity of the conductor. (5 Marks)
Ans. We know, Vd = (eE/m) \(\tau\)……..(i)
where;
e = charge of electron
E = Intensity of electric field
m = mass of electron
\(\tau\) = relaxation time
I = neAVd ………(ii)
Putting value of Vd from eq (i) in (ii),
I = neA (eE/m)\(\tau\)
I = (ne2A/m) \(\tau\)E
Current Density, J = I/A
J = (ne2A/mA) \(\tau\)E
J = (ne2/m) \(\tau\)E
J = (1/p) E (∵p = m/ne2\(\tau\)) ( p- resistivity of the conductor)
Ques 10. A conductor of length L is connected to a DC source of emf E. If this conductor is replaced by another conductor of the same material and same area of cross-section but of length 3L, how will the drift velocity change? (5 Marks)
Ans. Let ‘m’ be the mass of an electron and ‘e’ be the charge on it. When an external electric field ‘E’ is applied, the acceleration gained by an electron is;
F = ma
a = F/m
a = eE/m
Let v1, v2, v3 …….vn be final velocities of electrons then average velocity of the electron is given by,
vd = v1 + v2 + v3 …. vn / n ……..(i)
vd = (u1 + at1) + (u2 + at2) + …. (un + atn) / n [∵ v = u + at]
vd = (u1 + u2 + … un)/n + a (t1 + t2 + … tn)/n
vd = 0 + a
vd = a
∴ vd = (eE/n)
(where = t1 + t2 + …. tn / n is the average time between two successive collisions called relaxation time)
Now, vd = e/m (E/L)
∴ vd ∝ 1/L
Hence, when length is tripled, the drift velocity becomes one-third.
Ques 11. Derive an expression for drift velocity of free electrons. (5 Marks)
Ans. When a potential difference is applied across a conductor, an electric field is produced and force acting on the electrons is -Ee.
Fe = -Ee = -(V/l) e (∵ E = V/l)
a = -F/m = -eV/m
And we know, v = u + at
u = 0 and t =\(\tau\)
vd = -a\(\tau\)
vd = (-eV/lm)\(\tau\)
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