Relation between Mobility and Electric Current
Mobility is correctly defined as the value of the drift velocity per unit of electric field strength. Hence, the quicker the particle moves at a specific electric field strength, the vast the mobility is. The mobility of any particular type of particle in a stated solid may differ with temperature. It is indicated by ‘μ’. It is expressed in m2/Vs.
Electric current
An electric current is a course of charged particles, like electrons or ions, operating through an electrical conductor. It is measured as the net rate of flow of electric charge through a plane or into a control capacity. The symbol for electric current is ‘I’. It is expressed in ampere or ampere.
From Ohm’s law
I = V/R
Where,
- I is the current flowing in the coil in amperes,
- V is the voltage in the coil in volts, and
- R is the resistance in ohms.
Relation between mobility and electric current
As we all know, the average velocity obtained by the charged particles in a conductor because of an electric field is called the Drift velocity. When a voltage is applied across the conductor, loose electrons acquire velocity in the opposite way of the electric field. Hence, there will be a small drift velocity. The equation to calculate drift velocity is,
Vd = I/neA
Where,
- I = current
- e = electron’s charge
- A = area
- n = free density of electron
The equation of mobility is given as,
μe = Vd/E
Where,
- μ = mobility
- E = electric field.
Derivation
From Newton’s second law,
F = ma
⇒ a = F/me*
Where
F = Force exerted by the electric field, a = acceleration between collisions, me* = effective mass of an electron.
We know the force on an electron is -eE
Therefore, a = -eE/m
And drift velocity Vd = aTc
= -eTcE/m
Here, Tc = mean free velocity.
We want to know how the drift velocity changes with the electric field, so we combine the loose terms together to get
Vd = -μeE
Where
μe = eTc/me*
For holes, Vd = μhE
Where μh = eTc/mh*
Note: Electron mobility and hole mobility both are positive. To account for the minus charge, a minus sign is added for electron drift velocity.
- In an electrolyte, the charge carriers are both positive and negative ions.
- In an ionized gas, the charge carriers are electrons and positive ions.
Key points
- The mobility of a charge carrier is the mean velocity through which the carrier passes towards the positive tip of the conductor beneath the related potential difference.
- The mobility of electrons, even as a charge carrier, is higher than in holes. Mobility is also defined as the capability to progress freely.
- The electron in a conductor moves with a Fermi velocity, followed by zero average velocity. If we apply voltage, which will add to this net velocity, drift is formed.
Sample Questions
Question 1: A current I flows through a uniform wire of diameter d when the electron drift velocity is V. the same current will flow through a wire of diameter d/4 made of the same material if the draft velocity of the electrons is:
Solution:
We know that,
Vd = I/neA
= I/[ne(πr2)]
From the given data
Vd‘ = 1/neπ(4r)2
= (1/16)(1/neπr2)
Vd‘ = Vd/16.
Question 2: Two wires, each having r as radius, with different materials, are connected together end to end. The densities of charge carriers in the two wires are in the ratio of 3:2, then the ratio of the drift velocity of electrons in the two wires is?
Answer:
We know that the drift velocity is inversely proportional to the density of charge carriers.
Hence, the ratio of drift velocity is 2:3.
Question 3: A Cu wire has a cross-sectional area of 8 × 10-7 m2. The density of Cu is 8.5 × 1028 m-3. Calculate the mean drift velocity of the electrons through the wire when the current is 2A?
Solution:
Given,
I = 2A
n = 8.5 × 1028 m-3
A = 8 × 10-7 m2
Charge of electron e=1.6×10-19 Coulombs
We know that
Vd = I/neA
Vd = 2/(8.5 × 1028 × 8 × 10-7 × 1.6 × 10-19)
Vd = 1.83 × 10-4 m/s.
Question 4: A charged particle having a drift velocity of 10*10-4 m/s in the electric field of 5 × 10-10 v/m, find the mobility?
Solution:
Given, Vd = 10 × 10-4 m/s
E = 5 × 10-10 v/m
Therefore, Mobility μ = |Vd|/E
μ = 10 × 10-4/5 × 10-10
μ = 2 × 106 m2/vs.
Question 5: A conducting wire has a radius of 15 mm, resistivity ρ = 1 × 10-8 ohm/m, and a current of 10 A is flowing. The drift velocity of a free electron is 1 × 10-3 m/s. Find the mobility of free electrons?
Solution:
We know that,
μ = Vd/E ⇒ Vd = μE
⇒ Vd = μ(V/l)
= μ.IR/l= μ.I.ρ.l/Al
⇒ Vd = μ.I.ρ/A
⇒ μ = Vd.A/Iρ
= (1 × 10-3 × π × 15 × 15)/(10 × 1 × 10-8)
Therefore, μ = 7.07 × 106 m2v/s.
Question 6: Cu contains 9 × 1012 free electrons/m3. A Cu wire of cross-sectional area 8 × 10-5 m2 carries a current of 2 A. Find the drift speed of the electron?
Solution:
From,
V = I/neA
V = 2/(9 × 1012 × 1.6 × 10-19 × 8 × 10-5)
⇒ V = 17.36 × 109 m/s.
Question 7: The drift velocity is 1 × 10-4 m/s when the current in a coil is 2 A. Find the drift velocity when the current becomes 4 A?
Solution:
Given, the initial current I = 2A
The initial drift velocity Vd = 1 × 10-4 m/s
Increased current I’ = 4 A
Therefore, increased drift velocity Vd‘ = ?
We know that,
I ∝ Vd
⇒ I’∝Vd‘
⇒ Vd‘/Vd = I’/I
⇒ Vd‘ = I’Vd/I
Hence, Vd‘ = 2 × 10-4 m/s.
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