# 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 m^{2}/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,

V_{d }= I/neAWhere,

- I = current
- e = electron’s charge
- A = area
- n = free density of electron

The equation of mobility is given as,

μe = V_{d}/EWhere,

- μ = mobility
- E = electric field.

**Derivation**

From Newton’s second law,

F = ma

⇒ a = F/m

_{e}*Where

F = Force exerted by the electric field, a = acceleration between collisions, m

_{e}* = effective mass of an electron.We know the force on an electron is -eE

Therefore, a = -eE/m

And drift velocity V

_{d }= aT_{c}= -eT

_{c}E/mHere, T

_{c }= 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

V_{d }= -μ_{e}EWhere

μ

_{e }= eT_{c}/m_{e}*For holes,

V_{d }= μ_{h}EWhere μ

_{h }= eT_{c}/m_{h}*

**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,

V

_{d }= I/neA= I/[ne(πr

^{2})]From the given data

V

_{d}^{‘ }= 1/neπ(4r)^{2}= (1/16)(1/neπr

^{2})

_{ }V_{d‘}= V_{d}/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} m^{2}. 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}m^{2}Charge of electron e=1.6×10

^{-19 }CoulombsWe know that

V

_{d}= I/neAV

_{d}= 2/(8.5 × 1028 × 8 × 10^{-7 }× 1.6 × 10^{-19})V

_{d }= 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, V

_{d }= 10 × 10^{-4 }m/sE = 5 × 10

^{-10}v/mTherefore, Mobility μ = |V

_{d}|/Eμ = 10 × 10

^{-4}/5 × 10^{-10}μ = 2 × 10

^{6}m^{2}/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 = V

_{d}/E ⇒ V_{d }= μE⇒ V

_{d }= μ(V/l)= μ.IR/l= μ.I.ρ.l/Al

⇒ V

_{d }= μ.I.ρ/A⇒ μ = V

_{d}.A/Iρ= (1 × 10

^{-3 }× π × 15 × 15)/(10 × 1 × 10^{-8})Therefore, μ = 7.07 × 10

^{6}m^{2}v/s.

**Question 6: Cu contains 9 × 10 ^{12 }free electrons/m^{3}. A Cu wire of cross-sectional area 8 × 10^{-5} m^{2} carries a current of 2 A. Find the drift speed of the electron?**

**Solution:**

From,

V = I/neA

V = 2/(9 × 10

^{12 }× 1.6 × 10^{-19 }× 8 × 10^{-5})⇒ V = 17.36 × 10

^{9}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 V

_{d }= 1 × 10^{-4 }m/sIncreased current I’ = 4 A

Therefore, increased drift velocity V

_{d}‘ = ?We know that,

I ∝ V

_{d}⇒ I’∝V

_{d}‘⇒ V

_{d}‘/V_{d }= I’/I⇒ V

_{d}‘ = I’V_{d}/IHence, V

_{d}‘ = 2 × 10^{-4}m/s.^{ }

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