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# Electrical Conductivity Assignment Help

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Electrostatics - Electrical Conductivity

**Electrical Conductivity**

Current density at a point in a conductor is defined as the amount of current flowing per unit area of the conductor around that point provided the area is held in a direction normal to the current.

Let l be the current distributed uniformly across a conductor of cross-sectional area A. the magnitude of the current density for all points on that cross- section of the conductor is

**J = I/A**

Current density is a vector quantity its direction is the direction of motion of positive charge.

We know** I = An e u d**

**∴ j = I / A = n e u d **

The unit of current density is ampere** (metre)**^{-2} or (A m ^{-2})

Note for a particular surface of conductor, the current is the flux of j over that surface s and is given by

**I = ∫ j. ds**

Where ds is elementary surface area vector of as element taken over the particular surface S and integral is taken over the particular surface S and integral is taken over the surface in question.

Conductance (G) the inverse of resistance ® is called conductance of a conductor

Conductance **G = I / R**

The unit of conductance is mho or sidemen (symbol S).

Electrical conductivity the inverse of resistivity (p) of a conductor is called its electrical conductivity (a)

**σ = I/p**

The unit of electrical conductivity is

Mho m^{-1 }or Sm

The dimensions of electrical conductivity are

**σ = 1 / p = n e**^{2} τ / m = L^{– 3} x (AT)^{2} AT/M = M^{-1} L^{-3} T^{3} A^{2 }

Relation between **j σ and E**

We know **I = n Ae us = n Ae (eE / m τ) = n Ae**^{2} τ E/m

**Or I / A = ne**^{2} τ E / m

**Or j = 1 / p E [ ∴j = 1 / A and p = m / ne**^{2} τ]

∴ J = σE

(∴σ= I / p)

Relating between resistivity and electron mobility

We know that I **= n A eud ; vd = μE**

And** j = I / A = σE = E / p **

**∴ E / p = I / A = n e u d = n e μ E **

P = 1 / ne μ

Potential difference of 100 V is applied to the ends of a copper wire one metre long. Calculate the average drift velocity of the electrons? Compare it with thermal velocity at** 27.C** consider there is one conduction electron per atom. The density of copper is** 9.0 x 10**^{3} kg /m^{3}; atomic mass of copper is** 36.5 g.** Avogadro’s number = **6.0x 10**^{23} per gram – mole. Conductivity of copper is **5.18 x 10**^{7}O^{-1} Boltzmann constant =** 1.38 x 10 **^{– 23} jk^{-1}

Sol: Here; **V = 100 V, I = 1m,**

**M = 63.5g = 63.5 x 10**^{-3 }g;

P = 9.0 x 10^{3} kg / m^{3};

N = 6.0 x 1023 per gram – mole;

σ = 5.81 x 107 O^{ - 1} m^{-1}

Since **6.1023** copper atoms have a mass of **36.5g** and there is one conduction electron per atom so number density of electrons,

**N = 6.0 x 10**^{23} / 63.5 x 9.0 = 8.5 x 10^{22} cm^{-3} = 8.5 x 10^{28} m^{-3}

Electric filed E = V / I = 100 / 1 = 100 Vm^{-1}

AS J = σ E = n e vd

∴ vd = σE / ne = (5.81 x 10^{7}) x (100) / (8.5 x 10^{28}) x (1. 61 x 10 ^{– 19}) = 0.43ms^{-1}

Thermal velocity

V ms 3k B T / me = 3 x 1.38 x 10^{-23} x 300 / 9.1 x 10^{-31} = 1.17 x 10^{5} m/s

Vd / vrms = 0.43 / 1.17 x 10^{5} = 3.67 x 10^{-6}

Electrical Conductivity Assignment Help, Electrical Conductivity Homework Help, Electrical Conductivity Tutors, Electrical Conductivity Solutions, Electrical Conductivity Tutors, Electrostatics Help, Physics Tutors, Electrical Conductivity Questions Answers

**Electrical Conductivity**

Let l be the current distributed uniformly across a conductor of cross-sectional area A. the magnitude of the current density for all points on that cross- section of the conductor is

**J = I/A**

Current density is a vector quantity its direction is the direction of motion of positive charge.

We know

**I = An e u d**

**∴ j = I / A = n e u d**

The unit of current density is ampere

**(metre)**

^{-2}or (A m^{-2})Note for a particular surface of conductor, the current is the flux of j over that surface s and is given by

**I = ∫ j. ds**

Where ds is elementary surface area vector of as element taken over the particular surface S and integral is taken over the particular surface S and integral is taken over the surface in question.

Conductance (G) the inverse of resistance ® is called conductance of a conductor

Conductance

**G = I / R**

The unit of conductance is mho or sidemen (symbol S).

Electrical conductivity the inverse of resistivity (p) of a conductor is called its electrical conductivity (a)

**σ = I/p**

The unit of electrical conductivity is

Mho m

^{-1 }or Sm

The dimensions of electrical conductivity are

**σ = 1 / p = n e**

^{2}τ / m = L^{– 3}x (AT)^{2}AT/M = M^{-1}L^{-3}T^{3}A^{2 }

Relation between

**j σ and E**

We know

**I = n Ae us = n Ae (eE / m τ) = n Ae**

^{2}τ E/m**Or I / A = ne**

^{2}τ E / m**Or j = 1 / p E [ ∴j = 1 / A and p = m / ne**

^{2}τ]

∴ J = σE

(∴σ= I / p)

∴ J = σE

(∴σ= I / p)

Relating between resistivity and electron mobility

We know that I

**= n A eud ; vd = μE**

And

**j = I / A = σE = E / p**

**∴ E / p = I / A = n e u d = n e μ E**

P = 1 / ne μ

P = 1 / ne μ

Potential difference of 100 V is applied to the ends of a copper wire one metre long. Calculate the average drift velocity of the electrons? Compare it with thermal velocity at

**27.C**consider there is one conduction electron per atom. The density of copper is

**9.0 x 10**

^{3}kg /m^{3}; atomic mass of copper is

**36.5 g.**Avogadro’s number =

**6.0x 10**– mole. Conductivity of copper is

^{23}per gram**5.18 x 10**

^{7}O^{-1}Boltzmann constant =

**1.38 x 10**

^{– 23}jk^{-1}

**Here;**

Sol:

Sol:

**V = 100 V, I = 1m,**

**M = 63.5g = 63.5 x 10**

P = 9.0 x 10

N = 6.0 x 1023 per gram – mole;

σ = 5.81 x 107 O

^{-3 }g;P = 9.0 x 10

^{3}kg / m^{3};N = 6.0 x 1023 per gram – mole;

σ = 5.81 x 107 O

^{ - 1}m^{-1}

Since

**6.1023**copper atoms have a mass of

**36.5g**and there is one conduction electron per atom so number density of electrons,

**N = 6.0 x 10**

Electric filed E = V / I = 100 / 1 = 100 Vm

AS J = σ E = n e vd

∴ vd = σE / ne = (5.81 x 10

Thermal velocity

V ms 3k B T / me = 3 x 1.38 x 10

Vd / vrms = 0.43 / 1.17 x 10

^{23}/ 63.5 x 9.0 = 8.5 x 10^{22}cm^{-3}= 8.5 x 10^{28}m^{-3}Electric filed E = V / I = 100 / 1 = 100 Vm

^{-1}AS J = σ E = n e vd

∴ vd = σE / ne = (5.81 x 10

^{7}) x (100) / (8.5 x 10^{28}) x (1. 61 x 10^{– 19}) = 0.43ms^{-1}Thermal velocity

V ms 3k B T / me = 3 x 1.38 x 10

^{-23}x 300 / 9.1 x 10^{-31}= 1.17 x 10^{5}m/sVd / vrms = 0.43 / 1.17 x 10

^{5}= 3.67 x 10^{-6}