Current Density

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Current Density

Current density reveals how electric current is distributed across a conductor. A higher current density signifies a greater concentration of current within a specific area. In practical applications, understanding and managing current density is essential for designing electrical components and ensuring their optimal functioning. For instance, in wiring and circuit design, controlling current density helps prevent overheating and guarantees the safe, efficient operation of electronic devices.

1.0Definition of Current Density

It is defined as the current flowing per unit cross-sectional area of a conductor.
It gives us information about magnitude and direction of current passing through an area.

2.0Types of Current

On the basis of source

Direct current-Source of Direct current is Battery: An electric current is classified as direct current (DC) if both its magnitude and direction remain constant over time.
Alternating current-Source of Alternating current is A.C Generator: An electric current is said to be alternating current if its magnitude changes with time and polarity reverses periodically.

Image depicting direct and alternating current

Average Electric Current -If q charge flows for t time interval.

$i_{avg} = \frac{Δ q}{Δ t}$

Instantaneous Current-The value of current at particular instant.

$i = lim_{Δ t \to 0} (\frac{Δ q}{Δ t}) = \frac{d q}{d t}$

$d q = i d t$

$q = \int i d t$

3.0Current Density Formula

Current density is a vector quantity having direction of electric field or direction of current flow,mathematically represented as,

$J = \frac{I}{A}$

Formula of current density when the vector quantity has direction of electric field

Current density formula when the direction of the vector quantity has different direction than that of the magnetic field

Current density(J)= $\frac{I}{A Cos θ}$

$I = J A Cos θ$

$I = J \cdot A$ ,

$θ$ is the angle between the area vector and direction of current flow.

For Non-Uniform Distribution of Current:

If dI is current flowing through small elementary area then current density

Derivation for current density when the current distribution is not uniform

$(J) = \frac{d I}{d A Cos θ}$

$d I = d A Cos θ$

$I = \int J (d A Cos θ)$

4.0Units of Current Density

$SI Unit of (J) = \frac{I}{A} = \frac{Ampere}{metre ^{2}} = \frac{A}{m ^{2}}$

5.0Examples of Current Density

In household and industrial electrical wiring, current density helps in determining the appropriate wire gauge. For instance, a copper wire carrying a high current might have a current density of 3 A/mm². This helps in selecting a wire that can handle the electrical load without overheating.
During electroplating processes, current density influences the rate of metal deposition.

6.0Relation between Current density and Electric Field

From $J = \frac{I}{A} = \frac{n e A V _{d}}{A} = n e V_{d}$

$J = n e V_{d}$

Also $V_{d} = \frac{e E τ}{m}$

$J = n e \frac{e E τ}{m} = \frac{n e ^{2} E τ}{m} = (\frac{n e ^{2} τ}{m}) E$

or

$J = σ E$

$(σ = \frac{n e ^{2} τ}{m}) = i sco n d u c t i v i t yo f co n d u c t or$

is constant for any conductor, is independent of shape, size, volume of conductor.
In vector form $J = σ E$ This relationship is called microscopic Ohm’s Law.
Conductivity measures how effectively a material can conduct electricity.

$ρ = \frac{1}{σ} = \frac{m}{n e ^{2} τ}$ ( $ρ$ is the resistivity of a material)

$σ$ and $ρ$ are independent of conductor’s shape and size but depend on nature of material and temperature.

7.0Key considerations for current density

Current density in a non uniform cross section

For a given conductor current does not change with change in its cross-section because current is simply the rate of flow of charge. If a steady current flows in a metallic conductor of non-uniform cross-section:

I is same along the wire
Current density, Electric Field Strength, Drift Velocity are inversely proportional to the Area.

Here $I_{1} = I_{2}$ but $A_{1} < A_{2}$

So $J_{1} > J_{2}, E_{1} > E_{2}, v_{d 1} > v_{d 2}$

8.0Sample Questions on Current Density

Q-1.The current density within a cylindrical conductor with radius R varies as $J = J_{0} (1 - \frac{r}{R})$ ,where r denotes the distance from the central axis. This variation results in a maximum current density $J_{0}$ at the axis (when r = 0 , decreasing linearly to zero at the surface when r = R) . Express the current in terms of $J_{0}$ and the cross-sectional area A of the conductor.

Solution:

Here we consider the elementary ring of radius r and thickness dr.

Area of elementary ring,

$d A = (2 π r) d r$

From

$I = \int J \cdot d A = \int Jd A Cos θ$

$I = \int_{0}^{R} J_{0} (1 - \frac{r}{R}) (2 π r d r)$

$I = 2 π J_{0} \int_{0}^{R} (r - \frac{r ^{2}}{R}) d r = 2 π J_{0} [\frac{r ^{2}}{2} - \frac{r ^{3}}{3 R}]_{0}^{R} = 2 π J_{0} [\frac{R ^{2}}{2} - \frac{R ^{3}}{3 R}] = 2 π J_{0} [\frac{R ^{2}}{2} - \frac{R ^{2}}{3}]$

$I = 2 π J_{0} [\frac{R ^{2}}{6}] = π J_{0} \frac{R ^{2}}{3}$

$I = \frac{J _{0} A}{3} (∴ A = π R^{2})$

Q-2.For Conductor of non-uniform cross sectional-area, compare current density at section 1, 2 and 3.

Solution:

Problems on current density with non uniform cross section

Current through each area of cross-section is the same.

$J_{1} = \frac{I}{A _{1}}; J_{2} = \frac{I}{A _{2}}; J = \frac{I}{A _{3}}$

$A_{1} > A_{2} > A_{3} \Rightarrow J_{3} > J_{2} > J_{1}$

Q-3.Figure shown above is representing a current carrying wire with non-uniform cross sectional area. Find out relationship between physical quantities on their two ends.

Solution:

Conductor with non uniform cross section

Electric current is the rate at which electric charge flows and is independent of the cross-sectional area of the conductor. Hence $i_{1} = i_{2} = i$
$i = n A e v_{d} \Rightarrow v_{d} = \frac{i}{n e A}$

$v_{d} \propto \frac{1}{A} \Rightarrow A_{1} > A_{2}$

$v_{d 2} > v_{d 1}$

$J = n e v_{d}$ , $J \propto v_{d}$

$J_{1} < J_{2}$

$J = σ E \Rightarrow J \propto E$

$E_{2} > E_{1}$

Q-4.Given a copper wire with a length of 10 m and a radius of $(\frac{1 0 ^{- 2}}{π}) m m$ , which has an electrical resistance of 10 , Calculates the current density in the wire when the electric field strength is 10 V/m.

Solution:

$l = 10 m, r = \frac{1 0 ^{- 2}}{π} m, R = 10Ω, E = 10 V / m$

$J = σ E$

$(∴ ρ = \frac{1}{σ})$

$J = \frac{E}{ρ}$

$(∴ ρ = \frac{R A}{l})$

$J = \frac{El}{R A} = \frac{10 \times 10}{10 \times π ( \frac{1 0 ^{- 2}}{π} ) ^{2}} = 1 0^{5} A / m^{2}$

Current Density

Current density reveals how electric current is distributed across a conductor. A higher current density signifies a greater concentration of current within a specific area. In practical applications, understanding and managing current density is essential for designing electrical components and ensuring their optimal functioning. For instance, in wiring and circuit design, controlling current density helps prevent overheating and guarantees the safe, efficient operation of electronic devices.

1.0Definition of Current Density

It is defined as the current flowing per unit cross-sectional area of a conductor.
It gives us information about magnitude and direction of current passing through an area.

2.0Types of Current

On the basis of source

Direct current-Source of Direct current is Battery: An electric current is classified as direct current (DC) if both its magnitude and direction remain constant over time.
Alternating current-Source of Alternating current is A.C Generator: An electric current is said to be alternating current if its magnitude changes with time and polarity reverses periodically.

Average Electric Current -If q charge flows for t time interval.

$i_{avg} = \frac{Δ q}{Δ t}$

Instantaneous Current-The value of current at particular instant.

$i = lim_{Δ t \to 0} (\frac{Δ q}{Δ t}) = \frac{d q}{d t}$

$d q = i d t$

$q = \int i d t$

3.0Current Density Formula

Current density is a vector quantity having direction of electric field or direction of current flow,mathematically represented as,

$J = \frac{I}{A}$

Current density(J)= $\frac{I}{A Cos θ}$

$I = J A Cos θ$

$I = J \cdot A$ ,

$θ$ is the angle between the area vector and direction of current flow.

For Non-Uniform Distribution of Current:

If dI is current flowing through small elementary area then current density

$(J) = \frac{d I}{d A Cos θ}$

$d I = d A Cos θ$

$I = \int J (d A Cos θ)$

4.0Units of Current Density

$SI Unit of (J) = \frac{I}{A} = \frac{Ampere}{metre ^{2}} = \frac{A}{m ^{2}}$

5.0Examples of Current Density

In household and industrial electrical wiring, current density helps in determining the appropriate wire gauge. For instance, a copper wire carrying a high current might have a current density of 3 A/mm². This helps in selecting a wire that can handle the electrical load without overheating.
During electroplating processes, current density influences the rate of metal deposition.

6.0Relation between Current density and Electric Field

From $J = \frac{I}{A} = \frac{n e A V _{d}}{A} = n e V_{d}$

$J = n e V_{d}$

Also $V_{d} = \frac{e E τ}{m}$

$J = n e \frac{e E τ}{m} = \frac{n e ^{2} E τ}{m} = (\frac{n e ^{2} τ}{m}) E$

or

$J = σ E$

$(σ = \frac{n e ^{2} τ}{m}) = i sco n d u c t i v i t yo f co n d u c t or$

is constant for any conductor, is independent of shape, size, volume of conductor.
In vector form $J = σ E$ This relationship is called microscopic Ohm’s Law.
Conductivity measures how effectively a material can conduct electricity.

$ρ = \frac{1}{σ} = \frac{m}{n e ^{2} τ}$ ( $ρ$ is the resistivity of a material)

$σ$ and $ρ$ are independent of conductor’s shape and size but depend on nature of material and temperature.

7.0Key considerations for current density

For a given conductor current does not change with change in its cross-section because current is simply the rate of flow of charge. If a steady current flows in a metallic conductor of non-uniform cross-section:

I is same along the wire
Current density, Electric Field Strength, Drift Velocity are inversely proportional to the Area.

Here $I_{1} = I_{2}$ but $A_{1} < A_{2}$

So $J_{1} > J_{2}, E_{1} > E_{2}, v_{d 1} > v_{d 2}$

8.0Sample Questions on Current Density

Q-1.The current density within a cylindrical conductor with radius R varies as $J = J_{0} (1 - \frac{r}{R})$ ,where r denotes the distance from the central axis. This variation results in a maximum current density $J_{0}$ at the axis (when r = 0 , decreasing linearly to zero at the surface when r = R) . Express the current in terms of $J_{0}$ and the cross-sectional area A of the conductor.

Solution:

Here we consider the elementary ring of radius r and thickness dr.

Area of elementary ring,

$d A = (2 π r) d r$

From

$I = \int J \cdot d A = \int Jd A Cos θ$

$I = \int_{0}^{R} J_{0} (1 - \frac{r}{R}) (2 π r d r)$

$I = 2 π J_{0} \int_{0}^{R} (r - \frac{r ^{2}}{R}) d r = 2 π J_{0} [\frac{r ^{2}}{2} - \frac{r ^{3}}{3 R}]_{0}^{R} = 2 π J_{0} [\frac{R ^{2}}{2} - \frac{R ^{3}}{3 R}] = 2 π J_{0} [\frac{R ^{2}}{2} - \frac{R ^{2}}{3}]$

$I = 2 π J_{0} [\frac{R ^{2}}{6}] = π J_{0} \frac{R ^{2}}{3}$

$I = \frac{J _{0} A}{3} (∴ A = π R^{2})$

Q-2.For Conductor of non-uniform cross sectional-area, compare current density at section 1, 2 and 3.

Solution:

Current through each area of cross-section is the same.

$J_{1} = \frac{I}{A _{1}}; J_{2} = \frac{I}{A _{2}}; J = \frac{I}{A _{3}}$

$A_{1} > A_{2} > A_{3} \Rightarrow J_{3} > J_{2} > J_{1}$

Q-3.Figure shown above is representing a current carrying wire with non-uniform cross sectional area. Find out relationship between physical quantities on their two ends.

Solution:

Electric current is the rate at which electric charge flows and is independent of the cross-sectional area of the conductor. Hence $i_{1} = i_{2} = i$
$i = n A e v_{d} \Rightarrow v_{d} = \frac{i}{n e A}$

$v_{d} \propto \frac{1}{A} \Rightarrow A_{1} > A_{2}$

$v_{d 2} > v_{d 1}$

$J = n e v_{d}$ , $J \propto v_{d}$

$J_{1} < J_{2}$

$J = σ E \Rightarrow J \propto E$

$E_{2} > E_{1}$

Q-4.Given a copper wire with a length of 10 m and a radius of $(\frac{1 0 ^{- 2}}{π}) m m$ , which has an electrical resistance of 10 , Calculates the current density in the wire when the electric field strength is 10 V/m.

Solution:

$l = 10 m, r = \frac{1 0 ^{- 2}}{π} m, R = 10Ω, E = 10 V / m$

$J = σ E$

$(∴ ρ = \frac{1}{σ})$

$J = \frac{E}{ρ}$

$(∴ ρ = \frac{R A}{l})$

$J = \frac{El}{R A} = \frac{10 \times 10}{10 \times π ( \frac{1 0 ^{- 2}}{π} ) ^{2}} = 1 0^{5} A / m^{2}$

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Current Density

1.0Definition of Current Density

2.0Types of Current

3.0Current Density Formula

4.0Units of Current Density

5.0Examples of Current Density

6.0Relation between Current density and Electric Field

7.0Key considerations for current density

8.0Sample Questions on Current Density

Table of Contents

Current Density

1.0Definition of Current Density

2.0Types of Current

3.0Current Density Formula

4.0Units of Current Density

5.0Examples of Current Density

6.0Relation between Current density and Electric Field

7.0Key considerations for current density

8.0Sample Questions on Current Density

Table of Contents