## What is Current?

In physics, the term current is expressed as the rate of flow of electric charges across the area per unit time (I=q/t). It propagates along in the direction of the motion of positive charges. When the current is caused by the motion of electrons, the direction of the current is opposite to the flow.

## Microscopic view of current

Generally, we all see the macroscopic form of all objects in nature. In order to understand the nature of the object or an atom, we can see them at a microscopic level. Current (I) is a macroscopic and scalar quantity. The related microscopic quantity is current density (J) and it is a vector product. It has both direction and scalar magnitude. In a conductor, the current density is higher, more than the electric current.

The amount of charge passing per second through a unit area is called current density.

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

Where,

q=Charge in coulomb

t=Time in second

I=Current in Ampere

A=Area of the cross-section in square meter

The unit is Am^{-2}.

### Expression for Current Density

Let A be the cross- sectional area of a wire whose length is L and an electric field $\overrightarrow{E}$ is given from, it travels in left to right direction. The free electrons go towards the right with a stable drift velocity. The number of free electron per cubic meter be n.

Number of electrons in the wire = $n\left(AL\right)$

The magnitude of charge q in the wire is,

$q=\left(nAL\right)e$ (1)

Where,

e=Charge of an electron.

t=Time taken by an electron to pass out from the wire.

It is given by,

$t=L{v}_{d}$ (2)

Where,

L is the length of the wire.

${v}_{d}$is the drift velocity.

Hence, the electric current is,

$I=\frac{q}{t}=\frac{\left(nAL\right)e}{L{v}_{d}}\phantom{\rule{0ex}{0ex}}I=nAevd$ (3)

Current density J is,

$J=\frac{I}{A}=\frac{nAe{v}_{d}}{A}\phantom{\rule{0ex}{0ex}}J=nAe{v}_{d}$ (4)

This equation is effective only when the current direction is horizontal to area A. The vector form of this equation is,

$\overrightarrow{J}=ne\overrightarrow{{v}_{d}}$ (5)

The factor (ne) is volume charge density ρ of carriers. For positive carriers (ne) is positive, J and ${v}_{d}$ also directed in positive direction. For electron (ne) is negative and J and ${v}_{d}$ faces alternate directions.

## Microscopic view of Ohm’s law

In this topic, we deduct the microscopic form of ohms law from the current density. It is experimentally proved that in a metal conductor at a constant temperature, the current density J is directly proportional to the electric field E.

$J=\sigma E$ (6)

$\sigma -$Conductivity of the conductor (mho/m).

The above equation is the vector form of ohm’s law. The reciprocal of resistivity ρ is called conductivity.

$E=\rho J$ (7)

Consider the simple circuit as shown in the figure. A rod of length L is connected to the battery. From the circuit diagram, the electrical field along the rod is traveling from point M to N.

$E=VL$

Where, V = potential applied between the points M and N. Thus,

$J=\sigma \frac{V}{L}$

The electric current I is,

$I=JA=\sigma \left(\frac{V}{L}\right)A\phantom{\rule{0ex}{0ex}}I=\frac{\sigma VA}{}L\phantom{\rule{0ex}{0ex}}\frac{V}{I}=\frac{L}{\sigma A}=\rho \frac{L}{A}$ (8)

The ratio of voltage by the current is the resistance of the rod. Therefore,

$R=\rho \frac{L}{A}$ (9)

From (8) and (9),

$\frac{V}{I}=R\phantom{\rule{0ex}{0ex}}V=IR$

The strength of the electric current is directly proportional to the potential difference and inversely proportional to the resistance of the circuit.

## Drude-Lorentz theory

Some postulates given by Lorentz Drude theory are:

- There are n numbers of free electrons present in a metal. These free electrons move all over the metal volume as the gas molecules move in a container.
- The electrons make collisions from time to time with the positive ion that is fixed in the atomic lattice and sometimes they collide with themselves. We may avoid the collisions between the electrons while comparing the collisions between ion core and electrons.
- When the electric field is absent, the free electrons move in an arbitrary manner, probably in all directions which makes the current density is zero.
- When an external field is supplied, the electrons drift slowly with some average velocity, known as average drift velocity, in the directions which is opposite to the electric field. This velocity of electrons superimposes over their random velocity.
- The mean distance traveled by free electrons between two successive collisions with the positive ions is called mean free path λ .

## Electrical conductivity

Let m and e be the mass and charge of an electron which is kept in the field E, it experienced an acceleration a is given by,

$A=\frac{eE}{m}$ (10)

Where,

e=charge of the electron

E=electric field

m=mass the electron

The average time gap between two consecutive collisions of an electron with positive ions τ.

Increase in drift velocity in time$\tau =a\tau =\frac{eE\tau}{m}$

On next collision the velocity becomes zero and again it gradually gets the value eEτm before getting into another collision and so on. It is given by,

${v}_{d}=\frac{eE\tau}{2m}$

The current density J is,

$J=ne{v}_{d}=\frac{n{e}^{2}E\tau}{2m}$ (12)

Where n is the number of atoms in the wire.

$T=timeperiod$

From equation(6), electrical conductivity is,

$\sigma =JE=\frac{n{e}^{2}\tau}{2m}$ (13)

Whereas, electrical resistivity is,

$\rho =\frac{1}{\sigma}=\frac{2m}{n{e}^{2}\tau}$

Let λ be the mean free path and v be the root mean square value of thermal velocity of electrons.

The time taken τ is given by,

$\tau =\frac{\lambda}{v}$

Equation (13) becomes,

Electrical conductivity, $\sigma =\frac{n{e}^{2}\lambda}{2mv}$ (14)

### Kinetic energy of free electrons

Due to the movement of the particles within the system, the system possesses some energy and that energy is called thermal energy. Whereas, thermal velocity depicts the velocity of a particle in the system as well as heat or temperature in it.

If the electric field is absent in a semiconductor, the electrons move in a random motion due to the thermal energy. The average velocity of these motions is called thermal velocity.

The formula of kinetic energy for an ideal gas is,

$\frac{1}{2}m{v}^{2}=\frac{3}{2}kT\phantom{\rule{0ex}{0ex}}v=\frac{3kT}{m}\phantom{\rule{0ex}{0ex}}\sigma =\frac{n{e}^{2}\lambda}{6kT}$

## Deduction of Ohms law from Drude-Lorentz theory

From equation (14),

$J=\sigma E=\frac{n{e}^{2}\lambda}{2mE}$

At a certain temperature, the mean free path and thermal velocity become constant.

J α E

Le I is the current flowing through the conductor whose length is l and cross-sectional area A and the potential difference is V. Therefore,

$J=\frac{I}{A}andE=\frac{V}{L}$

$\frac{I}{A}\alpha \frac{V}{L}$

At a particular temperature, I and A of the conductor are constants.

I α V

That is at a given temperature, the electric current flowing through a conductor is directly proportional to the potential difference between its two ends, which is ohms law.

## Formulas

The formula of current density is,

$J=ne{v}_{d}$

Electrical conductivity of the wire is,

$\sigma =\frac{n{e}^{2}\lambda}{2mv}$

## Context and Applications

It is a basic and important topic in physics for all students, undergraduates, and postgraduates mainly for, Bachelors and masters in science (physics), Bachelors in technology (electrical engineering).

## Practice Problems

**Question 1**: Compute the drift velocity of free electrons of a copper wire whose cross sectional area is 0.7 mm^{2 }and carries a electric current of 0.5 A. Free electron density of copper is $8.5\times {10}^{28}{m}^{-3}.$

$a.5.2\times {10}^{-3}m/s\phantom{\rule{0ex}{0ex}}b.7.5\times {10}^{-3}m/s\phantom{\rule{0ex}{0ex}}c.0.052\times {10}^{-3}m/s\phantom{\rule{0ex}{0ex}}d.1.6\times {10}^{-3}m/s$

**Answer**: The correct option is (c).

**Given data**:

Area $A=0.7m{m}^{2}=0.7\times {10}^{-6}{m}^{2}$

Current I=0.5 A

Free electron density of copper n=$8.5\times {10}^{28}{m}^{-3}$

**Explanation:**

The relation between the drift velocity and the current is,

${v}_{d}=\frac{I}{neA}\phantom{\rule{0ex}{0ex}}{v}_{d}=\frac{0.5}{8.5x{10}^{28}{m}^{-3}\times 1.6\times {10}^{-19}C\times 0.7\times {10}^{-6}{m}^{2}}\phantom{\rule{0ex}{0ex}}{v}_{d}=0.052\times {10}^{-3}m/s.$

The drift velocity of the copper wire is $0.052\times {10}^{-3}m/s$.

**Question 2:** Ohm's law explains the relationship between________.

- Magnetic field and electric current
- Electric current and potential difference
- Magnetic field and resistance
- None of the above options

**Answer:** The correct option is (b).

**Explanation:** Ohm's law states the strength of the electric current is directly proportional to the potential difference and inversely proportional to the resistance of the circuit.

V α I

where V is the potential difference and I is the electric current.

**Question 3:** The resistance of a wire of length 5 m is 2 ohm. If the area of cross section of the wire is $2\times {10}^{-7}{m}^{2}$, find its resistivity.

$a.6\times {10}^{-6}\Omega m\phantom{\rule{0ex}{0ex}}b.0.08\times {10}^{-8}\Omega m\phantom{\rule{0ex}{0ex}}c.5\times {10}^{-8}\Omega m\phantom{\rule{0ex}{0ex}}d.8\times {10}^{-8}\Omega m$

**Answer: **The correct option (d).

**Given data:**

Length of the wire l=5 m

Resistance R=2 Ω

Area of cross section $A=2\times {10}^{-7}{m}^{2}$

**Explanation:**

The resistivity of the wire is given by,

$R=\rho \frac{L}{A}\phantom{\rule{0ex}{0ex}}2=\rho \frac{5}{2\times {10}^{-7}}\phantom{\rule{0ex}{0ex}}\rho =\frac{2\times 2\times {10}^{-7}}{5}\phantom{\rule{0ex}{0ex}}\rho =8\times {10}^{-8}\Omega m$

Therefore, the resistivity of the wire is $8\times {10}^{-8}\Omega m.$

**Question 4**: A potential difference created across a 15 Ω resistor is 20 V. Find the current through the resistor.

- 2A
- 10A
- 1.33A
- 5A

**Answer: **The correct option is (c).

**Given data:**

Potential difference V=20 V

Resistance R=$15\Omega $

**Explanation:**

From ohms law,

$V=IR\phantom{\rule{0ex}{0ex}}20V=I\times 15\Omega \phantom{\rule{0ex}{0ex}}I=\frac{20V}{15\Omega}=1.33\phantom{\rule{0ex}{0ex}}I=1.33A$

The current flow across the resistor is 1.33 A.

**Question 5:** Determine the current density if a 50 A current is flowing throwing a conductor of area of $5{m}^{2}$.

$a.10A/{m}^{2}\phantom{\rule{0ex}{0ex}}b.5A/{m}^{2}\phantom{\rule{0ex}{0ex}}c.8A/{m}^{2}\phantom{\rule{0ex}{0ex}}d.0.1A/{m}^{2}$

**Answer**: The correct option (a).

**Given data:**

Area $A=5{m}^{2}$

Current I=50 A

**Explanation:**

The current density J is given by,

$J=\frac{I}{A}\phantom{\rule{0ex}{0ex}}J=\frac{50A}{5{m}^{2}}\phantom{\rule{0ex}{0ex}}J=10A/{m}^{2}$

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