Electric Current

Electric current is one of the fundamental concepts in electricity and magnetism. Benjamin Franklin's experiments in the 18th century established the notion of electric charge flow, though the actual direction of conventional current differs from electron flow direction.

Definition of Electric Current: Electric current is defined as the amount of electric charge flowing through a cross-section per unit time. Mathematically: I=dQdtI = \frac{dQ}{dt}

where:

  • II is the electric current (measured in amperes, A)
  • dQdQ is the charge flowing through the conductor in time dtdt
  • For constant current: I=QtI = \frac{Q}{t}

Unit: 1 ampere (A) = 1 coulomb per second (C/s)

[JEE Tip] Current is a scalar quantity. Always distinguish between conventional current direction (positive charge flow) and electron flow direction (opposite).

Conventional Current vs Electron Flow

One of the most important distinctions in circuit analysis is between conventional current and electron flow.

Conventional Current

  • Defined as the flow of positive charges from positive terminal to negative terminal
  • Historical convention established before electrons were discovered
  • Used in all circuit analysis and Kirchhoff's laws
  • Direction: + to − through external circuit

Actual Electron Flow

  • Electrons are negatively charged and actually move from − to + terminal inside the conductor
  • Electrons drift slowly through the conductor (drift velocity of order mm/s), despite signals traveling at nearly the speed of light
  • In metallic conductors, only electrons move; positive ions remain fixed in the lattice
  • Direction: − to + through external circuit, opposite to conventional current

Why the Convention?

When conventional current was defined, the nature of charge carriers was unknown. We now know that in metals, electrons carry current, but the conventional direction is retained for theoretical consistency.

[NEET Important] In semiconductors, both electrons and holes carry current. In electrolytes, both positive and negative ions carry current. In ionized gases, both electrons and ions carry current.

Current Density

To understand current at the microscopic level, we introduce the concept of current density.

Macroscopic Definition: j=IAj = \frac{I}{A}

where:

  • jj is current density (A/m2^2)
  • II is total current through the conductor
  • AA is cross-sectional area perpendicular to current flow

Microscopic Definition: If nn is the number of free charge carriers per unit volume, each with charge ee, drifting with velocity vdv_d: j=nevdj = n e v_d

Derivation: Consider a conductor segment with cross-section AA and length l=vdΔtl = v_d \Delta t drifting in time Δt\Delta t. The charge contained is: Q=n⋅A⋅l⋅e=nAvdΔt⋅eQ = n \cdot A \cdot l \cdot e = n A v_d \Delta t \cdot e

Current is: I=QΔt=nevdAI = \frac{Q}{\Delta t} = n e v_d A

Therefore: j=IA=nevdj = \frac{I}{A} = n e v_d

[JEE Tip] The relation j=nevdj = n e v_d connects macroscopic current with microscopic properties of the conductor.

Metallic Conductors and Drift Velocity

Structure of Metallic Conductors

Metals consist of:

  • Positive metal ions fixed in a crystalline lattice
  • Free electrons (valence electrons) that move through the lattice
  • In conductors like copper, each atom contributes about one free electron

Random Thermal Motion without Electric Field

At room temperature, free electrons move randomly in all directions with thermal velocity vthv_{th}: 12mvth2=32kBT\frac{1}{2} m v_{th}^2 = \frac{3}{2} k_B T

where:

  • mm is electron mass
  • kBk_B is Boltzmann constant
  • TT is absolute temperature

Because of random motion and frequent collisions, the net displacement over time is zero. Therefore, no net current flows without an external electric field.

Effect of Electric Field

When an electric field EE is applied:

  1. Electrons experience a force F⃗=−eE⃗\vec{F} = -e\vec{E}
  2. They accelerate opposite to the field direction
  3. They collide with lattice atoms after traveling a short distance
  4. The average time between collisions is the relaxation time τ\tau
  5. A small drift velocity is superimposed on the random thermal motion

Drift Velocity

The drift velocity vdv_d is the average velocity of electrons in the direction opposite to the applied field.

Typical drift velocity: vd≈10−4 to 10−3 m/sv_d \approx 10^{-4} \text{ to } 10^{-3}\,\text{m/s}

Although this is very small compared with thermal velocity, the enormous number of free electrons (n≈1028 m−3n \approx 10^{28}\,\text{m}^{-3}) produces substantial current.

Example 1: Current from Number of Electrons

A wire carries electrons such that 6.25×10186.25 \times 10^{18} electrons pass through its cross-section in 10 seconds. Calculate the electric current.

Given:

  • N=6.25×1018N = 6.25 \times 10^{18}
  • t=10 st = 10\,\text{s}
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}

Formula: Q=Ne,I=QtQ = Ne, \qquad I = \frac{Q}{t}

Solution: Q=6.25×1018×1.6×10−19=1 CQ = 6.25 \times 10^{18} \times 1.6 \times 10^{-19} = 1\,\text{C} I=110=0.1 AI = \frac{1}{10} = 0.1\,\text{A}

Answer: 0.1 A0.1\,\text{A}

Example 2: Current Density

A copper wire of diameter 2 mm carries a current of 5 A. Calculate the current density.

Given:

  • I=5 AI = 5\,\text{A}
  • Diameter d=2 mm=2×10−3 md = 2\,\text{mm} = 2 \times 10^{-3}\,\text{m}
  • Radius r=1×10−3 mr = 1 \times 10^{-3}\,\text{m}

Formula: A=πr2,j=IAA = \pi r^2, \qquad j = \frac{I}{A}

Solution: A=π(10−3)2=π×10−6 m2A = \pi (10^{-3})^2 = \pi \times 10^{-6}\,\text{m}^2 j=5π×10−6≈1.59×106 A/m2j = \frac{5}{\pi \times 10^{-6}} \approx 1.59 \times 10^6\,\text{A/m}^2

Answer: 1.59×106 A/m21.59 \times 10^6\,\text{A/m}^2

Example 3: Drift Velocity

In a copper conductor, n=8.5×1028 m−3n = 8.5 \times 10^{28}\,\text{m}^{-3} and the current is I=2 AI = 2\,\text{A}. If the wire has cross-sectional area A=1 mm2A = 1\,\text{mm}^2, find the drift velocity.

Given:

  • n=8.5×1028 m−3n = 8.5 \times 10^{28}\,\text{m}^{-3}
  • I=2 AI = 2\,\text{A}
  • A=1 mm2=10−6 m2A = 1\,\text{mm}^2 = 10^{-6}\,\text{m}^2
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}

Formula: I=neAvd  ⟹  vd=IneAI = n e A v_d \implies v_d = \frac{I}{n e A}

Solution: vd=28.5×1028×1.6×10−19×10−6v_d = \frac{2}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}} vd=1.47×10−4 m/sv_d = 1.47 \times 10^{-4}\,\text{m/s}

Answer: 1.47×10−4 m/s1.47 \times 10^{-4}\,\text{m/s}

Example 4: Thermal Velocity vs Drift Velocity

For a free electron at room temperature (T=300 KT = 300\,\text{K}), compare the drift velocity from the previous example with thermal velocity.

Given:

  • vd=1.47×10−4 m/sv_d = 1.47 \times 10^{-4}\,\text{m/s}
  • T=300 KT = 300\,\text{K}
  • m=9.109×10−31 kgm = 9.109 \times 10^{-31}\,\text{kg}
  • kB=1.38×10−23 J/Kk_B = 1.38 \times 10^{-23}\,\text{J/K}

Formula: vth=3kBTmv_{th} = \sqrt{\frac{3k_B T}{m}}

Solution: vth=3×1.38×10−23×3009.109×10−31≈1.17×105 m/sv_{th} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{9.109 \times 10^{-31}}} \approx 1.17 \times 10^5\,\text{m/s}

Ratio: vdvth=1.47×10−41.17×105≈1.26×10−9\frac{v_d}{v_{th}} = \frac{1.47 \times 10^{-4}}{1.17 \times 10^5} \approx 1.26 \times 10^{-9}

Answer: The drift velocity is about 10−910^{-9} times the thermal velocity.

Example 5: Electron Density from Current and Drift Velocity

An unknown metal wire carries 10 A at a measured drift velocity of 5×10−5 m/s5 \times 10^{-5}\,\text{m/s}. The wire has radius 0.5 mm0.5\,\text{mm}. Calculate the electron density.

Given:

  • I=10 AI = 10\,\text{A}
  • vd=5×10−5 m/sv_d = 5 \times 10^{-5}\,\text{m/s}
  • r=0.5×10−3 mr = 0.5 \times 10^{-3}\,\text{m}
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}

Formula: n=IeAvd=Ieπr2vdn = \frac{I}{e A v_d} = \frac{I}{e \pi r^2 v_d}

Solution: A=π(0.5×10−3)2=π×0.25×10−6 m2A = \pi (0.5 \times 10^{-3})^2 = \pi \times 0.25 \times 10^{-6}\,\text{m}^2 n=101.6×10−19×π×0.25×10−6×5×10−5≈1.59×1030 m−3n = \frac{10}{1.6 \times 10^{-19} \times \pi \times 0.25 \times 10^{-6} \times 5 \times 10^{-5}} \approx 1.59 \times 10^{30}\,\text{m}^{-3}

Answer: 1.59×1030 m−31.59 \times 10^{30}\,\text{m}^{-3}

Example 6: Current Ratio in Two Wires

Two copper rods are made from the same material. Rod A has radius 1 mm and Rod B has radius 2 mm. Both have the same length and are connected to the same voltage source. Find the ratio of currents.

Solution: For same material and same length under same voltage, current is proportional to area: I∝A=πr2I \propto A = \pi r^2 IAIB=rA2rB2=1222=14\frac{I_A}{I_B} = \frac{r_A^2}{r_B^2} = \frac{1^2}{2^2} = \frac{1}{4}

Answer: IA:IB=1:4I_A : I_B = 1 : 4