Axiom Physics | Class 12 - Electric Current, Drift Velocity & Ohm's Law

⚡ 01. The Spark (Analogy)

Socho ek pipe me paani bhara hua hai: jab tak dono ends par pressure barabar hai, molecules randomly takrayenge par pipe ke kisi cross-section se net flow zero hoga. Jaise hi pump ek pressure difference (voltage) create karta hai, paani aage flow karna shuru kar deta hai. Theek waise hi, conductor ke andar Electric Current kisi surface se per second pass hone wale net charge ki rate hai[cite: 13]. Par dhyaan rahe: electrons bohot tez random speed ($\approx 10^5\text{ m/s}$) par chalte hain, lekin unka actual directional crawl yani Drift Velocity ($v_d \approx 10^{-4}\text{ m/s}$) bohot hi dheema hota hai[cite: 13]!

📐 02. The Core Law & Formulae

1. Electric Current ($I$)

Rate of net charge flow through any cross-sectional area of a conductor[cite: 13]:

$$I_{\text{avg}} = \frac{\Delta Q}{\Delta t}, \quad I = \lim_{\Delta t \to 0}\frac{\Delta Q}{\Delta t} = \frac{dQ}{dt}$$

SI Unit: $\text{Ampere (A)} = \text{Coulomb/second (C/s)}$. Scalar quantity (follows algebraic addition, not vector law)[cite: 13].

2. Current & Drift Velocity Relation

Macroscopic current linked to microscopic charge carriers[cite: 13]:

$$I = n e A v_d, \quad \vec{J} = \frac{I}{A}\hat{n} = n e \vec{v}_d$$

Where $n$ is electron density (number of free electrons per unit volume) and $\vec{J}$ is current density vector[cite: 13].

03. Interactive Visualizer: Electron-Lattice Drift Engine

Electric Field ($E$) 120 V/m

Temperature (Thermal Speed $v_{\text{th}}$) 300 K (Normal)

Lattice Density (Collisions) Standard Copper Lattice

Drift Speed ($v_d \propto E$):

0.42 mm/s

Relaxation Time ($\tau$):

2.50 × 10⁻¹⁴ s

Net Current Flow ($I$):

Left to Right (Conventional)

📝 04. Stepwise Derivations Blueprint[cite: 13]

Derivation A

Mathematical Expression for Drift Velocity ($v_d = \frac{eE\tau}{m}$)[cite: 13]

1. In the absence of an electric field, free electrons move randomly with high thermal speeds ($\vec{u}_i$). Average thermal velocity is zero[cite: 13]:

$$\vec{u}_{\text{avg}} = \frac{1}{N}\sum_{i=1}^N \vec{u}_i = 0$$

2. When an external field $\vec{E}$ is established, each electron of charge $-e$ and mass $m$ experiences acceleration[cite: 13]:

$$\vec{a} = \frac{\vec{F}}{m} = -\frac{e\vec{E}}{m}$$

3. Velocity of electron $i$ after elapsed time $t_i$ since its last collision[cite: 13]:

$$\vec{v}_i = \vec{u}_i + \vec{a}t_i = \vec{u}_i - \frac{e\vec{E}}{m}t_i$$

4. Average velocity over all electrons defines drift velocity $\vec{v}_d$[cite: 13]:

$$\vec{v}_d = \frac{1}{N}\sum_{i=1}^N \vec{v}_i = \underbrace{\frac{1}{N}\sum \vec{u}_i}_{0} - \frac{e\vec{E}}{m}\left(\frac{1}{N}\sum t_i\right)$$

5. Denoting average time between collisions as mean relaxation time $\tau$[cite: 13]:

$$\vec{v}_d = -\frac{e\vec{E}}{m}\tau \implies v_d = \frac{eE\tau}{m} \text{[cite: 13]}$$

Derivation B

Relation between Electric Current and Drift Velocity ($I = n e A v_d$)[cite: 13]

1. Consider a uniform conductor of cross-sectional area $A$ and length $l$ connected across potential difference $V$[cite: 13].

2. Let $n$ be the number density of free electrons (number of electrons per unit volume)[cite: 13].

3. In time interval $\Delta t$, electrons drifting with average velocity $v_d$ cover an axial distance $\Delta x = v_d \Delta t$[cite: 13].

4. All electrons present inside this volume $\Delta V$ will cross the reference cross-section in time $\Delta t$[cite: 13]:

$$\text{Volume of cylinder } \Delta V = A \cdot \Delta x = A (v_d \Delta t)$$

5. Total number of electrons inside this volume element $N = n \cdot \Delta V = n A v_d \Delta t$[cite: 13].

6. Total charge $\Delta Q$ carried by these electrons across the area[cite: 13]:

$$\Delta Q = N e = (n A v_d \Delta t) e$$

7. Dividing by time interval $\Delta t$ yields the fundamental relation[cite: 13]:

$$I = \frac{\Delta Q}{\Delta t} = n e A v_d \text{[cite: 13]}$$

Vector Current Density: $\vec{J} = \frac{I}{A} = n e \vec{v}_d$. Since electrons carry negative charge ($-e$), their physical drift velocity $\vec{v}_d$ is opposite to current density $\vec{J}$ and electric field $\vec{E}$[cite: 13].

Derivation C

Deduction of Ohm's Law from Microscopic Principles ($V = IR$)[cite: 13]

1. Substitute $v_d = \frac{eE\tau}{m}$ into the current expression $I = neAv_d$[cite: 13]:

$$I = n e A \left(\frac{e E \tau}{m}\right) = \left(\frac{n e^2 \tau A}{m}\right) E$$

2. For a conductor of length $l$ with applied potential difference $V$, uniform electric field is $E = \frac{V}{l}$[cite: 13]:

$$I = \left(\frac{n e^2 \tau A}{m}\right) \frac{V}{l} = \left(\frac{n e^2 \tau A}{m l}\right) V$$

3. Rearranging for potential difference $V$[cite: 13]:

$$V = \left( \frac{m}{n e^2 \tau} \frac{l}{A} \right) I = \left(\rho \frac{l}{A}\right) I = R I \text{[cite: 13]}$$

Resistance & Resistivity: $R = \rho \frac{l}{A}$, where intrinsic resistivity $\rho = \frac{m}{n e^2 \tau}$ depends solely on material and temperature[cite: 13].

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: Is Electric Current a Vector or Scalar?

Even though electric current has both magnitude and direction along the wire, it is strictly a SCALAR quantity[cite: 13]. It does not obey vector addition laws ($\vec{A} + \vec{B}$); currents at a junction add algebraically ($I_{\text{total}} = I_1 + I_2$) regardless of the angle between incoming wires[cite: 13]. In contrast, Current Density ($\vec{J}$) is a true vector[cite: 13].

Trap 2: Variable Wire Cross-Section ($v_d \propto 1/A$)

In a non-uniform wire with steady current: Current $I$ is strictly constant at all cross-sections[cite: 13]. However, current density $J = I/A$ and drift speed $v_d = \frac{I}{neA}$ are inversely proportional to area ($v_d \propto 1/A$)[cite: 13]. Electrons move faster through narrow portions of a wire!

Trap 3: Drift Speed vs Current Signal Speed

Do not confuse the slow snail-like drift speed of electrons ($v_d \approx 10^{-4}\text{ m/s}$) with the speed of electric current propagation[cite: 13]. The signal travels at the speed of electromagnetic waves in conductors ($\approx 10^8\text{ m/s}$), creating an instantaneous field that pushes all electrons simultaneously.

Trap 4: Mobility ($\mu$) Sign Convention

While drift velocity $\vec{v}_d$ is a vector directed antiparallel to $\vec{E}$ for electrons[cite: 13], Mobility $\mu = \frac{|\vec{v}_d|}{E}$ is by definition always a positive scalar[cite: 13]. For electrons: $\mu_e = \frac{e\tau_e}{m_e}$; for holes: $\mu_h = \frac{e\tau_h}{m_h}$.

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: A copper wire of cross-sectional area $1.0 \times 10^{-7}\text{ m}^2$ carries a current of $1.5\text{ A}$. Assuming free electron density $n = 8.5 \times 10^{28}\text{ m}^{-3}$, calculate the drift speed of electrons. If wire length is $3.0\text{ m}$, find the time taken by an electron to drift from one end to the other[cite: 13].

Solution:

1. Formula: $I = n e A v_d \implies v_d = \frac{I}{n e A}$[cite: 13]:

$$v_d = \frac{1.5}{(8.5 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (1.0 \times 10^{-7})} = \frac{1.5}{1.36 \times 10^3} \approx 1.10 \times 10^{-3}\text{ m/s} = 1.10\text{ mm/s}$$

2. Time taken across distance $l = 3.0\text{ m}$:

$$t = \frac{l}{v_d} = \frac{3.0\text{ m}}{1.10 \times 10^{-3}\text{ m/s}} \approx 2727\text{ seconds} \approx 45.5\text{ minutes}$$

JEE/NEET Competitive Trap

Problem: Two wires of the same material have lengths in the ratio $1:2$ and diameters in the ratio $2:3$. If they are connected in series across a battery, find the ratio of (i) drift velocities of electrons in them, (ii) potential differences across them[cite: 13].

Solution:

1. In series connection, current $I$ is identical in both wires[cite: 13]: $I_1 = I_2 = I$.

2. Area ratio: $A \propto d^2 \implies \frac{A_1}{A_2} = \left(\frac{d_1}{d_2}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$.

3. Drift speed ratio: $I = neAv_d \implies v_d = \frac{I}{neA} \implies v_d \propto \frac{1}{A}$ (since material is same, $n$ is same)[cite: 13]:

$$\frac{v_{d1}}{v_{d2}} = \frac{A_2}{A_1} = \frac{9}{4}$$

4. Resistance ratio: $R = \rho \frac{l}{A} \implies \frac{R_1}{R_2} = \left(\frac{l_1}{l_2}\right) \times \left(\frac{A_2}{A_1}\right) = \left(\frac{1}{2}\right) \times \left(\frac{9}{4}\right) = \frac{9}{8}$[cite: 13].

5. In series, $V = IR \implies V \propto R$[cite: 13]:

$$\frac{V_1}{V_2} = \frac{R_1}{R_2} = \frac{9}{8}$$

🗂️ 07. Quick Revision Flashcard Core Formulae

Current Definition

$I = \frac{dQ}{dt}$[cite: 13]

Current & $v_d$

$I = n e A v_d$[cite: 13]

Drift Velocity

$v_d = \frac{eE\tau}{m}$[cite: 13]

Mobility

$\mu = \frac{v_d}{E} = \frac{e\tau}{m}$[cite: 13]