Chapter 3 · Current Electricity

3.1 Electric Current, Drift Velocity and Ohm's Law

Charges in motion: the current $I=dq/dt$, the current density $\mathbf j$, Ohm's law $V=IR$ — and the surprising picture behind it, in which free electrons crawl along a wire at a fraction of a millimetre per second, $I=neAv_d$.

Flick a switch and the ceiling light comes on at once, even though the switch may be ten metres of wire away from the lamp. A phone charger pushes a couple of amperes into a battery for an hour; a lightning stroke moves a similar amount of charge in a thousandth of a second; a nerve fibre carries tiny currents of sodium and potassium ions. Chapters 1 and 2 dealt with charges at rest. From now on the charges move, and the first questions are simple ones: how do we measure how much charge flows, what pushes it, and what holds it back?

This section defines electric current and current density, states Ohm's law and introduces the material property behind resistance, the resistivity. Then we look inside a metal wire. Its free electrons rush about at around $10^5$ m/s, yet when a current flows they drift along the wire at well under a millimetre per second. The drift velocity, the relaxation time between collisions and the mobility of the electrons explain where Ohm's law comes from and what sets the resistivity of copper.

Electric Current and Current Density

Imagine a surface cutting across a wire, or across a beam of ions, or across the gap in a spark. If positive charge crosses the surface in one direction (or negative charge in the other), there is a current through it. What matters is the net charge crossing per unit time.

📘 Definition 3.1 — Electric current
If a net charge $\Delta q$ crosses a surface in time $\Delta t$, the average electric current through the surface is $\Delta q/\Delta t$, and the current at an instant is $$I=\lim_{\Delta t\to0}\frac{\Delta q}{\Delta t}=\frac{dq}{dt}.$$ The SI unit of current is the ampere: $1\ \text{A}=1\ \text{C/s}$ (it is one of the SI base units; Chapter 4 explains how it is fixed using the force between currents). The direction of the current is, by convention, the direction in which positive charge flows; electrons, being negative, move against the current. A current that does not change with time is a steady current; then $q=It$.

Current is a scalar: it is a rate through a surface, and it obeys ordinary arithmetic, not vector addition (the arrow on a wire only tells you which way round the circuit charge flows). Typical sizes: a few microamperes in a nerve fibre, about $0.3$ A in a household LED bulb, $5$–$15$ A in an electric heater, $200$ A in a car's starter motor, and tens of kiloamperes in a lightning stroke.

Currents in different conductors. In a metal the moving charges are the free electrons, one or two per atom, which are not bound to any particular atom. In an electrolyte such as salt water both positive and negative ions move, in opposite directions — and both contribute to a current in the same sense, because negative charge moving left carries charge just as positive charge moving right does. In a gas discharge (a neon sign, lightning) ions and electrons both move. With no field applied, the free electrons in a metal still move, very fast, but in random directions: as many cross any surface one way as the other, so the net current is zero. A current needs a steady push, an electric field inside the conductor, maintained by a cell or another source.

The current tells us the total flow through the whole cross-section. To describe how concentrated the flow is, we divide by the area.

📘 Definition 3.2 — Current density
When a current $I$ flows uniformly and perpendicularly through a cross-section of area $A$, the current density has magnitude $$j=\frac{I}{A},$$ measured in $\text{A/m}^2$. Current density is a vector $\mathbf j$ pointing in the direction of flow of positive charge at that point. If the area $\Delta\mathbf A$ is tilted, only the component of $\mathbf j$ along its normal counts: $\Delta I=\mathbf j\cdot\Delta\mathbf A$.
✏️ Example 1 — Charging a phone
A phone charger supplies a steady current of $2.0$ A for $30$ minutes through a copper lead of cross-sectional area $1.0\ \text{mm}^2$. (a) How much charge flows? (b) How many electrons is that? (c) Find the current density in the lead.
Concept For a steady current $q=It$; each electron carries $e=1.6\times10^{-19}$ C; $j=I/A$.
(a) Charge $t=30\times60=1800$ s, so $q=It=(2.0)(1800)=3.6\times10^{3}\ \text{C}$.
(b) Electrons $N=\dfrac qe=\dfrac{3.6\times10^3}{1.6\times10^{-19}}=2.25\times10^{22}$ — about $2.3\times10^{22}$ electrons.
(c) Current density $A=1.0\ \text{mm}^2=1.0\times10^{-6}\ \text{m}^2$, so $j=\dfrac{2.0}{1.0\times10^{-6}}=2.0\times10^{6}\ \text{A/m}^2$, along the wire in the direction of the current.
Check Units: $\text{A}\times\text{s}=\text{C}$ ✓. The same electrons are not "used up": as many leave the charger as return to it.

Ohm's Law, Resistance and Resistivity

In 1827 Georg Simon Ohm measured the current through metal wires connected to different numbers of cells. For a given wire kept at a fixed temperature he found that the current was proportional to the potential difference across it. (Potential difference — the work done per unit charge in moving charge between two points — is developed fully in Chapter 2; here we only need it in volts, as you met it in Class 10.)

📐 Law 3.1 — Ohm's law

For a metallic conductor kept at a constant temperature (and in the same physical state), the current $I$ through it is proportional to the potential difference $V$ across it: $$V=IR,$$ where the constant $R$ is the resistance of the conductor. The SI unit of resistance is the ohm, $1\ \Omega=1\ \text{V/A}$.

The ratio $V/I$ can be worked out for any device, and it is always called its resistance at that operating point. What Ohm's law adds is that for metals this ratio is a constant, independent of $V$ — the graph of $I$ against $V$ is a straight line through the origin. Many devices do not behave like this; Section 3.2 deals with them.

What resistance depends on. Take a uniform wire. Put two identical wires end to end: the same current now has to be pushed through twice the length, needing twice the potential difference — so $R\propto l$. Put them side by side: each carries half the current for the same $V$ — so $R\propto1/A$. The constant of proportionality depends only on the material and its temperature.

📘 Definition 3.3 — Resistivity and conductivity
The resistance of a uniform conductor of length $l$ and cross-sectional area $A$ is $$R=\rho\,\frac{l}{A},$$ where $\rho$ (rho) is the resistivity of the material, measured in ohm metres ($\Omega\ \text{m}$). Resistivity depends on the material and its temperature, but not on the size or shape of the sample. Its reciprocal $\sigma=1/\rho$ is the conductivity, in siemens per metre ($\text{S/m}$, with $1\ \text{S}=1\ \Omega^{-1}$).

Ohm's law can be rewritten so that it refers to a single point inside the material rather than to a whole wire. In a uniform wire the field is $E=V/l$ and the current density is $j=I/A$. Then $V=IR$ becomes $El=jA\cdot\rho l/A$, that is $E=\rho j$. Both are vectors pointing the same way, so $$\mathbf E=\rho\,\mathbf j\qquad\text{or}\qquad \mathbf j=\sigma\,\mathbf E.$$ This point form of Ohm's law says: the field at a point inside a conductor drives a current density proportional to it. It holds for any shape of conductor, which is why physicists prefer it.

⚠️ Common mistakes with resistance and resistivity

1. Resistance is not resistivity. A thick copper bar and a fine copper wire have the same $\rho$ but very different $R$. $\rho$ belongs to the material; $R$ belongs to the object.

2. Forgetting that stretching changes the area too. When a wire is drawn out, its volume $lA$ stays the same. Doubling $l$ halves $A$, so $R=\rho l/A=\rho l^2/(\text{volume})$ goes up by a factor of $4$, not $2$.

3. Using the diameter as the radius. $A=\pi r^2=\pi d^2/4$. A wire of diameter $1.0$ mm has $A=7.85\times10^{-7}\ \text{m}^2$, not $3.14\times10^{-6}\ \text{m}^2$.

4. Mixing units. Convert $\text{mm}^2$ to $\text{m}^2$ ($1\ \text{mm}^2=10^{-6}\ \text{m}^2$) before using $R=\rho l/A$ with $\rho$ in $\Omega\ \text{m}$.

✏️ Example 2 — Resistance of a copper lead
An extension lead contains a copper wire of length $20$ m and diameter $1.0$ mm. The resistivity of copper is $1.7\times10^{-8}\ \Omega\ \text{m}$. (a) Find the resistance of the wire. (b) Find the potential difference across it when it carries $2.0$ A. (c) The same wire is drawn out to twice its length. What is its new resistance?
Area $r=0.50$ mm $=5.0\times10^{-4}$ m, so $A=\pi r^2=\pi(5.0\times10^{-4})^2=7.85\times10^{-7}\ \text{m}^2$.
(a) Resistance $$R=\rho\frac lA=\frac{(1.7\times10^{-8})(20)}{7.85\times10^{-7}}\approx0.43\ \Omega.$$
(b) Ohm's law $V=IR=(2.0)(0.433)\approx0.87$ V — a small loss, which is why leads are made of copper.
(c) Stretched The volume is unchanged, so $l\to2l$ means $A\to A/2$: $R'=\rho\dfrac{2l}{A/2}=4R\approx1.7\ \Omega$.
Check Point form: $j=2.0/7.85\times10^{-7}=2.55\times10^6\ \text{A/m}^2$, $E=\rho j=0.0433$ V/m, and $El=0.87$ V ✓.

Drift of Electrons and the Origin of Resistivity

Why should a metal obey Ohm's law at all? Picture the free electrons in a copper wire as a gas of tiny particles inside a fixed lattice of positive ions. At room temperature they move in random directions with speeds of the order of $10^5$ m/s (a classical estimate, $\sqrt{3k_BT/m}$; quantum theory gives even larger speeds). Every so often an electron collides with a vibrating ion or an impurity and flies off in a new, random direction. Averaged over all the electrons, the velocity is zero: there is no net flow.

Now apply a field $\mathbf E$ along the wire. Between collisions every electron feels the force $-e\mathbf E$ and accelerates at $$\mathbf a=-\frac{e\mathbf E}{m}.$$ An electron whose last collision was a time $t$ ago, and which left that collision with velocity $\mathbf v_0$, now has velocity $\mathbf v=\mathbf v_0-\dfrac{e\mathbf E}{m}t$. Average this over all $N$ electrons at one instant. The velocities $\mathbf v_0$ just after collisions point in random directions, so they average to zero. The times $t$ since the last collision vary from electron to electron, and their average is a property of the metal called the relaxation time $\tau$. So the average velocity of the electrons is not zero any more: $$\mathbf v_d=-\frac{e\mathbf E}{m}\,\tau.$$ The random motion is still there, enormous and unchanged; on top of it the whole electron gas slowly drifts opposite to the field.

📘 Definition 3.4 — Drift velocity and relaxation time
The drift velocity $\mathbf v_d$ of the free electrons in a conductor is their average velocity when a field is applied — the small, steady velocity superposed on their random thermal motion. The relaxation time $\tau$ is the average time between successive collisions of a free electron (equivalently, the average time since an electron's last collision). For metals at room temperature $\tau\sim10^{-14}$ s.

Now count the charge that crosses a cross-section of area $A$ in a time $\Delta t$. Every electron moves a distance $|\mathbf v_d|\Delta t$ along the wire on average, so all the electrons within that distance of the cross-section — those in a volume $A|\mathbf v_d|\Delta t$ — get across. If there are $n$ free electrons per unit volume, the charge crossing is $e\,n\,A|\mathbf v_d|\Delta t$, and dividing by $\Delta t$ gives the current. Putting in $v_d=eE\tau/m$ and comparing with $j=\sigma E$ identifies the conductivity.

📐 Result 3.2 — Drift velocity, current and resistivity

For a conductor with $n$ free electrons per unit volume, each of charge $-e$ and mass $m$, and relaxation time $\tau$, in a field $\mathbf E$:

Drift velocity $$\mathbf v_d=-\frac{e\tau}{m}\,\mathbf E,\qquad v_d=\frac{eE\tau}{m}.$$

Current and current density $$I=neAv_d,\qquad \mathbf j=-ne\,\mathbf v_d=\frac{ne^2\tau}{m}\,\mathbf E.$$

Conductivity and resistivity $$\sigma=\frac{ne^2\tau}{m},\qquad \rho=\frac{m}{ne^2\tau}.$$

Because $n$ and $\tau$ do not depend on $E$ (at a fixed temperature), $\mathbf j$ is proportional to $\mathbf E$: this is Ohm's law, derived from the motion of electrons.

The model explains more than Ohm's law. It tells us that a metal with many free electrons ($n$ large) and long free flights ($\tau$ large) is a good conductor; it will explain in Section 3.2 why the resistivity of a metal rises when it is heated (faster-vibrating ions make collisions more frequent, so $\tau$ falls), and why a semiconductor behaves differently.

✏️ Example 3 — How fast do the electrons drift?
A copper wire of cross-sectional area $1.0\ \text{mm}^2$ carries a current of $2.0$ A. Copper has $n=8.5\times10^{28}$ free electrons per cubic metre. (a) Find the drift speed. (b) How long does an electron take, on average, to drift $1.0$ m along the wire? (c) Compare the drift speed with the random thermal speed, about $1\times10^5$ m/s.
Concept $I=neAv_d$, so $v_d=\dfrac{I}{neA}$.
(a) Drift speed $$v_d=\frac{2.0}{(8.5\times10^{28})(1.6\times10^{-19})(1.0\times10^{-6})}=\frac{2.0}{1.36\times10^{4}}\approx1.5\times10^{-4}\ \text{m/s},$$ i.e. about $0.15$ mm/s, opposite to the current.
(b) Time $t=\dfrac{1.0}{1.47\times10^{-4}}\approx6.8\times10^{3}$ s — nearly two hours to drift one metre.
(c) Comparison $\dfrac{v_d}{v_\text{thermal}}\approx\dfrac{1.5\times10^{-4}}{1\times10^{5}}\sim10^{-9}$. The drift is a tiny bias on a furious random motion.
Why the lamp lights at once The wire is already full of free electrons. When the switch closes, the field is set up along the whole circuit at nearly the speed of light, and electrons everywhere start drifting at the same moment — including those in the lamp's filament. No electron has to travel from the switch to the lamp. Check: Interactive 1 opens on this wire and reads $v_d=1.47\times10^{-4}$ m/s.
Interactive 1 — Electrons Drifting in a Wire

Each dot is a free electron in a short piece of wire, moving with random thermal motion and scattering off the lattice. Switch the field on and every electron is pushed to the left between collisions — the whole cloud drifts, opposite to the field and the current. Everything is slowed down, and the drift is exaggerated about $10^{8}$ times compared with the thermal motion (really it is a billion times smaller). The readouts give the real values from $v_d=I/(neA)$; the last one is measured from the animated cloud itself and settles near $v_d$ after half a minute or so.

2.01.0
$n$ (m⁻³)—
drift speed $v_d$ (m/s)—
current density $j$ (A/m²)—
time to drift 1 m—
measured cloud drift (animation)—

The relaxation time and the drift speed are hidden inside the wire, but the resistivity is easy to measure. Result 3.2 lets us run the argument backwards and find $\tau$ from $\rho$. The drift speed per unit field is a useful number in its own right, especially for semiconductors, where it is measured directly.

📘 Definition 3.5 — Mobility
The mobility $\mu$ of a charge carrier is the magnitude of its drift velocity per unit electric field: $$\mu=\frac{v_d}{E}=\frac{e\tau}{m}.$$ Its SI unit is $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ (that is, (m/s) per (V/m)). In terms of mobility, $\sigma=ne\mu$ and $\rho=\dfrac{1}{ne\mu}$.
✏️ Example 4 — Relaxation time and mobility of copper
Copper has resistivity $1.7\times10^{-8}\ \Omega\ \text{m}$ and $n=8.5\times10^{28}\ \text{m}^{-3}$. (a) Find the relaxation time of its free electrons. (b) Find their mobility. (c) Find the field inside the wire of Example 3 and check the drift speed found there. Take $m=9.1\times10^{-31}$ kg.
(a) Relaxation time From $\rho=\dfrac{m}{ne^2\tau}$: $$\tau=\frac{m}{ne^2\rho}=\frac{9.1\times10^{-31}}{(8.5\times10^{28})(1.6\times10^{-19})^2(1.7\times10^{-8})}\approx2.5\times10^{-14}\ \text{s}.$$
(b) Mobility $\mu=\dfrac{e\tau}{m}=\dfrac{(1.6\times10^{-19})(2.46\times10^{-14})}{9.1\times10^{-31}}\approx4.3\times10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$.
(c) Field In Example 3, $j=2.0\times10^6\ \text{A/m}^2$, so $E=\rho j=(1.7\times10^{-8})(2.0\times10^6)=0.034$ V/m — a very weak field. Then $v_d=\mu E=(4.33\times10^{-3})(0.034)\approx1.5\times10^{-4}$ m/s, the same as before ✓.
Picture In $25$ femtoseconds an electron moving at $10^5$ m/s covers only a few nanometres — about ten atomic spacings — before it is knocked into a new direction. Interactive 2 shows one electron's velocity being built up and wiped out again and again.
✏️ Example 5 — Thick wire and thin wire in series
Two copper wires $A$ and $B$ are joined end to end and carry the same steady current. Wire $A$ has diameter $1.0$ mm and wire $B$ has diameter $2.0$ mm. Find the ratios (a) $j_A/j_B$, (b) $v_{dA}/v_{dB}$, (c) $E_A/E_B$. (d) Which wire has the larger resistance per metre, and by what factor?
Concept In series the current $I$ is the same (charge cannot pile up at the joint in a steady state). The areas are in the ratio $A_A/A_B=(1.0/2.0)^2=1/4$.
(a), (b) $j=I/A$, so $j_A/j_B=A_B/A_A=4$. The same material has the same $n$, and $v_d=j/(ne)$, so $v_{dA}/v_{dB}=4$ as well: the electrons speed up in the thin wire, like water in a narrow pipe.
(c) Field $E=\rho j$ with the same $\rho$: $E_A/E_B=4$. A larger field is needed to drive the larger current density.
(d) Resistance $R/l=\rho/A$, so the thin wire $A$ has $4$ times the resistance per metre, and $4$ times the potential drop per metre.
Interactive 2 — One Electron Between Collisions

The graph follows the extra velocity that the field gives one electron (its random thermal part is left out). Between collisions it grows steadily at $a=eE/m$; at each collision it is wiped out. The free times are random, with average $\tau$. The dashed line is the time-average of the velocity over all the collisions simulated so far (a few thousand, and counting) — it sits on the green line $eE\tau/m$, the drift speed, even though no single flight looks like it. The defaults are those of Examples 3 and 4.

0.03424.6
acceleration $eE/m$ (m/s²)—
$v_d=eE\tau/m$ (m/s)—
running average so far (m/s)—
collisions counted—
mobility $e\tau/m$ (m²/V s)—
⚠️ Common mistakes about drift

1. Thinking electrons race round the circuit. The signal (the field) travels at nearly the speed of light; the electrons themselves drift at fractions of a millimetre per second.

2. Thinking the thermal speed carries the current. Random velocities point every way and average to zero. Only the small drift gives a net flow of charge.

3. Expecting the drift speed to keep increasing. Each electron accelerates only between collisions; the collisions keep the average velocity steady. That steady average is why the current is steady and why there is resistance.

4. Assuming the same drift speed everywhere in a circuit. In series the current is the same, not the drift speed: $v_d=I/(neA)$ is larger where the wire is thinner or where $n$ is smaller.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — current density that varies across a wire

Not required by the CBSE syllabus. Electric current is listed by JEE Main, NEET and JEE Advanced; a current density that varies across the wire is a JEE-style application (it uses integration) that no syllabus names.

So far $j$ has been the same at every point of a cross-section. In some JEE problems it is not — for example, at high frequencies the current crowds towards the surface of a wire. If $j$ depends only on the distance $r$ from the axis of a wire of radius $R$, split the cross-section into thin rings of radius $r$ and width $dr$. Each ring has area $2\pi r\,dr$ and carries $dI=j(r)\,2\pi r\,dr$, so (uses integration) $$I=\int_0^R j(r)\,2\pi r\,dr.$$ For $j=j_0\,r/R$ this gives $I=\dfrac{2\pi j_0}{R}\cdot\dfrac{R^3}{3}=\dfrac23\pi R^2j_0$: two-thirds of what a uniform density $j_0$ would carry. For $j=j_0(1-r/R)$, $I=2\pi j_0\left(\dfrac{R^2}{2}-\dfrac{R^2}{3}\right)=\dfrac13\pi R^2j_0$. The board examination uses only uniform current density.

Exercises

I
Current and current density. Use $e=1.6\times10^{-19}$ C.
  • 1 A steady current of $0.50$ A flows for $2.0$ minutes. How much charge passes?
  • 2 How many electrons per second pass a point in a wire carrying $1.0$ mA?
  • 3 $3.0\times10^{20}$ electrons pass through a cross-section of a wire in one minute. Find the current.
  • 4 The charge through a conductor varies as $q=3t^2+2t+1$, with $q$ in coulombs and $t$ in seconds. Find the current at $t=2.0$ s.
  • 5 A wire of diameter $2.0$ mm carries $5.0$ A. Find the current density.
  • 6 In a simple model of the hydrogen atom an electron goes round the nucleus $6.6\times10^{15}$ times per second. What current does this orbit represent?
  • 7 In an electrolyte, $2.0\times10^{18}$ sodium ions $\text{Na}^+$ move to the right and $1.5\times10^{18}$ chloride ions $\text{Cl}^-$ move to the left across a plane each second. Find the current and its direction.
  • 8 Electrons in a wire drift towards the west. Which way is the current? Is current a vector?
1 $q=It=(0.50)(120)=60$ C.
2 $\dfrac Ie=\dfrac{1.0\times10^{-3}}{1.6\times10^{-19}}=6.25\times10^{15}$ electrons per second (about $6.3\times10^{15}$).
3 $I=\dfrac{Ne}{t}=\dfrac{(3.0\times10^{20})(1.6\times10^{-19})}{60}=0.80$ A.
4 $I=\dfrac{dq}{dt}=6t+2=14$ A at $t=2.0$ s.
5 $A=\pi(1.0\times10^{-3})^2=3.14\times10^{-6}\ \text{m}^2$, $j=\dfrac{5.0}{3.14\times10^{-6}}\approx1.6\times10^{6}\ \text{A/m}^2$.
6 Charge $e$ passes any point of the orbit $6.6\times10^{15}$ times per second: $I=ef=(1.6\times10^{-19})(6.6\times10^{15})\approx1.1\times10^{-3}$ A $=1.1$ mA, opposite to the electron's motion.
7 Both flows carry charge in the same sense (positive to the right $\equiv$ negative to the left): $I=(2.0+1.5)\times10^{18}\times1.6\times10^{-19}=0.56$ A, to the right (the direction of the $\text{Na}^+$ ions).
8 Towards the east, opposite to the electrons. No: current has a direction along the wire but adds as a scalar (at a junction we simply add currents; they do not obey the parallelogram law). Current density $\mathbf j$ is a vector.
II
Ohm's law, resistance and resistivity. Copper: $\rho=1.7\times10^{-8}\ \Omega\ \text{m}$; nichrome: $\rho=1.1\times10^{-6}\ \Omega\ \text{m}$.
  • 1 A potential difference of $12$ V drives $0.30$ A through a resistor. Find its resistance.
  • 2 Find the resistance of a copper wire $50$ m long with cross-sectional area $1.0\ \text{mm}^2$.
  • 3 What length of nichrome wire of diameter $0.50$ mm has a resistance of $10\ \Omega$?
  • 4 A wire of resistance $R$ is drawn out to three times its length. Find its new resistance.
  • 5 A wire of resistance $R$ is cut into four equal pieces, which are then joined side by side (in parallel). Find the resistance of the combination.
  • 6 Two wires of the same material have the same mass. Their lengths are in the ratio $1:2$. Find the ratio of their resistances.
  • 7 A $1.0$ m length of wire of cross-sectional area $0.50\ \text{mm}^2$ has a resistance of $2.0\ \Omega$. Find the resistivity and conductivity of the material.
  • 8 The field inside a copper conductor is $0.010$ V/m. Find the current density.
1 $R=V/I=12/0.30=40\ \Omega$.
2 $R=\rho l/A=(1.7\times10^{-8})(50)/(1.0\times10^{-6})=0.85\ \Omega$.
3 $A=\pi(2.5\times10^{-4})^2=1.96\times10^{-7}\ \text{m}^2$; $l=\dfrac{RA}{\rho}=\dfrac{(10)(1.96\times10^{-7})}{1.1\times10^{-6}}\approx1.8$ m.
4 Volume fixed: $l\to3l$, $A\to A/3$, so $R\to9R$.
5 Each piece has $R/4$. Four equal resistors in parallel: $\dfrac{R/4}{4}=\dfrac{R}{16}$.
6 Same mass and material means the same volume $V$: $R=\rho l/A=\rho l^2/V\propto l^2$, so $R_1:R_2=1:4$.
7 $\rho=\dfrac{RA}{l}=\dfrac{(2.0)(0.50\times10^{-6})}{1.0}=1.0\times10^{-6}\ \Omega\ \text{m}$; $\sigma=1/\rho=1.0\times10^{6}$ S/m.
8 $j=\sigma E=\dfrac{E}{\rho}=\dfrac{0.010}{1.7\times10^{-8}}\approx5.9\times10^{5}\ \text{A/m}^2$, along $\mathbf E$.
III
Drift velocity. Copper: $n=8.5\times10^{28}\ \text{m}^{-3}$; $m=9.1\times10^{-31}$ kg.
  • 1 A copper wire of cross-section $2.0\ \text{mm}^2$ carries $3.0$ A. Find the drift speed of its free electrons.
  • 2 How long does an electron take to drift $5.0$ m along that wire?
  • 3 If the electrons drift so slowly, explain why a lamp at the end of a $5.0$ m lead lights as soon as the switch is closed.
  • 4 The current in a wire is kept the same while its diameter is halved. What happens to the drift speed?
  • 5 A conductor with $n=5.0\times10^{28}\ \text{m}^{-3}$ and cross-section $1.0\ \text{mm}^2$ has a drift speed of $2.0\times10^{-4}$ m/s. Find the current.
  • 6 A fixed potential difference is applied to a wire at constant temperature. What happens to the drift speed if (a) $V$ is doubled, (b) the length of the wire is doubled with $V$ fixed, (c) the diameter is doubled with $V$ and $l$ fixed?
  • 7 Aluminium has $\rho=2.7\times10^{-8}\ \Omega\ \text{m}$ and $n=1.8\times10^{29}\ \text{m}^{-3}$. Find the relaxation time of its free electrons.
  • 8 Find the mobility of the free electrons in aluminium.
1 $v_d=\dfrac{I}{neA}=\dfrac{3.0}{(8.5\times10^{28})(1.6\times10^{-19})(2.0\times10^{-6})}\approx1.1\times10^{-4}$ m/s.
2 $t=\dfrac{5.0}{1.10\times10^{-4}}\approx4.5\times10^{4}$ s, about $12.6$ hours.
3 The lead is full of free electrons. Closing the switch sets up the field throughout the circuit at nearly the speed of light, so the electrons in the lamp filament start drifting almost at once; none needs to come from the switch.
4 $A$ falls to a quarter, so $v_d=I/(neA)$ becomes $4$ times larger.
5 $I=neAv_d=(5.0\times10^{28})(1.6\times10^{-19})(1.0\times10^{-6})(2.0\times10^{-4})=1.6$ A.
6 $v_d=\dfrac{e\tau}{m}E=\dfrac{e\tau}{m}\dfrac Vl$. (a) Doubles. (b) Halves ($E$ halves). (c) Unchanged — $E=V/l$ is the same; the current goes up $4$ times, but because the area does too, the drift speed stays the same.
7 $\tau=\dfrac{m}{ne^2\rho}=\dfrac{9.1\times10^{-31}}{(1.8\times10^{29})(2.56\times10^{-38})(2.7\times10^{-8})}\approx7.3\times10^{-15}$ s.
8 $\mu=\dfrac{e\tau}{m}=\dfrac{(1.6\times10^{-19})(7.31\times10^{-15})}{9.1\times10^{-31}}\approx1.3\times10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$.
IV
The electron model and its consequences.
  • 1 Free electrons with relaxation time $2.5\times10^{-14}$ s are in a field of $0.050$ V/m. Find their drift speed.
  • 2 Electrons of mobility $4.0\times10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$ are in a field of $0.020$ V/m. Find their drift speed.
  • 3 Show that $\rho=\dfrac{1}{ne\mu}$. A sample of doped silicon has $n=1.0\times10^{21}$ conduction electrons per cubic metre with mobility $0.13\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$. Find its resistivity.
  • 4 When a metal is heated, its relaxation time falls to half while $n$ stays the same. What happens to its resistivity?
  • 5 Estimate the average distance a free electron travels between collisions if its random speed is $1\times10^5$ m/s and $\tau=2.5\times10^{-14}$ s.
  • 6 The random thermal speed of the electrons is about $10^9$ times their drift speed. Why does it produce no current?
  • 7 Modelling. A household copper cable of cross-section $2.5\ \text{mm}^2$ carries $16$ A to a geyser. Find (a) the drift speed and the distance an electron drifts in one hour, (b) the field inside the copper, and (c) the potential drop along $20$ m of the cable.
  • 8 Copper wires $A$ (diameter $1.0$ mm) and $B$ (diameter $0.50$ mm), each $1.0$ m long, are joined in series to a cell. Find $v_{dB}/v_{dA}$, $E_B/E_A$ and the ratio of the potential differences across them.
1 $v_d=\dfrac{eE\tau}{m}=\dfrac{(1.6\times10^{-19})(0.050)(2.5\times10^{-14})}{9.1\times10^{-31}}\approx2.2\times10^{-4}$ m/s.
2 $v_d=\mu E=(4.0\times10^{-3})(0.020)=8.0\times10^{-5}$ m/s.
3 $\sigma=\dfrac{ne^2\tau}{m}=ne\cdot\dfrac{e\tau}{m}=ne\mu$, so $\rho=\dfrac{1}{ne\mu}$. Silicon: $\rho=\dfrac{1}{(1.0\times10^{21})(1.6\times10^{-19})(0.13)}\approx0.048\ \Omega\ \text{m}$ — about three million times that of copper, though each carrier is $30$ times more mobile; the difference is the far smaller $n$.
4 $\rho=m/(ne^2\tau)\propto1/\tau$: it doubles.
5 $\lambda\approx v\tau=(1\times10^5)(2.5\times10^{-14})=2.5\times10^{-9}$ m, about $2.5$ nm.
6 The thermal velocities point in random directions; at any instant as many electrons cross a section one way as the other, so they average to zero. Only the small, common drift gives a net transfer of charge.
7 (a) $v_d=\dfrac{16}{(8.5\times10^{28})(1.6\times10^{-19})(2.5\times10^{-6})}\approx4.7\times10^{-4}$ m/s; in $3600$ s it drifts about $1.7$ m. (b) $j=6.4\times10^6\ \text{A/m}^2$, $E=\rho j=(1.7\times10^{-8})(6.4\times10^6)\approx0.11$ V/m. (c) $V=El\approx(0.109)(20)\approx2.2$ V.
8 Same current; $A_B=A_A/4$. So $v_{dB}/v_{dA}=4$, $E_B/E_A=4$ and (equal lengths) $V_B/V_A=4$ — equivalently $R_B=4R_A$.
🎯 Exam corner — current, drift and Ohm's law BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Define drift velocity and relaxation time. Derive the relation $I=neAv_d$ and hence show that the resistivity of a metal is $\rho=\dfrac{m}{ne^2\tau}$. Calculate $\rho$ for a metal with $n=8.5\times10^{28}\ \text{m}^{-3}$ and $\tau=2.5\times10^{-14}$ s.

Definitions Drift velocity: the average velocity of the free electrons when a field is applied. Relaxation time: the average time between successive collisions.
Drift Between collisions $\mathbf a=-e\mathbf E/m$, so $\mathbf v_i=\mathbf u_i-e\mathbf Et_i/m$. Averaging, $\langle\mathbf u_i\rangle=0$ and $\langle t_i\rangle=\tau$: $\mathbf v_d=-e\mathbf E\tau/m$.
Current In time $\Delta t$ the electrons in a volume $Av_d\Delta t$ cross a section: charge $neAv_d\Delta t$, so $I=neAv_d$. Then $j=nev_d=\dfrac{ne^2\tau}{m}E$; comparing with $j=E/\rho$ gives $\rho=\dfrac{m}{ne^2\tau}$.
Numbers $\rho=\dfrac{9.1\times10^{-31}}{(8.5\times10^{28})(1.6\times10^{-19})^2(2.5\times10^{-14})}\approx1.7\times10^{-8}\ \Omega\ \text{m}$ (copper). Exam tip: state that the initial velocities average to zero — it is the step examiners look for.

E2. Board · 1 mark Assertion (A): A lamp lights almost the instant its switch is closed, although free electrons drift along the wire at less than a millimetre per second. Reason (R): The electric field is set up throughout the circuit at nearly the speed of light, so electrons everywhere in the circuit start drifting together. (A) Both A and R are true and R is the correct explanation of A. (B) Both are true but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.

Concept Both statements are true, and R is exactly why the filament's own electrons respond at once.
Answer (A). Exam tip: the speed of the signal and the speed of the charge carriers are different things — keep them apart.

E3. JEE Main A wire is drawn out so that its length increases by $10\%$, its volume staying the same. Its resistance increases by (A) $10\%$ (B) $20\%$ (C) $21\%$ (D) $5\%$.

Concept With fixed volume $V$, $R=\rho l/A=\rho l^2/V\propto l^2$.
Numbers $R'/R=(1.10)^2=1.21$.
Answer (C) $21\%$. Exam tip: for small stretches $\Delta R/R\approx2\,\Delta l/l$ gives $20\%$ — option (B) is the trap set for that approximation.

E4. NEET A conductor of cross-section $2.0\ \text{mm}^2$ with $1.0\times10^{29}$ free electrons per cubic metre carries $1.6$ A. The drift speed is (A) $5.0\times10^{-5}$ m/s (B) $5.0\times10^{-4}$ m/s (C) $2.5\times10^{-5}$ m/s (D) $1.0\times10^{-4}$ m/s.

Equation $v_d=\dfrac{I}{neA}=\dfrac{1.6}{(1.0\times10^{29})(1.6\times10^{-19})(2.0\times10^{-6})}=5.0\times10^{-5}$ m/s.
Answer (A). Exam tip: $\text{mm}^2\to\text{m}^2$ is a factor $10^{-6}$; forgetting it moves the answer by six powers of ten.

E5. JEE Advanced (One or more correct.) Wires $A$ and $B$ of the same metal, of radii $r$ and $2r$, are joined in series and carry a steady current. Then (A) the drift speed in $A$ is $4$ times that in $B$ (B) the electric field in $A$ is $4$ times that in $B$ (C) the current densities in $A$ and $B$ are equal (D) the number of free electrons per unit volume in $A$ is $4$ times that in $B$.

Setup Series: same $I$. Same metal: same $n$ and $\rho$. $A_A/A_B=1/4$.
Options (A) $v_d=I/(neA)\propto1/A$: true. (B) $E=\rho j=\rho I/A$: true. (C) $j\propto1/A$, so $j_A=4j_B$: false. (D) $n$ is a property of the metal: false.
Answer (A), (B). Exam tip: in series, list what is shared ($I$) and what belongs to the material ($n$, $\rho$, $\tau$); everything else follows from $A$.

E6. JEE Main The current density in a long wire of radius $R$ falls linearly from $j_0$ on the axis to zero at the surface: $j=j_0(1-r/R)$. The current in the wire is (A) $\pi R^2j_0$ (B) $\tfrac12\pi R^2j_0$ (C) $\tfrac13\pi R^2j_0$ (D) $\tfrac23\pi R^2j_0$.

Concept Non-uniform current density (beyond the board syllabus — see the 🚀 box): add up rings, $dI=j\,2\pi r\,dr$ (uses integration).
Integral $I=2\pi j_0\displaystyle\int_0^R\left(r-\frac{r^2}{R}\right)dr=2\pi j_0\left(\frac{R^2}{2}-\frac{R^2}{3}\right)=\frac{\pi R^2j_0}{3}$.
Answer (C). Exam tip: sanity check — the answer must be less than $\pi R^2j_0$, the current if $j$ were $j_0$ everywhere.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How far does a free electron travel between collisions in copper, silver and aluminium — and why does that distance matter for the wiring inside a computer chip?

Why it matters. The copper "interconnects" that link the billions of transistors in a modern processor are now only a few tens of nanometres wide. Engineers at chip foundries in Taiwan, Korea, the USA and Europe have found that copper's resistivity rises sharply when a wire becomes narrower than the distance its electrons travel between collisions, because the electrons start bouncing off the wire's surfaces. That is one reason for research into new interconnect metals such as cobalt and ruthenium. Result 3.2 lets you estimate the key distance yourself.

What to do.

  1. Use $\tau=\dfrac{m}{ne^2\rho}$ with: copper $n=8.5\times10^{28}\ \text{m}^{-3}$, $\rho=1.7\times10^{-8}\ \Omega\ \text{m}$; silver $n=5.9\times10^{28}$, $\rho=1.6\times10^{-8}$; aluminium $n=1.8\times10^{29}$, $\rho=2.7\times10^{-8}$. Find $\tau$ and the mobility of each.
  2. Estimate the mean free path $\lambda=v\tau$ using the classical speed $v\approx1.2\times10^5$ m/s.
  3. Quantum theory says the electrons that carry current move much faster, at the "Fermi speed": $1.6\times10^6$ m/s (copper), $1.4\times10^6$ m/s (silver), $2.0\times10^6$ m/s (aluminium). Recalculate $\lambda$.
  4. Compare both sets of answers with the spacing of atoms in copper, about $0.26$ nm, and with a $20$ nm wide chip interconnect. Which estimate suggests surface scattering should matter?

Tools: Calculator or spreadsheet.

Hand in: A table of $\tau$, $\mu$ and both estimates of $\lambda$ for the three metals, and a paragraph on what the comparison tells you about the classical picture.

Going further. If the electrons in a narrow wire lose some of their free time to surface collisions, how would that change $\tau$ and hence $\rho$? Sketch how you expect $\rho$ to vary with wire width.

Instructor note: $\tau\approx2.5\times10^{-14}$ s (Cu), $3.8\times10^{-14}$ s (Ag), $7.3\times10^{-15}$ s (Al); $\mu\approx4.3\times10^{-3},\ 6.7\times10^{-3},\ 1.3\times10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$. Classical $\lambda\approx2.9,\ 4.4,\ 0.85$ nm; with Fermi speeds $\lambda\approx39,\ 53,\ 15$ nm. The classical estimate is only about ten atomic spacings, but the quantum estimate is comparable to a modern interconnect width — so surface scattering shortens the effective $\tau$ and raises $\rho$, as observed. (Electrons in a perfect lattice are not scattered by the ions at all; collisions are with lattice vibrations, impurities and defects.)

🧪 Self-check quiz

0/5 correct

Q1. The SI unit of current density is

Q2. A copper wire carrying a few amperes has electrons drifting at roughly

Q3. The potential difference across a wire is kept fixed and its length is doubled (same material, area and temperature). The drift speed

Q4. The SI unit of mobility is

Q5. The resistivity of a metal wire depends on

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