Mobility

Two wires sit side by side in a circuit — one copper, one a lightly doped semiconductor — carrying the same current under the same applied field. Inside each, billions of charge carriers are drifting, colliding, restarting, drifting again, in the same random zig-zag dance you met while deriving drift velocity. Yet the two materials behave utterly differently: copper conducts a thousand times better than the semiconductor even though the semiconductor might have carriers that respond just as eagerly to the field. What single number captures "how eagerly a carrier responds to a volt per metre pushing on it," independent of how many carriers there are? That number is mobility. Mobility is the quiet bridge between the microscopic collision picture of conduction — carriers of charge $e$ and mass $m$ , scattering every relaxation time $\tau$ — and the macroscopic, meter-readable world of current, current density and conductivity. It answers a sharper question than drift velocity alone: not just "how fast do carriers drift," but "how fast do they drift per unit field," a property of the material and the carrier that does not care how hard you are pushing. Once you have mobility, current density $J = ne\mu E$ and conductivity $\sigma = ne\mu$ fall out in two lines, and an entire menu of comparisons — why copper beats iron, why doped silicon beats pure silicon, why heating a metal hurts conduction while heating a semiconductor helps it — becomes a one-line argument about $n$ and $\mu$ . This article builds mobility from the drift-velocity relation you already have, derives its two workhorse formulas, walks two worked numbers (a copper wire and a semiconductor sample, in the pattern of your Pass-1 node), and closes with the mark-losing traps examiners watch for: mixing up $\mu$ with permeability, forgetting mobility is a material property rather than a function of $E$ , and misreading how conductivity behaves as temperature rises.

Key takeaways

Definition. Mobility $\mu = v_d/E$ is drift velocity acquired per unit applied electric field; SI unit $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ .
Microscopic origin. From $v_d = eE\tau/m$ , mobility is $\mu = e\tau/m$ — it depends only on the carrier's charge, mass and average time between collisions $\tau$ , not on $E$ itself.
Link to current. Substituting $v_d = \mu E$ into $I = nAev_d$ gives $I = nAe\mu E$ , so current density $J = I/A = ne\mu E$ .
Link to conductivity. Comparing $J = ne\mu E$ with $J = \sigma E$ gives $\sigma = ne\mu$ , and hence resistivity $\rho = 1/(ne\mu)$ .
Two knobs, not one. A material conducts well only if it has enough carriers ( $n$ ) and those carriers are mobile ( $\mu$ ); either one alone is not enough.
Temperature tells opposite stories. In a metal, $\tau$ (and so $\mu$ ) falls as temperature rises, raising resistivity; in a semiconductor, a much larger rise in $n$ usually wins, so resistivity falls overall even though $\mu$ itself may also dip slightly.
Key idea: mobility mu = v_d/E = etau/m is the field-independent "responsiveness" of a charge carrier, and multiplying it by carrier density n gives conductivity sigma = ne*mu, so how well a material conducts depends on both how many carriers it has and how mobile each one is.

The core insight

Mobility $\mu = v_d/E = e\tau/m$ is a material-and-carrier property (not a function of the applied field) that converts carrier density into conductivity via $\sigma = ne\mu$ , so conduction quality depends jointly on carrier count and carrier responsiveness.

Where it fits in the syllabus

Mobility sits inside the Current Electricity chapter of CBSE Class 12 Physics, immediately after drift velocity and the microscopic model of current, and just before resistivity and its temperature dependence — the prescribed relation is $\mu = v_d/E$ , together with the current expression $I = nAev_d$ that mobility plugs into. It is fully within the 2025-26 board scope: no potentiometer, no carbon-code, no superconductivity content is needed here, and this section stays entirely within the drift-velocity-to-resistivity thread the syllabus lays out.

The physical picture

Recall the free-electron model of conduction: between collisions with the lattice, a free electron of charge magnitude $e$ and mass $m$ accelerates under the local field $E$ , and on average survives a time $\tau$ (the relaxation time) before its momentum is randomised again. Averaging over the population gives the familiar drift-velocity result:

v_d = \dfrac{eE\tau}{m}

where: $v_d$ is the magnitude of drift velocity (m/s), $e$ is the elementary charge ( $1.6\times10^{-19}\ \text{C}$ ), $E$ is the applied electric field (V/m), $\tau$ is the relaxation time (s), and $m$ is the carrier's mass (kg, for an electron $9.1\times10^{-31}\ \text{kg}$ ). Physically, $eE/m$ is the acceleration a free carrier feels, and $\tau$ is how long it gets to accelerate before the next collision wipes out its gained velocity — the product is the average drift speed that survives. Mobility is defined as the ratio of this drift velocity to the field that caused it:

\mu = \dfrac{v_d}{E}

where: $\mu$ is mobility, with SI unit $\dfrac{\text{m/s}}{\text{V/m}} = \text{m}^2\text{V}^{-1}\text{s}^{-1}$ . Substituting the drift-velocity expression for $v_d$ shows what mobility is really made of:

\mu = \dfrac{e\tau}{m}

where: every quantity on the right — $e$ , $\tau$ , $m$ — is a property of the carrier and the material, and the field $E$ has cancelled out completely. This is the physically important reading: mobility does not care how hard you push, only how the carrier and its environment respond to a push. A carrier with a long relaxation time (few collisions) or a small mass is intrinsically more mobile; doubling the applied field simply doubles $v_d$ , leaving $\mu$ unchanged, exactly as Ohm's law demands for an ohmic conductor.

Figure

Two identical wires side by side under the same applied field with electron paths shown, one wire scattering electrons frequently with short zig-zag steps and low drift speed labelled low mobility, the other scattering rarely with long smooth steps and high drift speed labelled high mobility, same field same charge but different responsiveness

Connecting mobility to current and conductivity

Two or three algebraic steps carry mobility from a defined ratio to the workhorse formula every board question needs, so it is worth working through carefully rather than quoting the result from memory.

Derivation ADeriving conductivity in terms of mobility

The goal is to start from the microscopic current expression $I = nAev_d$ , replace the drift velocity with mobility using $v_d = \mu E$ , and arrive at the two most-used results of this topic — current density $J = ne\mu E$ and conductivity $\sigma = ne\mu$ — so that any current-electricity problem giving mobility, carrier density and field (or asking for any one of these three given the other two plus current) can be solved in a couple of lines rather than re-deriving drift velocity from scratch each time.

Step 1. Start from the microscopic current formula already established for a conductor of cross-section $A$ , free carrier density $n$ and carrier charge magnitude $e$ : $I = nAev_d$ , where $v_d$ is the drift speed.

Step 2. Substitute the mobility relation $v_d = \mu E$ (rearranged from $\mu = v_d/E$ ) directly into that expression: $I = nAe(\mu E) = nAe\mu E$ . Nothing new has been assumed — this is a pure substitution.

Step 3. Divide both sides by the cross-sectional area $A$ to get current density $J = I/A$ : $J = ne\mu E$ . This is the current per unit area flowing under a field $E$ , expressed entirely in terms of carrier density and mobility.

Step 4. Compare with the general (Ohm's-law) relation between current density and field, $J = \sigma E$ , valid for any ohmic conductor. Matching the two expressions for $J$ term by term gives $\sigma E = ne\mu E$ , and since this must hold for any $E$ , the field cancels, leaving $\sigma = ne\mu$ . Taking the reciprocal gives resistivity $\rho = \dfrac{1}{ne\mu}$ , tying mobility directly to the resistivity topic that follows this one in the chapter. Result: $J = ne\mu E$ , $\sigma = ne\mu$ , $\rho = 1/(ne\mu)$ — conductivity is the product of how many carriers there are and how mobile each one is.

The compact result $\sigma = ne\mu$ is worth sitting with. It says conduction quality is genuinely a product of two independent ideas: carrier density $n$ (a counting question — how many free carriers per cubic metre) and mobility $\mu$ (a responsiveness question — how well each carrier converts field into drift speed). A material can fail to conduct for either reason alone: an insulator has almost no free carriers ( $n$ tiny) even though whatever carriers exist might respond fine to a field, while a material with many carriers but very frequent collisions (small $\tau$ , hence small $\mu$ ) still conducts poorly. Neither $n$ nor $\mu$ by itself tells the whole story — only their product does. For materials with more than one carrier type — semiconductors, which conduct via both free electrons and holes — the contributions simply add: $\sigma = n_e e \mu_e + n_h e \mu_h$ , where $\mu_e$ and $\mu_h$ are electron and hole mobilities. This composite picture is useful background for understanding why semiconductors behave so differently from metals, though the syllabus does not ask for quantitative two-carrier numericals here. Temperature dependence is where mobility connects forward to the resistivity topic. Since $\mu = e\tau/m$ , and since $\tau$ in a metal shrinks as temperature rises (a hotter lattice vibrates more, so free electrons collide more often), electron mobility in a metal falls as temperature increases — this is the same microscopic fact that makes a metal's resistance rise with heating, because $\rho = 1/(ne\mu)$ and $n$ is essentially fixed in a metal while $\mu$ drops. In a semiconductor, heating does the opposite to $n$ : far more electron-hole pairs become available for conduction as temperature rises, and this increase in $n$ is large enough to overwhelm any small drop in $\mu$ , so a semiconductor's resistivity falls overall with temperature even though its mobility, like a metal's, tends to dip slightly.

Worked examples

Board level - drift velocity of electrons in a copper wire from mobility and applied field

The mobility of free electrons in copper is $\mu = 4.3\times10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$ . A copper wire of length $2\ \text{m}$ has a potential difference of $10\ \text{V}$ applied across it. Find the drift velocity of the electrons.

Step 1. Find the electric field inside the wire from the applied potential difference and the length: $E = \dfrac{V}{L} = \dfrac{10}{2} = 5\ \text{V/m}$ . This treats the wire as uniform, so the field is the same everywhere along it.

Step 2. Apply the mobility relation $v_d = \mu E$ directly, since mobility is already given: $v_d = (4.3\times10^{-3})(5) = 2.15\times10^{-2}\ \text{m/s}$ .

Step 3. Interpret the size of the answer. $2.15\times10^{-2}\ \text{m/s}$ is about $2.15\ \text{cm/s}$ — this is noticeably larger than the millimetre-per-second drift speeds you see in ordinary household-current problems, because the field here ( $5\ \text{V/m}$ , from a $10\ \text{V}$ source over a short $2\ \text{m}$ wire) is comparatively large. Answer: $v_d \approx 2.15\times10^{-2}\ \text{m/s} = 2.15\ \text{cm/s}$ .

Competitive level - conductivity and current density of a semiconductor sample, then the equivalent resistivity

A semiconductor sample has free electron density $n = 2\times10^{19}\ \text{m}^{-3}$ and electron mobility $\mu = 0.14\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$ (holes neglected for simplicity). Find (a) the conductivity of the sample, (b) the current density when a field of $200\ \text{V/m}$ is applied, and (c) the resistivity of the sample.

Step 1. Compute conductivity from $\sigma = ne\mu$ : $\sigma = (2\times10^{19})(1.6\times10^{-19})(0.14)$ . Multiply the first two factors: $2\times10^{19}\times1.6\times10^{-19} = 3.2$ ; then $3.2\times 0.14 = 0.448$ . So $\sigma \approx 0.448\ \Omega^{-1}\text{m}^{-1}$ .

Step 2. Compute current density from $J = \sigma E$ : $J = 0.448\times 200 = 89.6\ \text{A/m}^2$ . This is the same result $J = ne\mu E$ would give directly, since $\sigma$ was itself built from $ne\mu$ .

Step 3. Compute resistivity as the reciprocal of conductivity, $\rho = 1/\sigma = 1/0.448 \approx 2.23\ \Omega\text{m}$ . This links the mobility calculation straight into the resistivity topic that follows this chapter section — a semiconductor's resistivity is set by the same $n$ and $\mu$ that fixed its conductivity here. Answer: $\sigma \approx 0.448\ \Omega^{-1}\text{m}^{-1}$ , $J = 89.6\ \text{A/m}^2$ , $\rho \approx 2.23\ \Omega\text{m}$ . Timing note: this is a two-minute, three-substitution problem once $\sigma = ne\mu$ is memorised — the only place to lose time is the power-of-ten multiplication in Step 1, so group the exponents first ( $10^{19}\times10^{-19} = 10^0$ ) before multiplying the decimal parts, and keep resistivity as a quick reciprocal at the end rather than recomputing from scratch.

Common traps

Symbol clash. $\mu$ also denotes permeability (Moving Charges and Magnetism) and magnetic moment context in other chapters; in Current Electricity, $\mu$ with units $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ always means mobility — check the units, not just the letter.
Thinking mobility depends on $E$ . For an ohmic conductor, $\mu = e\tau/m$ is fixed by the material; it is drift velocity $v_d = \mu E$ that scales with the field, not mobility itself. A question that changes the applied voltage is testing whether you keep $\mu$ constant and only recompute $v_d$ .
Unit mix-ups in $\sigma = ne\mu$ . Three different unit systems collide here — $n$ in $\text{m}^{-3}$ , $e$ in coulombs, $\mu$ in $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ — and a wrong power of ten in any one factor silently wrecks the final answer. Always group the powers of ten first, as in the worked example above.
Assuming carrier density alone decides conduction. A sample with a large $n$ but small $\mu$ (frequent collisions) can conduct worse than one with smaller $n$ but much larger $\mu$ ; $\sigma = ne\mu$ is a genuine product, and an exam question can test either factor in isolation.
Reversing the temperature story for semiconductors. Do not say "mobility increases with temperature in a semiconductor" — mobility itself typically still falls (or is roughly flat); it is the much larger rise in carrier density $n$ that drives resistivity down overall. Attribute the effect to $n$ , not to $\mu$ rising.

Common trapMobility is a material property, not a knob you turn with voltage

The single biggest mark-loser on this topic is treating mobility as if it changes when you change the applied field or the current in a circuit. It does not, for an ohmic conductor at fixed temperature: $\mu = e\tau/m$ depends only on the carrier's charge, mass and average collision time, all of which are properties of the material, not of the field you happen to apply. If a problem doubles the voltage across a wire, the correct response is that drift velocity $v_d = \mu E$ doubles while $\mu$ stays exactly the same — writing " $\mu$ doubles" instead of " $v_d$ doubles" is a direct sign of confusing the definition $\mu = v_d/E$ with a claim that $\mu$ and $E$ are proportional, when in fact $\mu$ is precisely the constant of proportionality that stays fixed.

You should be able to…

State the definition $\mu = v_d/E$ and give its SI unit, $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ .
Derive $\mu = e\tau/m$ starting from the drift-velocity expression $v_d = eE\tau/m$ .
Derive $J = ne\mu E$ and $\sigma = ne\mu$ starting from $I = nAev_d$ , and take the reciprocal to get $\rho = 1/(ne\mu)$ .
Calculate drift velocity, current density, conductivity or resistivity given mobility and the other relevant quantities, keeping powers of ten grouped to avoid arithmetic slips.
Explain, using $\sigma = ne\mu$ , why conduction quality depends jointly on carrier density and carrier mobility rather than on either alone.
Explain qualitatively why metal mobility falls with temperature while semiconductor conductivity rises with temperature despite a similar drop in mobility.

Ask the tutor: Want a few more mixed numericals moving between mobility, conductivity and resistivity, or a side-by-side comparison of how metals and semiconductors respond to heating?

Why does mobility not depend on the electric field if drift velocity clearly does?

Because mobility is defined as the ratio $v_d/E$ , not $v_d$ itself. Drift velocity really does grow when you apply a stronger field — that part is true and matches everyday intuition, a harder push gives a faster average speed. But for an ohmic material, $v_d$ grows in direct proportion to $E$ , so the ratio $v_d/E$ stays constant. Mobility is that constant of proportionality: it captures how efficiently a given carrier converts field strength into drift speed, a property fixed by $\tau$ and $m$ , both material facts unrelated to how hard you happen to be pushing at the moment.

Is mobility the same thing as conductivity? They seem to measure similar things.

No, though they are closely related. Mobility $\mu$ is a property of a single carrier type — how responsive one electron or hole is to a field. Conductivity $\sigma = ne\mu$ combines that responsiveness with carrier density $n$ , the population of carriers available. A material could have very high mobility but still conduct poorly if it has almost no free carriers (like a pure semiconductor at low temperature), and conversely a material with only moderate mobility but a huge carrier density (like a metal) can still conduct extremely well. Conductivity is the macroscopic outcome; mobility is one of the two microscopic ingredients that produce it.

Why does heating a metal make it conduct worse, but heating a semiconductor make it conduct better, if both involve mobility dropping?

Because the two materials differ in what temperature does to $n$ . In a metal, the number of free electrons per unit volume is already essentially fixed (roughly one or two per atom, regardless of temperature) — so when heating shrinks $\tau$ and hence $\mu$ , resistivity $\rho = 1/(ne\mu)$ has nowhere to go but up. In a semiconductor, very few electrons are free to conduct at room temperature to begin with, so heating dramatically increases $n$ by promoting many more electrons into the conduction band — that increase in $n$ is so large it swamps the small drop in $\mu$ , and $\sigma = ne\mu$ rises overall even though mobility itself is behaving the same way it does in a metal.

Do I need to memorise the two-carrier formula sigma equals n_e e mu_e plus n_h e mu_h for the CBSE board exam?

It is useful conceptual background for understanding semiconductors, but the CBSE Class 12 syllabus does not require quantitative numericals combining separate electron and hole mobilities — that level of detail belongs to a dedicated semiconductor-physics treatment. For board purposes, focus on the single-carrier relations $\mu = v_d/E = e\tau/m$ , $J = ne\mu E$ and $\sigma = ne\mu$ , and be ready to reason qualitatively about why a semiconductor's resistivity falls with temperature even though a metal's rises.