Q.The number density of free electrons in a copper conductor estimated in Example 3.1 is $8.5 \times 10^{28}\ \text{m}^{-3}$. How long does an electron take to drift from one end of a wire $3.0\ \text{m}$ long to its other end? The area of cross-section of the wire is $2.0 \times 10^{-6}\ \text{m}^2$ and it is carrying a current of $3.0\ \text{A}$.

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.In the diagram, the current through the part of the wire with 1 mm diameter is 1.6 A and the drift speed is $2\times10^{-4}\ ms^{-1}$. The drift speed in the part of wire with 2 mm diameter is (A) $0.5\times10^{-4}\ ms^{-1}$ (B) $1\times10^{-4}\ ms^{-1}$ (C) $4\times10^{-4}\ ms^{-1}$ (D) $2\times10^{-4}\ ms^{-1}$

   ›Reveal solutionSolution

   Current continuity across a wire of varying cross-section fixes $v_d\propto 1/d^2$; the answer is $0.5\times10^{-4}\ ms^{-1}$.

   Concept and Intuition

   In a single current-carrying wire, the current $I$ must be identical everywhere along its length — charge can't pile up anywhere in steady state. Since $I=nAev_d$ and both the free-electron density $n$ and charge $e$ are the same material property throughout, the product $Av_d$ must stay constant along the wire: $A_1v_1=A_2v_2$.

   Step-by-Step Solution

   1. Cross-sectional area $A\propto d^2$, so $d_1^2v_1=d_2^2v_2$.
   2. $v_2 = v_1\left(\dfrac{d_1}{d_2}\right)^2 = 2\times10^{-4}\times\left(\dfrac{1}{2}\right)^2$. …

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2. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If current of 80 A is passing through a straight conductor of length 10 m, then the total momentum of electrons in the conductor is (mass of electron = $9.1\times10^{-31}$ kg and charge of electron = $1.6\times10^{-19}$ C) (A) $910\times10^{-9}$ N s (B) $910\times10^{-11}$ N s (C) $455\times10^{-9}$ N s (D) $455\times10^{-11}$ N s

   ›Reveal solutionSolution

   Total electron momentum in a current-carrying wire simplifies neatly to $p = mIL/e$ once the electron density and cross-sectional area cancel out — the answer is $455\times10^{-11}$ N·s.

   Concept and Intuition

   Each conduction electron drifts with a small average velocity $v_d$ superposed on its random thermal motion; only the drift contributes a net momentum since random velocities cancel in the average. The current is related to drift velocity by $I = nAev_d$, where $n$ is electron number density and $A$ is the wire's cross-sectional area. The total number of conduction electrons in the wire is $N = nAL$ (density × volume). Multiplying $N$ by the electron mass $m$ and $v_d$ gives total momentum — and remarkably, the microscopic details $n$ and $A$ drop out, leaving a clean formula only in terms of measurable quantities $I$, $L$, $m$, $e$.

   Step-by-Step Solution

   1. Drift velocity from current: $I = nAev_d \Rightarrow v_d = \dfrac{I}{nAe}$.
   2. Total number of electrons in the wire: $N = nAL$ (n = number density, A = cross-section, L = length).
   3. Total momentum: $p = Nmv_d = nAL \cdot m \cdot \dfrac{I}{nAe} = \dfrac{mIL}{e}$ (n and A cancel).
   4. Substitute: $m=9.1\times10^{-31}$ kg, $I=80$ A, $L=10$ m, $e=1.6\times10^{-19}$ C.
   5. Numerator: $9.1\times10^{-31}\times80\times10 = 7.28\times10^{-28}$. …

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3. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.In a metal, the charge carrier density is $9.1 \times 10^{28}\ \mathrm{m^{-3}}$ and its electrical conductivity is $6.4 \times 10^7\ \mathrm{S\,m^{-1}}$. When an electric field of $10\ \mathrm{N\,C^{-1}}$ is applied to the metal, then the average time between two successive collisions of electrons in the metal is (Mass of electron = $9.1 \times 10^{-31}$ kg; charge of electron = $1.6 \times 10^{-19}$ C) (A) $4.6 \times 10^{-14}$ s (B) $2.5 \times 10^{-13}$ s (C) $4.6 \times 10^{-13}$ s (D) $2.5 \times 10^{-14}$ s

   ›Reveal solutionSolution

   The Drude-model relation $\sigma = ne^2\tau/m$ connects conductivity to the average collision time; solving for $\tau$ (a material property, independent of the applied field) gives $2.5\times10^{-14}$ s.

   Concept and Intuition

   In the free-electron (Drude) picture of conduction, electrons accelerate under the field between collisions with the lattice, and $\tau$ — the relaxation time — is the average time between successive collisions. Microscopically it depends on the charge-carrier density $n$, the metal's conductivity $\sigma$, and the electron mass $m$, through $\sigma = \dfrac{ne^2\tau}{m}$. Notice this formula has no explicit $E$-field dependence: $\tau$ characterises how frequently an electron scatters off the lattice, a property of the material's microstructure, not of how hard we push it with an external field. The applied field of $10\ \mathrm{N\,C^{-1}}$ in the question is extra information not needed to find $\tau$ (though it would be needed to find drift velocity or current).

   Step-by-Step Solution

   1. Start from $\sigma = \dfrac{ne^2\tau}{m}$ and solve for $\tau$: $\tau = \dfrac{\sigma m}{ne^2}$.
   2. Compute $ne^2 = (9.1\times10^{28})\times(1.6\times10^{-19})^2 = 9.1\times10^{28}\times2.56\times10^{-38} = 2.3296\times10^{-9}$. …

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4. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.The potential difference across a conducting wire of length 20 cm is 30 V. If the electron mobility is $2 \times 10^{-6} \mathrm{~m^2~V^{-1}~s^{-1}}$, then the drift velocity of the electrons is (A) $3 \times 10^{-3} \mathrm{~m~s^{-1}}$ (B) $1.5 \times 10^{-3} \mathrm{~m~s^{-1}}$ (C) $1.5 \times 10^{-4} \mathrm{~m~s^{-1}}$ (D) $3 \times 10^{-4} \mathrm{~m~s^{-1}}$

   ›Reveal solutionSolution

   Drift velocity is mobility times electric field; converting length to metres and dividing gives $3\times10^{-4}\ \mathrm{m/s}$.

   Concept and Intuition

   Electron mobility $\mu$ is defined as the drift velocity acquired per unit applied electric field: $v_d = \mu E$. The field inside a uniform conducting wire under a steady potential difference is simply $E = V/L$ — the voltage spread evenly over the wire's length.

   Step-by-Step Solution

   1. Convert length: $L = 20\ \mathrm{cm} = 0.20\ \mathrm m$.
   2. Electric field: $E = \dfrac{V}{L} = \dfrac{30}{0.20} = 150\ \mathrm{V/m}$.
   3. Drift velocity: $v_d = \mu E = 2\times10^{-6}\times150 = 3\times10^{-4}\ \mathrm{m/s}$. …

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5. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.The area of cross-section of a copper wire is $4\times10^{-7}\ \mathrm{m^2}$ and the electrons per cubic metre in copper is $8\times10^{28}$. If the wire carries a current of 6.4 A, then the drift velocity of the electrons (in $10^{-3}\ \mathrm{m\,s^{-1}}$) is (A) 0.25 (B) 2.5 (C) 0.125 (D) 1.25

   ›Reveal solutionSolution

   This tests the microscopic current relation $I = nAev_d$; solving for $v_d$ with the given wire and current data gives $1.25\times10^{-3}\,\mathrm{m\,s^{-1}}$.

   Concept and Intuition

   Current in a conductor is carried by free electrons drifting slowly under the applied field, superimposed on their fast random thermal motion. The relation $I = nAev_d$ connects the macroscopic current $I$ to the microscopic drift velocity $v_d$, through the free-electron density $n$, the cross-sectional area $A$, and the electron charge $e$.

   Step-by-Step Solution

   1. Rearranging $I = nAev_d$: $v_d = \dfrac{I}{nAe}$.
   2. Compute the denominator: $nAe = (8\times10^{28})(4\times10^{-7})(1.6\times10^{-19})$.
   3. Multiply the numbers: $8\times4\times1.6 = 51.2$; multiply the powers of ten: $10^{28-7-19}=10^{2}$. So $nAe = 51.2\times10^{2} = 5120$.
   4. $v_d = \dfrac{6.4}{5120} = 1.25\times10^{-3}\,\mathrm{m\,s^{-1}}$. …

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6. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.Drift speed

   (v) varies with the intensity of electric field (E) as per the relation (A) $v \propto E$ (B) $v \propto \dfrac{1}{E}$ (C) $v \propto E^2$ (D) $v \propto E^{-2}$

   ›Reveal solutionSolution

   Drift speed is directly proportional to the applied electric field through the carrier mobility.

   Concept and Intuition

   Electrons in a conductor accelerate between collisions; averaged over many collisions this gives a steady drift velocity $v_d=\mu E$, where mobility $\mu=e\tau/m$ is a material constant (independent of $E$ for ohmic conductors).

   Step-by-Step Solution

   1. $v_d = \mu E$, with $\mu$ constant.
   2. Hence $v_d \propto E$ — a linear, first-power relationship.

   Common Mistakes …

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7. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.A steady current is flowing in a metallic conductor of non-uniform cross section. The physical quantity which remains constant is (A) Electricity current density (B) Drift velocity (C) Electricity current density and drift velocity (D) Electric current

   ›Reveal solutionSolution

   This tests the continuity principle for steady currents: the current itself is conserved along a conductor of varying cross-section, not the current density or drift velocity.

   Concept and Intuition

   In steady state, charge cannot pile up anywhere in the conductor, so the same amount of charge must pass every cross-section per second — that is exactly the statement that current $I$ is constant along the wire. But current density $J=I/A$ and drift velocity $v_d=J/(ne)$ both depend on the local area $A$, so they must change as the conductor's cross-section changes.

   Step-by-Step Solution

   1. Steady current + charge conservation $\Rightarrow$ $I$ entering any cross-section = $I$ leaving it, for every cross-section along the conductor.
   2. Current density $J=I/A$: since $A$ varies (non-uniform cross-section) but $I$ is constant, $J$ must vary inversely with $A$. …

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8. AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
   Q.If n, e, $\tau$ and m represent the concentration, the charge, the relaxation time and the mass of electron in a metal, then the resistance of a wire made of the metal of length '$l$' and area of cross-section A is (A) $\dfrac{ml}{ne^2\tau A}$ (B) $\dfrac{m\tau^2 A}{ne^2 l}$ (C) $\dfrac{ne^2\tau A}{2ml}$ (D) $\dfrac{ne^2 A}{2m\tau l}$

   ›Reveal solutionSolution

   The standard free-electron (Drude) derivation gives resistivity $\rho = m/(ne^2\tau)$, so $R = \rho l/A = ml/(ne^2\tau A)$.

   Concept and Intuition

   In the free-electron model of conduction, electrons accelerate under an applied field between collisions (average time $\tau$), giving a drift velocity proportional to $eE\tau/m$. This leads to a current density $J = ne^2\tau E/m$, i.e. conductivity $\sigma = ne^2\tau/m$ and resistivity $\rho = 1/\sigma = m/(ne^2\tau)$.

   Step-by-Step Solution

   1. Drift velocity: $v_d = \dfrac{eE\tau}{m}$.
   2. Current density: $J = nev_d = \dfrac{ne^2\tau}{m}E$.
   3. So conductivity $\sigma = \dfrac{ne^2\tau}{m}$ and resistivity $\rho = \dfrac{1}{\sigma} = \dfrac{m}{ne^2\tau}$. …

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9. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.The drift velocity of electrons in a conducting wire connected to a cell is V$_d$. If the length of the wire is doubled and area of cross-section is halved then the drift velocity of electrons becomes (A) V$_d$ (B) $\frac{V_d}{2}$ (C) 2V$_d$ (D) 4V$_d$

   ›Reveal solutionSolution

   Since drift velocity is set by the electric field ($v_d=\mu E$, $E=V/L$) and the cell's voltage is unchanged, doubling the wire's length halves the drift velocity; the area change doesn't matter for this quantity.

   Concept and Intuition

   Drift velocity of electrons is related to the electric field inside the conductor by $v_d = \dfrac{eE\tau}{m}$, where $\tau$ (relaxation time) and other constants are material properties independent of the wire's geometry. The electric field in a wire of length $L$ connected to a fixed-voltage source is $E = V/L$. So $v_d$ depends only on $V$ and $L$ — not on the cross-sectional area directly.

   Step-by-Step Solution

   1. Original: $v_d = \dfrac{eE\tau}{m}$, with $E = \dfrac{V}{L}$, so $v_d \propto \dfrac{V}{L}$.
   2. The wire remains connected to the same cell, so $V$ (EMF/terminal voltage) is unchanged.
   3. New length $L' = 2L$. New field: $E' = \dfrac{V}{2L} = \dfrac{E}{2}$.
   4. New drift velocity: $v_d' = \dfrac{eE'\tau}{m} = \dfrac{1}{2}\cdot\dfrac{eE\tau}{m} = \dfrac{V_d}{2}$. …

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10. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.Charge passing through a conductor of cross-section $0.3\ \mathrm{m}^2$ is given by $q=(3t^2+5t+2)\ \mathrm{C}$ where 't' is in seconds. The drift velocity at $t = 2$ s is (Concentration of electrons in the conductor $= 2\times10^{25}\ \mathrm{m}^{-3}$) (A) $0.77\times10^{-5}\ \mathrm{ms}^{-1}$ (B) $0.93\times10^{-5}\ \mathrm{ms}^{-1}$ (C) $1.77\times10^{-5}\ \mathrm{ms}^{-1}$ (D) $2.08\times10^{-5}\ \mathrm{ms}^{-1}$

    ›Reveal solutionSolution

    Differentiating the given charge-time relation gives the instantaneous current at $t=2$s, and dividing by $nAe$ (from $I=nAev_d$) gives the drift velocity, $\approx1.77\times10^{-5}\ \mathrm{ms^{-1}}$.

    Concept and Intuition

    Current is the rate of flow of charge, $I=\dfrac{dq}{dt}$. Microscopically, this current arises from the drift of free electrons, related by $I=nAev_d$, where $n$ is the number density of charge carriers, $A$ the conductor's cross-sectional area, $e$ the electronic charge, and $v_d$ the drift velocity. Combining these lets us extract the drift velocity from a time-varying charge function.

    Step-by-Step Solution

    1. Charge: $q(t)=3t^2+5t+2$. Current: $I(t)=\dfrac{dq}{dt}=6t+5$.
    2. At $t=2\ \mathrm s$: $I=6(2)+5=12+5=17\ \mathrm A$.
    3. Drift velocity relation: $I=nAev_d\Rightarrow v_d=\dfrac{I}{nAe}$.
    4. Substitute $n=2\times10^{25}\ \mathrm{m^{-3}}$, $A=0.3\ \mathrm{m^2}$, $e=1.6\times10^{-19}\ \mathrm C$: …

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11. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.Choose the correct option with respect to the statements A and B (A): When no electric field is applied across a conductor, the path of free electrons between two successive collisions in it is straight (B): When an electric field is applied across a conductor, the drift velocity of electrons is independent of time (A) A and B are true (B) A is true and B is false (C) A is false and B is true (D) A and B are false

    ›Reveal solutionSolution

    This tests two basic facts about electron motion in conductors from the free-electron/Drude picture: straight-line paths without a field, and a time-independent (steady) drift velocity with a field. Both statements are correct.

    Concept and Intuition

    In a conductor with no electric field, free electrons move randomly at high thermal speeds, travelling in straight lines between successive collisions (no net force acts on them between collisions, so Newton's first law keeps their path straight). When an electric field is switched on, each electron additionally accelerates uniformly between collisions; averaging over the very large number of electrons and their very short relaxation time $\tau$, one obtains a drift velocity $v_d = \dfrac{eE\tau}{m}$ that depends only on the field and the material's relaxation time — not on how long the field has been applied, since the steady-state drift is established almost instantaneously.

    Step-by-Step Solution

    1. Statement A: with no external field, the net force on a free electron between collisions is zero, so by Newton's first law its path is a straight line. This is standard free-electron theory — A is TRUE. …

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12. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.An electron takes $40\times10^3$ s to drift from one end of a metal wire of length 2 m to its other end. The area of cross-section of the wire is $4\,\mathrm{mm^2}$ and it is carrying a current of 1.6 A. The number density of free electrons in the metal wire is (A) $8\times10^{28}\,\mathrm{m^{-3}}$ (B) $6\times10^{28}\,\mathrm{m^{-3}}$ (C) $4\times10^{28}\,\mathrm{m^{-3}}$ (D) $5\times10^{28}\,\mathrm{m^{-3}}$

    ›Reveal solutionSolution

    Use the drift-velocity relation $I = neAv_d$ with $v_d=L/t$ to find $n = 5\times10^{28}\,\mathrm{m^{-3}}$.

    Concept and Intuition

    Current is carried by free electrons drifting slowly through the wire; $I = n e A v_d$ links the microscopic picture (electron number density and drift speed) to the macroscopic current we measure.

    Step-by-Step Solution

    1. Drift velocity: $v_d = \dfrac{L}{t} = \dfrac{2\,\text{m}}{40\times10^{3}\,\text{s}} = 5\times10^{-5}$ m/s.
    2. From $I = n e A v_d$: $n = \dfrac{I}{eAv_d}$.
    3. Substitute: $n = \dfrac{1.6}{(1.6\times10^{-19})(4\times10^{-6})(5\times10^{-5})}$.
    4. Denominator $= 1.6\times10^{-19}\times4\times10^{-6}\times5\times10^{-5} = 3.2\times10^{-29}$. …

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