Q.A copper wire has a free-electron number density $n = 8.5 \times 10^{28}\ \text{m}^{-3}$ and a cross-sectional area $A = 1 \times 10^{-6}\ \text{m}^2$. It carries a steady current of $1.5\ \text{A}$. Calculate the drift velocity of the free electrons.

Showing the 12 most recent of 31 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Drift velocity Vd varies with the intensity of electric field E as per the relation

   (a) Vd is proportional to E^2

   (b) Vd is proportional to 1/E

   (c) Vd is proportional to sqrt(E)

   (d) Vd is proportional to E

   ›Reveal solutionSolution

   Drift velocity is the (small) average velocity electrons gain between collisions due to the electric field, and it comes out directly proportional to E.

   When an electric field E is applied to a conductor, each free electron experiences a force F = eE, giving it an acceleration a = eE/m between collisions with the lattice ions. If tau is the average time between collisions (relaxation time), the average extra velocity gained (the drift velocity) is

   vd = a * tau = (eE/m) * tau = (e*tau/m) * E

   Since e, tau, and m are constants for a given conductor at a given temperature, vd is directly proportional to E. This is exactly why Ohm's law (V = IR, i.e. current density J proportional to E) holds for ohmic conductors.

   ✓Final answer

   (d) Vd is proportional to E.

2. CBSE 2026Set ANNUAL1 mark
   Q.If the current flowing in a copper wire be allowed to flow in another copper wire of same length but of doubled the radius then what will be the effect on the drift velocity of the electron?

   ›Reveal solutionSolution

   For the same current, $v_d \propto 1/A$, and doubling the radius quadruples the cross-sectional area.

   Current is related to drift velocity by $I = nAev_d$, so for the same current $I$ (and the same material, hence the same $n$), $v_d = \dfrac{I}{nAe} \propto \dfrac{1}{A}$. If the radius is doubled, the cross-sectional area $A=\pi r^2$ becomes 4 times larger. So the drift velocity becomes

   $v_d' = \frac{v_d}{4}$

   ✓Final answer

   The drift velocity decreases to $\dfrac14$ of its original value, since area (and hence $v_d\propto1/A$) changes by a factor of 4 when the radius doubles.

3. CBSE 2026Set ANNUAL1 mark
   Q.State Ohm's law in terms of current density, specific conductance and electric field intensity.

   ›Reveal solutionSolution

   Microscopic Ohm's law: current density J = σE (σ = conductivity, E = field).

   The usual Ohm's law is V = IR. In microscopic (vector) form, it relates the current density J (current per unit cross-sectional area) to the electric field E inside the conductor through the material's specific conductance (conductivity) σ:

   J = σ E.

   Here σ is the conductivity of the material. Since resistivity ρ = 1/σ, this can equally be written E = ρ J. This form shows Ohm's law is a property of the material at each point, independent of geometry.

   ✓Final answer

   J = σE (current density = conductivity × electric field intensity).

4. CBSE 2026Set SEM31 markMCQ
   Q.Which of the following statement(s) is/are true ? A potential difference of V is applied at the two ends of a conductor of length l and area of cross-section A. Statement I : When potential difference is doubled, current density also gets doubled. Statement II : When potential difference is doubled, drift velocity gets halved. Statement III : When area of cross-section is doubled, current density decreases.

   (a) I and II are true

   (b) Only I is true

   (c) Only III is true

   (d) II and III are true

   ›Reveal solutionSolution

   Doubling V doubles E, so J = σE and v_d = μE both double — Statement I true, Statement II (drift velocity halved) false. J = V/(ρl) is independent of area, so Statement III (J decreases when A doubles) is also false. Only I is true → option (b).

   Statement I: J = σE and E = V/l, so doubling V doubles E and hence doubles the current density J. TRUE.

   Statement II: drift velocity v_d = (eE/m)τ ∝ E ∝ V. Doubling V doubles v_d, it does not halve it. FALSE.

   Statement III: with V fixed, I = V/R = VA/(ρl), so J = I/A = V/(ρl) — independent of the cross-sectional area. Doubling A does not decrease J. FALSE.

   Hence only Statement I is correct. These drift-velocity and current-density relations are core NCERT/CBSE Class 12 Physics (Current Electricity).

   ✓Final answer

   (b) Only I is true

5. CBSE 2025Set D1 markMCQ
   Q.The relation between drift velocity v of free electrons in conductor in electric conduction and potential difference V between ends of conductor is (A) proportional to V (B) inversely proportional to V (C) proportional to V^2 (D) inversely proportional to V^2

   ›Reveal solutionSolution

   Drift velocity is directly proportional to the potential difference V.

   In a conductor of length L across which a potential difference V is applied, the electric field is E = V/L. Free electrons acquire a drift velocity

   $v_d = \frac{eE\tau}{m} = \frac{eV\tau}{mL}$

   where τ is the mean free time and m the electron mass. For a fixed conductor (fixed L, τ, m), v_d ∝ V.

   ✓Final answer

   (A) proportional to V.

6. CBSE 2025Set D1 markMCQ
   Q.If the length of a conductor is doubled while keeping the potential difference across it constant, then the drift velocity of electron will (A) remain the same (B) be double (C) be halved (D) increase fourfold

   ›Reveal solutionSolution

   Doubling the length at constant V halves the drift velocity.

   The drift velocity is

   $v_d = \frac{eE\tau}{m} = \frac{eV\tau}{mL}$

   With the potential difference V held constant, v_d is inversely proportional to the length L (because the field E = V/L weakens as L increases). Doubling L → E halves → v_d halves.

   ✓Final answer

   (C) be halved.

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.A thick wire is stretched so that its length becomes two times. What is the ratio of change in resistance of the wire to the initial resistance of the wire?

   (i) 2 : 1

   (ii) 4 : 1

   (iii) 3 : 1

   (iv) 1 : 4

   ›Reveal solutionSolution

   New resistance is 4 times the old, so the change is 3 times the original: ratio 3 : 1.

   Resistance $R = \rho L/A$. Stretching keeps the volume $AL$ constant, so if length doubles ($L \to 2L$) the area halves ($A \to A/2$). Then $R' = \rho (2L)/(A/2) = 4\rho L/A = 4R$. The change in resistance is $R' - R = 4R - R = 3R$. Hence $\frac{\Delta R}{R} = \frac{3R}{R} = 3$, i.e. 3 : 1.

   ✓Final answer

   (iii) 3 : 1.

8. CBSE 2025Set ANNUAL1 markMCQ
   Q.Calculate the amount of charge flowing in 2 minutes in a wire of resistance 10 ohm when a potential difference of 20 volts is applied between its ends.

   (i) 120 C

   (ii) 240 C

   (iii) 20 C

   (iv) 4 C

   ›Reveal solutionSolution

   Q = It = (V/R) x t = 2 A x 120 s = 240 C.

   Current $I = \frac{V}{R} = \frac{20}{10} = 2\,A$. Time $t = 2\,min = 120\,s$. Charge $Q = I t = 2 \times 120 = 240\,C$.

   ✓Final answer

   (ii) 240 C.

9. CBSE 2024Set 55/2/11 markMCQ
   Q.Electrons drift with speed $v_d$ in a conductor with potential difference $V$ across its ends. If $V$ is reduced to $\dfrac{V}{2}$, their drift speed will become : (A) $\dfrac{v_d}{2}$ (B) $v_d$ (C) $2v_d$ (D) $4v_d$

   ›Reveal solutionSolution

   Drift speed is directly proportional to the applied potential difference for a given conductor, so halving $V$ halves $v_d$. The new drift speed is $\dfrac{v_d}{2}$, which is option (A).

   The key here is understanding what drift speed actually depends on. Many students memorise the formula $v_d = \frac{I}{n e A}$ and then try to relate $I$ to $V$ via Ohm's law — that works, but it's easy to lose track of which quantities stay constant. Let's build it from the physics up.

   Drift speed is the average velocity electrons acquire due to an electric field inside the conductor. That field is $E = V/L$, where $L$ is the length of the conductor. The force on each electron is $eE$, and in the steady state, this force is balanced by collisions with the lattice, giving a constant drift speed proportional to the field. So the fundamental proportionality is:

   $v_d \propto E \quad \text{and since} \quad E = \frac{V}{L}, \quad \text{we get} \quad v_d \propto V$

   for a fixed conductor (fixed $L$, fixed material properties like relaxation time $\tau$, mass $m$, charge $e$).

   Now let's walk through it step by step.

   1. Start with the microscopic relation. The drift speed is given by $v_d = \frac{eE\tau}{m}$, where $\tau$ is the average time between collisions (relaxation time). This comes from $F = eE = m a$, and then $v_d = a\tau$. For a given conductor at a fixed temperature, $\tau$, $m$, and $e$ are constants.

   2. Express the electric field in terms of the applied voltage. For a conductor of length $L$, the uniform electric field inside is $E = V/L$. So:

   $v_d = \frac{e (V/L) \tau}{m} = \left(\frac{e\tau}{mL}\right) V$

   The quantity in parentheses is constant for a given conductor. So $v_d$ is directly proportional to $V$.

   3. Apply the change. If $V$ becomes $V/2$, then:

   $v_d' = \left(\frac{e\tau}{mL}\right) \cdot \frac{V}{2} = \frac{1}{2} \left(\frac{e\tau}{mL}\right) V = \frac{v_d}{2}$

   The drift speed halves.

   Watch out

   A common mistake is to think $v_d \propto I$ and then use $I = V/R$ to get $v_d \propto V$, which is correct — but only if $R$ is constant. For a metallic conductor at constant temperature, $R$ is indeed constant, so the proportionality holds. The danger is when students blindly apply $v_d = I/(n e A)$ without realising that $I$ itself changes with $V$.

   Tip

   You can also think of it this way: drift speed is the "terminal velocity" of electrons under an electric field. Just like a ball falling through a fluid reaches a terminal speed proportional to the driving force, here the driving force is $eE$, so halving $V$ halves $E$, which halves the force, and thus halves the drift speed.

   ✓Final answer

   The new drift speed is $\boxed{\dfrac{v_d}{2}}$, which corresponds to option (A).

10. CBSE 2024Set FS1 markMCQ
    Q.If drift velocity of electron be $v_d$ and intensity of electric field E, then which relation among the following obeys Ohm's law?

    (i) $v_d \propto E^2$

    (ii) $v_d = \text{Constant}$

    (iii) $v_d \propto E$

    (iv) $v_d \propto \sqrt{E}$

    ›Reveal solutionSolution

    Drift velocity $v_d=\dfrac{eE\tau}{m}$ is directly proportional to the field $E$, which is exactly what Ohm's law requires — option (iii).

    Concept. Under an electric field $E$, free electrons acquire a steady drift velocity

    $v_d=\frac{eE\tau}{m},$

    where $\tau$ is the mean relaxation time.

    Why. Ohm's law states $V\propto I$, i.e. current density $J=nev_d$ is proportional to $E$ ($J=\sigma E$). Since $n$, $e$ are fixed, $J\propto v_d$, so Ohm's law holds only when $v_d\propto E$.

    ✓Final answer

    (iii) $v_d\propto E$

11. CBSE 2024Set ANNUAL1 mark
    Q.Define the relaxation time of the free electrons drifting in a conductor.

    ›Reveal solutionSolution

    The average free time an electron travels between collisions with the lattice.

    Free electrons in a conductor are in continuous random thermal motion, frequently colliding with the fixed lattice ions. The relaxation time $\tau$ is defined as the average time interval between two successive collisions of a free electron with the ions of the conductor. It is a key quantity in the microscopic model of conduction, since the drift velocity acquired under an applied field is $v_d = eE\tau/m$, and the conductivity $\sigma = ne^2\tau/m$ depends directly on $\tau$.

    ✓Final answer

    Relaxation time is the average time between two successive collisions of a free electron with the conductor's lattice ions.

12. CBSE 2023Set 55/1/11 markMCQ
    Q.A steady current flows through a metallic wire whose area of cross-section $(A)$ increases continuously from one end of the wire to the other. The magnitude of drift velocity $(v_d)$ of the free electrons as a function of $A$ can be represented by :

    (a)

    (b)

    (c)

    (d)

    ›Reveal solutionSolution

    For a steady current, the product $A v_d$ is constant because $I = n e A v_d$ is fixed. Therefore $v_d \propto 1/A$, which is a rectangular hyperbola — option (a).

    

    The key to this question is understanding why drift velocity changes when the wire's cross-section changes — and that comes from the definition of steady current itself.

    When we say "a steady current flows", we mean that the same amount of charge passes through every cross-section of the wire per second. The wire is in series with itself: whatever charge flows past the thin end must also flow past the thick end in the same time. If it didn't, charge would pile up somewhere — and that would violate the steady-state condition.

    Now, current $I$ is given by:

    $I = n e A v_d$

    where $n$ is the free electron density (number per unit volume), $e$ is the electron charge, $A$ is the cross-sectional area at that point, and $v_d$ is the drift velocity.

    For a metallic wire, $n$ and $e$ are constants (same material throughout). And for a steady current, $I$ is constant along the wire. So:

    $n e A v_d = \text{constant}$

    which means:

    $A v_d = \text{constant}$

    Therefore:

    $v_d \propto \frac{1}{A}$

    This is the relationship we need.

    1. Identify the mathematical form. $v_d \propto 1/A$ is an inverse proportion. Its graph is a rectangular hyperbola — a curve that falls steeply when $A$ is small and flattens out as $A$ grows large. It never touches either axis (asymptotic behaviour).

    2. Check the options against this.

       • Option (a) shows exactly this: a curve that drops as $A$ increases, shaped like a hyperbola.
       • Option (b) is a straight line — that would mean $v_d \propto -A + \text{constant}$, which is wrong.
       • Option (c) is a horizontal line — that would mean $v_d$ is independent of $A$, which contradicts $A v_d = \text{constant}$.
       • Option (d) is a straight line through the origin — that would mean $v_d \propto A$, which is the opposite of what we have.

    3. A common pitfall. Students sometimes think that because the wire gets thicker, electrons have more room and therefore drift faster. That's the opposite of what happens. Think of it this way: if the same number of cars per minute must pass through both a narrow tunnel and a wide tunnel, the cars must move faster through the narrow tunnel to maintain the flow rate. In the wide tunnel, they can spread out and move slower. Electrons behave the same way.

    Watch out

    Do not confuse drift velocity with the speed of individual electrons. Drift velocity is the average net velocity of the electron cloud. A larger area means more parallel paths for the same total current, so each electron's net drift is slower.

    Tip

    If you ever forget the relationship, just remember: current is flow per unit time per unit area. More area with the same flow means slower drift. The formula $I = n e A v_d$ is your anchor — keep $I$ constant and everything else follows.

    ✓Final answer

    The correct option is (a) — a rectangular hyperbola showing $v_d \propto 1/A$.