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 13 on this concept.

1. KCET 2026Set C21 markMCQ
   Q.The number of electrons moving per second through the filament of a lamp of 60W operating at 120V is nearly (e = $1.6 \times 10^{-19}$ C) _______ (A) $6.2 \times 10^{18}$ (B) $6.2 \times 10^{19}$ (C) $3.1 \times 10^{18}$ (D) $3.1 \times 10^{19}$

   ›Reveal solutionSolution

   Find the operating current from $I = P/V$, then divide by the electronic charge $e$ to get the number of electrons flowing per second.

   Step 1 — Find the current through the filament

   $I = \frac{P}{V} = \frac{60\ \text{W}}{120\ \text{V}} = 0.5\ \text{A}$

   Step 2 — Convert current to number of electrons per second

   Since $I = \dfrac{n e}{t}$, the number of electrons per second is …

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2. KCET 2026Set C21 markMCQ
   Q.In a conducting region, $10^{19}$ electrons and $10^{19}$ protons move to the left, while $10^{19}$ $\alpha$-particles move to the right per second. The resulting electric current is (e = $1.6 \times 10^{-19}$ C) _______ (A) 3.2 A towards left (B) 3.2 A towards right (C) 1.6 A towards left (D) 1.6 A towards right

   ›Reveal solutionSolution

   Conventional current follows the direction of positive charge flow; a negative charge moving one way is equivalent to a positive current flowing the opposite way. Convert each species and add up the net current.

   Step 1 — Current due to protons (moving left)

   Protons carry charge $+e$ and move left, so they contribute a conventional current to the LEFT:

   $I_p = n e = (10^{19})(1.6\times10^{-19}) = 1.6\ \text{A (left)}$

   Step 2 — Current due to electrons (moving left)

   Electrons carry charge $-e$ and move left; a negative charge moving left is equivalent to a positive current flowing RIGHT:

   $I_e = n e = (10^{19})(1.6\times10^{-19}) = 1.6\ \text{A (right, equivalent)}$

   Step 3 — Current due to $\alpha$-particles (moving right) …

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3. KCET 2025Set D-41 markMCQ
   Q.The variations of resistivity $\rho$ with absolute temperature $T$ for three different materials X, Y and Z are shown in the graph below. Identify the materials X, Y and Z. (A) X – copper, Y – semiconductor, Z – nichrome (B) X – semiconductor, Y – nichrome, Z – copper (C) X – nichrome, Y – copper, Z – semiconductor (D) X – copper, Y – nichrome, Z – semiconductor

   ›Reveal solutionSolution

   Use the sign of $d\rho/dT$ to pick out the semiconductor, then use the magnitude of $\rho$ and of the temperature coefficient to separate pure copper from the alloy nichrome.

   Step 1 — The physics behind $\rho(T)$.

   From the free-electron model,

   $\rho = \frac{m}{n e^2 \tau}$

   where $n$ is the free-carrier density and $\tau$ the mean time between collisions. Temperature acts on these two quantities in opposite ways in metals and semiconductors:

   • In a metal, $n$ is essentially fixed (one or two free electrons per atom, regardless of $T$). Raising $T$ makes the lattice ions vibrate harder, so electrons collide more often, $\tau$ falls, and hence $\rho$ rises. Over a normal range this is close to linear: $\rho = \rho_0[1 + \alpha(T - T_0)]$ with $\alpha > 0$.

   • In a semiconductor, there is an energy gap. Raising $T$ promotes many more electrons across it, so $n$ rises exponentially — an effect that overwhelms the modest fall in $\tau$. Hence $\rho$ falls sharply with $T$: $\alpha$ is negative.

   Step 2 — Identify Z.

   Curve Z starts high at low $T$ and decreases rapidly as $T$ rises — a negative temperature coefficient. Only a semiconductor behaves this way.

   $\boxed{Z = \text{semiconductor}}$

   This already eliminates options (A) and (B), which assign Z to nichrome and copper respectively.

   Step 3 — X and Y are both metallic.

   Both X and Y increase with $T$, so both have $\alpha > 0$: both are metals/alloys. The remaining choice is between (C) X–nichrome, Y–copper and (D) X–copper, Y–nichrome. Two independent clues settle it.

   Clue (i) — the intercept, i.e. the size of $\rho$.

   At room temperature:

   $\rho_{\text{copper}} \approx 1.7\times10^{-8}\ \Omega\,\text{m}, \qquad \rho_{\text{nichrome}} \approx 100\times10^{-8}\ \Omega\,\text{m}$

   Nichrome's resistivity is roughly 60 times that of copper — the very reason nichrome is used for heating elements and copper for wiring. The graph shows X starting low and Y starting higher. Therefore X = copper and Y = nichrome.

   Clue (ii) — the slope, i.e. the temperature coefficient. …

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4. COMEDK 2025Set 2025-A1 markMCQ
   Q.Two wires A and B made of same material having length 10 cm and 40 cm respectively are connected in parallel to the same source of emf 10 V . What will be the ratio of the drift velocity of the electrons in the wire $A$ to the drift velocity in the wire $B$ ? (A) $\frac{V_{d A}}{V_{d B}}=1: 4$ (B) $\frac{V_{d A}}{V_{d B}}=1: 2$ (C) $\frac{V_{d A}}{V_{d B}}=4: 1$ (D) $\frac{V_{d A}}{V_{d B}}=2: 1$

   ›Reveal solutionSolution

   The drift velocity ratio depends only on the lengths when wires of the same material are in parallel (same voltage). Since $v_d \propto 1/L$, the ratio $v_{dA}:v_{dB} = L_B:L_A = 40:10 = 4:1$, so option (C) is correct.

   Concept & Intuition

   Drift velocity $v_d$ is the average speed electrons gain under an electric field. For a wire of length $L$ with a voltage $V$ across it, the electric field inside is $E = V/L$. Since $v_d = \frac{eE}{m}\tau$ (where $\tau$ is the relaxation time, same for same material), we get $v_d \propto E \propto 1/L$ when $V$ is fixed. In a parallel connection, each wire gets the same voltage, so the shorter wire has a stronger field and thus faster drift velocity. The ratio is simply the inverse ratio of lengths.

   Step-by-step reasoning

   1. Drift velocity formula

      For a conductor, drift velocity is $v_d = \frac{eE\tau}{m}$, where $e$ is electron charge, $m$ is mass, $\tau$ is average time between collisions. For the same material, $\tau$ is identical.

   2. Electric field in each wire

      Both wires are connected to the same 10 V source in parallel, so each has $V = 10\,\text{V}$ across its ends.

      Electric field: $E = V/L$.

      For wire A: $L_A = 10\,\text{cm} = 0.1\,\text{m}$, so $E_A = 10/0.1 = 100\,\text{V/m}$.

      For wire B: $L_B = 40\,\text{cm} = 0.4\,\text{m}$, so $E_B = 10/0.4 = 25\,\text{V/m}$.

   3. Relating drift velocity to length

      Since $v_d \propto E$ and $E \propto 1/L$, we have $v_d \propto 1/L$ (for fixed voltage and same material).

      Therefore:

      $$$$

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5. COMEDK 2025Set 2025-M1 markMCQ
   Q.Two identical conductors of lengths 1 and 31 respectively are maintained at the same temperature. They are given potential differences in the ratio $1: 3$. The ratio of their drift velocities is (A) $1: 9$ (B) $9: 1$ (C) $1: 1$ (D) $1: 3$

   ›Reveal solutionSolution

   The drift velocity depends only on the applied electric field and the material properties, not on the length of the conductor. Since the electric field is the same for both conductors (potential difference divided by length gives the same ratio), the drift velocities are equal. The ratio is $1:1$, so the correct option is (C).

   Concept and Intuition

   Drift velocity $v_d$ is the average speed of charge carriers in a conductor under an electric field. For a given material at constant temperature, $v_d$ is directly proportional to the electric field $E$ inside the conductor:

   $v_d = \mu E$

   where $\mu$ is the mobility (constant for a given material and temperature). The electric field $E$ is the potential difference $V$ divided by the length $L$ of the conductor:

   $E = \frac{V}{L}$

   So the drift velocity becomes:

   $v_d = \mu \frac{V}{L}$

   The problem gives two conductors of lengths $L_1 = 1$ and $L_2 = 31$ (units arbitrary), with potential differences in the ratio $V_1 : V_2 = 1 : 3$. The key insight: we must compare the electric fields, not just the voltages, because the lengths differ.

   Step-by-step reasoning

   1. Write the drift velocity formula for each conductor

      For conductor 1: $v_{d1} = \mu \frac{V_1}{L_1}$

      For conductor 2: $v_{d2} = \mu \frac{V_2}{L_2}$

      (Mobility $\mu$ is the same because the conductors are identical and at the same temperature.)

   2. Take the ratio of drift velocities

   $\frac{v_{d1}}{v_{d2}} = \frac{V_1 / L_1}{V_2 / L_2} = \frac{V_1}{V_2} \cdot \frac{L_2}{L_1}$

   3. Plug in the given values $V_1 : V_2 = 1 : 3$ means $\frac{V_1}{V_2} = \frac{1}{3}$. $L_1 = 1$, $L_2 = 31$, so $\frac{L_2}{L_1} = \frac{31}{1} = 31$. Thus:

   $\frac{v_{d1}}{v_{d2}} = \frac{1}{3} \times 31 = \frac{31}{3}$

   That is not one of the given options — something is off.

   4. Re-examine the problem statement The lengths are given as "1 and 31" — but these are likely in the same units. However, the ratio $1:3$ for potential differences might be intended to match the length ratio so that the electric fields become equal. Let’s check: if $L_1 = 1$ and $L_2 = 3$ (not 31), then: $\frac{v_{d1}}{v_{d2}} = \frac{1}{3} \times \frac{3}{1} = 1$ …

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6. KCET 2024Set D-21 markMCQ
   Q.A body has a charge of $-3.2~\mu\text{C}$. The number of excess electrons it has is (A) $5.12 \times 10^{25}$ (B) $5 \times 10^{12}$ (C) $2 \times 10^{13}$ (D) $5.12 \times 10^{13}$

   ›Reveal solutionSolution

   Use quantisation of charge, $q = ne$, and divide the magnitude of the body's charge by the electronic charge.

   Step 1 — The concept: charge is quantised

   Any charge on a body is an integral multiple of the elementary charge:

   $q = n e, \qquad e = 1.6\times10^{-19}\ \text{C}$

   A negative body carries excess electrons; a positive one has a deficit. So the sign tells us the electrons are extra, and the magnitude tells us how many.

   Step 2 — Convert the charge to coulombs

   $|q| = 3.2\ \mu\text{C} = 3.2 \times 10^{-6}\ \text{C}$

   Step 3 — Solve for $n$

   $n = \frac{|q|}{e} = \frac{3.2\times10^{-6}}{1.6\times10^{-19}}$

   $n = \frac{3.2}{1.6}\times 10^{-6-(-19)} = 2 \times 10^{13}$ …

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7. COMEDK 2024Set 2024-A1 markMCQ
   Q.The mobility of the charge carriers increases with (A) The average collision time (B) Increase in the mass of the charge (C) Increase in the electric field (D) Decrease in the charge of the mobile carriers

   ›Reveal solutionSolution

   Mobility is defined as drift velocity per unit electric field, and it depends inversely on mass and directly on the average collision time (relaxation time). The correct option is (A).

   Concept & Intuition

   Mobility ($\mu$) tells us how quickly a charge carrier (like an electron) can move through a material when an electric field is applied. The key microscopic picture: carriers accelerate under the field, but constantly collide with atoms/impurities, which resets their velocity. The average time between collisions is called the relaxation time ($\tau$). A longer $\tau$ means the carrier accelerates for a longer stretch before being scattered, so it gains a higher drift speed — hence higher mobility. Mass and charge also matter: a heavier carrier accelerates less for the same force, and a larger charge feels a stronger force, but the question asks which increases mobility.

   Step-by-step reasoning

   1. Recall the formula for mobility From the Drude model of conduction, mobility is given by

   $\mu = \frac{e \tau}{m}$

   where $e$ is the charge of the carrier, $\tau$ is the average collision time (relaxation time), and $m$ is the mass of the carrier. This formula shows the direct proportionalities.

   2. Analyze each option
      • (A) The average collision time — $\mu \propto \tau$. If $\tau$ increases, mobility increases. This matches the intuition: longer time between collisions → higher drift velocity.
      • (B) Increase in the mass of the charge — $\mu \propto 1/m$. Heavier carriers move less readily, so mobility decreases. …

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8. COMEDK 2024Set 2024-M1 markMCQ
   Q.In a given semiconductor, the ratio of the number density of electron to number density of hole is $2: 1$. If $\frac{1}{7}$th of the total current is due to the hole and the remaining is due to the electrons, the ratio of the drift velocity of holes to the drift velocity of electrons is : (A) $\frac{2}{3}$ (B) $\frac{3}{1}$ (C) $\frac{3}{2}$ (D) $\frac{1}{3}$

   ›Reveal solutionSolution

   The key idea is to relate current contributions to charge carrier densities, drift velocities, and charges. Using the given ratios, the ratio of hole drift velocity to electron drift velocity is found to be $\frac{1}{3}$, which corresponds to option (D).

   We start with the fundamental concept: In a semiconductor, the total current is the sum of the electron current and the hole current. Each current depends on the number density of the carrier, its charge, and its drift velocity. The problem gives us ratios for densities and for current contributions, so we can set up equations that relate these quantities and solve for the unknown drift velocity ratio.

   Step-by-step reasoning:

   1. Define variables and given ratios

      Let $n_e$ be the number density of electrons and $n_h$ the number density of holes.

      Given: $\frac{n_e}{n_h} = \frac{2}{1}$, so $n_e = 2n_h$.

      Let $v_e$ and $v_h$ be the drift velocities of electrons and holes respectively.

      Let $I_e$ and $I_h$ be the currents due to electrons and holes.

      Given: $\frac{1}{7}$ of total current is due to holes, so $I_h = \frac{1}{7} I_{\text{total}}$ and $I_e = \frac{6}{7} I_{\text{total}}$.

      Hence, $\frac{I_h}{I_e} = \frac{1/7}{6/7} = \frac{1}{6}$.

   2. Write current expressions

      For a semiconductor, the current due to each carrier type is proportional to its charge magnitude, number density, and drift velocity. Since electrons and holes have the same magnitude of charge $e$, we have:

   $I_e \propto n_e e v_e \quad \text{and} \quad I_h \propto n_h e v_h.$

   Therefore, the ratio of hole current to electron current is:

   $\frac{I_h}{I_e} = \frac{n_h v_h}{n_e v_e}.$

   3. Substitute known ratios …

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9. KCET 2023Set A-31 markMCQ
   Q.The ratio of the magnitudes of electric field to the magnetic field of an electromagnetic wave is of the order of (A) $10^5~\mathrm{ms}^{-1}$ (B) $10^{-5}~\mathrm{ms}^{-1}$ (C) $10^8~\mathrm{ms}^{-1}$ (D) $10^{-8}~\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   In an EM wave $E = cB$, so the ratio of the field magnitudes is just the speed of light, of order $10^8\ \text{m s}^{-1}$.

   1. The concept — the two fields of an EM wave are not independent.

   Maxwell's equations force the electric and magnetic fields of a plane electromagnetic wave to be in phase, mutually perpendicular, perpendicular to the direction of propagation, and — crucially — to have magnitudes in a fixed ratio:

   $E_0 = c\,B_0 \qquad\Longrightarrow\qquad \frac{E_0}{B_0} = c$

   where

   $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 3\times 10^{8}\ \text{m s}^{-1}$

   2. A dimensional check confirms it.

   $\frac{[E]}{[B]} = \frac{\text{V m}^{-1}}{\text{T}} = \frac{\text{N C}^{-1}}{\text{N A}^{-1}\text{m}^{-1}} = \text{m s}^{-1}$

   The ratio has the dimensions of a speed — which is why every option is quoted in m s⁻¹. Only a value of order $10^8$ m s⁻¹ can be the speed of light.

   3. Order of magnitude. …

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10. KCET 2021Set B-21 markMCQ
    Q.A wire of resistance $3~\Omega$ is stretched to twice its original length. The resistance of the new wire will be (A) $1.5~\Omega$ (B) $3~\Omega$ (C) $6~\Omega$ (D) $12~\Omega$

    ›Reveal solutionSolution

    When a wire is stretched, its volume stays constant. Doubling the length reduces the cross-sectional area by half, and since resistance $R = \rho \frac{L}{A}$, the new resistance becomes $4$ times the original — so $12~\Omega$.

    The key here is drift velocity and the microscopic origin of resistance. Resistance depends on how hard it is for electrons to move through the wire. That difficulty comes from collisions with atoms, and it scales with the length (more distance to travel) and inversely with the cross-sectional area (more space means fewer collisions per electron). So $R \propto \frac{L}{A}$.

    When you stretch a wire, you're not adding or removing material — the volume stays the same. That's the crucial constraint. Let's use it.

    1. Volume conservation

       Original volume: $V = A_1 L_1$

       New volume: $V = A_2 L_2$

       Since $L_2 = 2 L_1$, we have $A_1 L_1 = A_2 (2 L_1)$, so $A_2 = \frac{A_1}{2}$.

    2. Resistance formula

       Original resistance: $R_1 = \rho \frac{L_1}{A_1} = 3~\Omega$

       New resistance: $R_2 = \rho \frac{L_2}{A_2} = \rho \frac{2 L_1}{A_1/2} = \rho \frac{2 L_1 \times 2}{A_1} = 4 \rho \frac{L_1}{A_1} = 4 R_1$ …

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11. KCET 2020Set A-11 markMCQ
    Q.A metal rod of length 10 cm and a rectangular cross-section of 1 cm $\times \frac{1}{2}$ cm is connected to a battery across opposite faces. The resistance will be (A) maximum when the battery is connected across 1 cm $\times \frac{1}{2}$ cm faces. (B) maximum when the battery is connected across 10 cm $\times \frac{1}{2}$ cm faces. (C) maximum when the battery is connected across 10 cm $\times$ 1 cm faces. (D) same irrespective of the three faces.

    ›Reveal solutionSolution

    Resistance depends on the length-to-area ratio $R = \rho \frac{L}{A}$. For a given volume, the longest path and smallest cross-section give the largest resistance — here, connecting across the $1 \text{ cm} \times \frac12 \text{ cm}$ faces yields the maximum.

    The key idea is deceptively simple: resistance is not a property of the object alone — it depends on which pair of opposite faces you connect the battery to. The rod has three distinct pairs of opposite faces, each giving a different length $L$ (the distance between the faces) and a different cross-sectional area $A$ (the area of the face itself). Since $R = \rho \frac{L}{A}$, you just need to compute $L/A$ for each case and compare.

    Let’s label the dimensions clearly: length = 10 cm, width = 1 cm, height = $\frac12$ cm. The three pairs of opposite faces are:

    1. Faces $1 \text{ cm} \times \frac12 \text{ cm}$ — these are the two smallest faces. The distance between them is the full length, 10 cm. The cross-sectional area is the area of that face: $1 \times \frac12 = 0.5 \text{ cm}^2$.

    2. Faces $10 \text{ cm} \times \frac12 \text{ cm}$ — these are the long, narrow side faces. The distance between them is the width, 1 cm. The cross-sectional area is $10 \times \frac12 = 5 \text{ cm}^2$.

    3. Faces $10 \text{ cm} \times 1 \text{ cm}$ — these are the large top and bottom faces. The distance between them is the height, $\frac12$ cm. The cross-sectional area is $10 \times 1 = 10 \text{ cm}^2$.

    Now compute $L/A$ for each:

    1. $L = 10 \text{ cm}$, $A = 0.5 \text{ cm}^2$ → $L/A = 10 / 0.5 = 20 \text{ cm}^{-1}$.
    2. $L = 1 \text{ cm}$, $A = 5 \text{ cm}^2$ → $L/A = 1 / 5 = 0.2 \text{ cm}^{-1}$.
    3. $L = 0.5 \text{ cm}$, $A = 10 \text{ cm}^2$ → $L/A = 0.5 / 10 = 0.05 \text{ cm}^{-1}$. …

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12. KCET 2020Set A-11 markMCQ
    Q.A rod of length $2$ m slides with a speed of $5 \text{ ms}^{-1}$ on a rectangular conducting frame as shown in figure. There exists a uniform magnetic field of $0.04$ T perpendicular to the plane of the figure. If the resistance of the rod is $3 \Omega$. The current through the rod is (A) $75 \text{ mA}$ (B) $133 \text{ mA}$ (C) $0.75 \text{ A}$ (D) $1.33 \text{ A}$

    ›Reveal solutionSolution

    Compute the motional emf $\varepsilon=B\ell v$ with the rod's length $\ell=2$ m, then apply Ohm's law with $R=3\,\Omega$.

    Step 1 — Why an emf appears (the concept).

    As the rod slides, the free charges inside it move with velocity $\vec v$ through the magnetic field $\vec B$ and feel the Lorentz force $q(\vec v\times\vec B)$. Positive charge accumulates at one end, negative at the other, setting up a potential difference — the motional emf. Equivalently (Faraday), the area of the circuit changes at a rate $\dfrac{dA}{dt}=\ell v$, so

    $\varepsilon=\left|\frac{d\Phi}{dt}\right|=B\,\frac{dA}{dt}=B\,\ell\,v$

    Step 2 — Identify the quantities.

    • Length of the sliding rod (the side perpendicular to $\vec v$, which is the effective length): $\ell=2\ \text{m}$
    • Speed: $v=5\ \text{m s}^{-1}$
    • Field (perpendicular to the plane, so $\sin\theta=1$): $B=0.04\ \text{T}$
    • Resistance of the rod (the only resistance stated): $R=3\ \Omega$

    Step 3 — Compute the emf.

    $\varepsilon=B\ell v=0.04\times 2\times 5$

    $\varepsilon=0.4\ \text{V}$

    Step 4 — Apply Ohm's law to the circuit.

    The rod is the seat of the emf and also the only resistor, so …

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