(a) Define 'current density'. Is it a scalar or a vector ? An electric field $\vec{E}$ is maintained in a metallic conductor. If $n$ be the number of electrons (mass $m$, charge $-e$) per unit volume in the conductor and $\tau$ its relaxation time, show that the current density $\vec{j} = \alpha\vec{E}$, where $\alpha = \dfrac{ne^2}{m}\tau$. OR (b) What is a Wheatstone bridge ? Obtain the necessary conditions under which the Wheatstone bridge is balanced.

> [!TLDR] > Part (a): current density is the current per unit area — a vector; from the drift-velocity (Drude) model $\vec j=\dfrac{ne^2\tau}{m}\vec E=\alpha\vec E$. Part (b): a Wheatstone bridge is four resistors with a galvanometer; at balance ($I_g=0$) the condition is $\dfrac{P}{Q}=\dfrac{R}{S}$. ## Part (a) ### Definition Current density $\vec j$ is the electric current flowing per unit area of cross-section held perpendicular to the flow. It is a **vector**, directed along the conventional current (opposite to the electron drift). ### Deriving $\vec j=\alpha\vec E$ (Drude model) 1. **Force and acceleration.** An electron of charge $-e$ and mass $m$ in field $\vec E$ experiences $\vec F=-e\vec E$, giving $$\vec a=\frac{\vec F}{m}=-\frac{e}{m}\vec E.$$ 2. **Drift velocity.** Between collisions the electron accelerates for the average relaxation time $\tau$; the mean velocity it acquires is $$\vec v_d=-\frac{e\tau}{m}\vec E.$$ 3. **Current density.** If there are $n$ electrons per unit volume, in unit time all electrons within a distance $|\vec v_d|$ cross unit area, so $$\vec j=n(-e)\vec v_d=n(-e)\left(-\frac{e\tau}{m}\vec E\right)=\frac{ne^2\tau}{m}\vec E.$$ 4. **Identify the constant.** Comparing with $\vec j=\alpha\vec E$, $$\boxed{\alpha=\frac{ne^2\tau}{m}},$$ …

(a) Define 'current density'. Is it a scalar or a vector ? An electric field vecE is maintained in a metallic

Question

\vec{E}

(OR)

(b) What is a Wheatstone bridge ? Obtain the necessary conditions under which the Wheatstone bridge is balanced.

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Part (a)Concept understanding — Drift Velocity

Drift Velocity: The Slow March of Electrons

Electrons in a metal are always moving — but randomly. At room temperature they zip around at roughly $10^6$ m/s, colliding with the lattice ions every few trillionths of a second. Without an electric field this motion cancels out: for every electron heading left another heads right, so the net velocity is zero.

Apply a battery and the field gives every electron a tiny, steady push in one direction. Between collisions the electron accelerates only briefly before smashing into an ion and losing its directed motion. What survives is a very small average velocity along the field — the drift velocity.

Note

The random thermal speed is about $10^5$ m/s, but the drift velocity is only about $10^{-4}$ m/s — about a billion times slower. An electron drifts slower than a snail, yet a lamp lights instantly, because the electric field (not the electrons) propagates at nearly the speed of light and starts every electron drifting almost at once.

The precise definition

Drift velocity ( $v_d$ ) is the average velocity acquired by the charge carriers in a conductor under an applied electric field:

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

where:

$e$ = electron charge ( $1.6 \times 10^{-19}$ C)
$E$ = electric field inside the conductor (V/m)
$\tau$ = average relaxation time — the mean time between collisions (s)
$m$ = electron mass ( $9.1 \times 10^{-31}$ kg)

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

Linking to current

Drift velocity connects the microscopic motion of electrons to the current an ammeter reads:

I = n e A v_d

where $n$ is the free-electron number density and $A$ the cross-sectional area. A larger $v_d$ means more current, but $v_d$ stays tiny because $\tau$ is tiny (about $10^{-14}$ s in copper).

Tip

For a copper wire carrying $1$ A with area $1$ mm² and $n \approx 8.5 \times 10^{28}$ m $^{-3}$ :

$v_d = \frac{I}{neA} \approx \frac{1}{(8.5\times10^{28})(1.6\times10^{-19})(10^{-6})} \approx 7 \times 10^{-5}\ \text{m/s}$ …

Why this formula?

Drift Velocity: Why the Formula Holds

Let's build this from first principles — understanding why electrons drift the way they do, not just memorizing the formula.

1. The Core Idea: What is Drift Velocity?

In a conductor, free electrons are constantly moving randomly (thermal motion, speeds ~ $10^5$ m/s). Without an electric field, their net displacement is zero — they're like a swarm of bees buzzing in all directions.

When we apply an electric field $\vec{E}$ , it gently nudges each electron in the opposite direction (since electrons are negatively charged). This small, steady net velocity superimposed on the random motion is drift velocity ( $v_d$ ).

Key insight: Drift velocity is not the speed of individual electrons — it's the average velocity of the entire electron cloud.

2. The Derivation: Step by Step

Step 1: Force on a single electron

An electron of charge $-e$ in an electric field $\vec{E}$ experiences:

\vec{F} = -e \vec{E}

The magnitude of acceleration (opposite to $\vec{E}$ ) is:

a = \frac{F}{m} = \frac{eE}{m}

where $m$ is the electron's mass.

Step 2: What happens between collisions?

Electrons don't accelerate forever — they keep colliding with atoms/ions in the metal lattice. Let the average time between collisions be $\tau$ (relaxation time). Just after a collision an electron's velocity is essentially random (zero average in the field direction); it then accelerates for time $\tau$ before the next collision.

Step 3: Drift velocity

Averaging the field-driven velocity over the relaxation time $\tau$ gives the net drift:

\boxed{v_d = \frac{eE\tau}{m}}

Here $\tau$ is the average time since the last collision, so this expression already averages over electrons at every stage between collisions — it is the standard result used in the NCERT treatment.

3. Connecting to Current: The Big Picture

Drift velocity directly gives us current density $J$ :

J = n e v_d

where $n$ = number of free electrons per unit volume. …

Part (b)Concept understanding — Wheatstone Bridge

The Wheatstone Bridge – From Intuition to Precision

Imagine you have a single unknown resistor and you want to find its value. You could use an ohmmeter, but those are not always accurate for very small or very large resistances. A more elegant method is to compare it against known resistances in a circuit that acts like a balance scale — that is the Wheatstone bridge.

The core idea is simple: make the voltage at two points equal, so no current flows between them. When that happens, you know the ratio of the resistances.

The Circuit Layout

The bridge has four resistors arranged in a diamond shape:

        A
       / \
      P   Q
     /     \
    B-------C
     \     /
      R   S
       \ /
        D

A battery is connected across A and D. A sensitive galvanometer (G) is connected between B and C. The four resistors are labelled P, Q, R, and S. Usually, three are known and one (say, S) is unknown.

The Intuition: Two Voltage Dividers

Look at the left side: from A to D through P and R. That is a voltage divider. The voltage at B is a fraction of the battery voltage, determined by the ratio of P to R.

Now look at the right side: from A to D through Q and S. That is another voltage divider. The voltage at C is a fraction of the battery voltage, determined by the ratio of Q to S.

If the voltage at B equals the voltage at C, then no current flows through the galvanometer — the bridge is balanced.

The Condition for Balance

When the bridge is balanced, the voltage drop across P equals the voltage drop across Q (since both start at A), and the voltage drop across R equals the voltage drop across S (since both end at D). From the voltage divider rule:

Voltage at B: $V_B = V_A \cdot \frac{R}{P+R}$
Voltage at C: $V_C = V_A \cdot \frac{S}{Q+S}$

Setting $V_B = V_C$ gives:

\frac{R}{P+R} = \frac{S}{Q+S}

Cross-multiply:

R(Q+S) = S(P+R)

RQ + RS = SP + SR

The $RS$ terms cancel, leaving:

RQ = SP

Or, rearranged:

\frac{P}{Q} = \frac{R}{S}

\frac{P}{Q} = \frac{R}{S}

That is the balance condition of the Wheatstone bridge. When this holds, the galvanometer shows zero deflection.

Measuring an Unknown Resistance

Suppose S is unknown. You set P, Q, and R to known values. You adjust R (or the ratio P/Q) until the galvanometer reads zero. Then you compute:

S = \frac{Q}{P} \cdot R

This is why the bridge is so useful: you do not need to measure current or voltage accurately — you only need to detect when current is zero. That is far more sensitive and precise.

Tip

In practice, P and Q are often made equal (a 1:1 ratio), so the unknown S simply equals R. This is the "equal-arm" bridge.

Why It Works So Well

The galvanometer is a null detector — it only tells you whether current is flowing, not how much. This eliminates errors from meter calibration, battery voltage fluctuations, and temperature effects. The accuracy depends only on the precision of the known resistors. …

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Part (a): current density is the current per unit area — a vector; from the drift-velocity (Drude) model $\vec j=\dfrac{ne^2\tau}{m}\vec E=\alpha\vec E$ . Part (b): a Wheatstone bridge is four resistors with a galvanometer; at balance ( $I_g=0$ ) the condition is $\dfrac{P}{Q}=\dfrac{R}{S}$ .

Part (a)

Definition

Current density $\vec j$ is the electric current flowing per unit area of cross-section held perpendicular to the flow. It is a vector, directed along the conventional current (opposite to the electron drift).

Deriving $\vec j=\alpha\vec E$ (Drude model)

Force and acceleration. An electron of charge $-e$ and mass $m$ in field $\vec E$ experiences $\vec F=-e\vec E$ , giving

\vec a=\frac{\vec F}{m}=-\frac{e}{m}\vec E.

Drift velocity. Between collisions the electron accelerates for the average relaxation time $\tau$ ; the mean velocity it acquires is

\vec v_d=-\frac{e\tau}{m}\vec E.

Current density. If there are $n$ electrons per unit volume, in unit time all electrons within a distance $|\vec v_d|$ cross unit area, so

\vec j=n(-e)\vec v_d=n(-e)\left(-\frac{e\tau}{m}\vec E\right)=\frac{ne^2\tau}{m}\vec E.

Identify the constant. Comparing with $\vec j=\alpha\vec E$ , $\boxed{\alpha=\frac{ne^2\tau}{m}},$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 40 on this concept.

CBSE 2026Set 55/3/11 markMCQ
$\dfrac{P}{Q} = \dfrac{R}{S}$
›Reveal solutionSolution
The assertion is true because the Wheatstone bridge is a reciprocal network, but the reason given (a trivial algebraic manipulation) does not explain why interchanging the cell and galvanometer preserves balance. The correct option is (B).

The Wheatstone bridge is a beautiful example of a reciprocal circuit. When balanced, no current flows through the galvanometer because the potential difference across it is zero. The question asks whether swapping the positions of the cell and galvanometer affects this balance condition.

The assertion claims the balance condition $\frac{P}{Q} = \frac{R}{S}$ remains unchanged after the swap. This is indeed true, and follows from the reciprocity theorem in circuit theory: in a linear, bilateral network (one with resistors only, no diodes or other one-way elements), interchanging a voltage source and a current-measuring device does not change the current through the measuring device.

The reason given, however, is just the algebraic statement that $\frac{P}{Q} = \frac{R}{S}$ implies $\frac{Q}{P} = \frac{S}{R}$ . While mathematically correct, this doesn't explain anything about the physical interchange of components. It's a red herring.

Let me show why the assertion is actually true:

Original configuration: The cell is connected between two opposite nodes (say A and C), and the galvanometer between the other two (B and D). At balance, the potentials at B and D are equal, so $V_B = V_D$ .

Deriving the balance condition: Using voltage dividers along the two arms:

$V_B = V_A + \frac{Q}{P+Q}(V_C - V_A), \quad V_D = V_A + \frac{S}{R+S}(V_C - V_A)$
Setting $V_B = V_D$ gives $\frac{Q}{P+Q} = \frac{S}{R+S}$ , which simplifies to $\frac{P}{Q} = \frac{R}{S}$ .

After interchange: Now the cell is between B and D, and the galvanometer between A and C. For balance, we need $V_A = V_C$ (no current through the galvanometer).

New balance condition: With the cell across B–D, we can write: …
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CBSE 2026Set V11 markMCQ
$Q. Consider the following statements about a balanced Wheatstone's bridge. Statement-I : The current through the galvanometer is zero. Statement-II : If the positions of the galvanometer and the battery are interchanged in the circuit, the current in the galvanometer will be zero. Among the above two statements :$
(a) Only Statement-I is true
(b) Only Statement-II is true
(c) Both the Statements are wrong
(d) Both the Statements are true
›Reveal solutionSolution

(d) Both the Statements are true …

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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

…

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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

…

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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.

…

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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.

…

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CBSE 2026Set SEM31 markMCQ
$Q. In which case will the null condition of a Wheatstone bridge change ?$
(a) If the resistances in different arms are changed
(b) If the positions of the battery and the galvanometer are interchanged
(c) If a battery of different emf is used
(d) If a galvanometer of different resistance is used
›Reveal solutionSolution
A Wheatstone bridge is balanced when P/Q = R/S — only the four arm resistances matter. Changing an arm's resistance upsets balance; interchanging the cell and galvanometer, or changing their values/emf, does not. Option (a).

Step 1 — balance condition (NCERT/CBSE Class 12 Physics, Current Electricity): P/Q = R/S for the four ratio arms.

Step 2 — evaluate each option:

(a) Changing arm resistances alters the P/Q or R/S ratio → balance changes. This is the correct choice. …
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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

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

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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

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

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CBSE 2025Set ANNUAL1 mark
$Q. What is balanced condition of Wheatstone bridge ?$
›Reveal solutionSolution

A Wheatstone bridge is balanced when the ratio of resistances in the two arms is equal on both sides, so the galvanometer carries no current.

A Wheatstone bridge has four resistances P, Q, R, S arranged in a diamond, with a galvanometer connected across the middle (between the P–Q junction and the R–S junction) and a battery driving current through the outer loop. The bridge is said to be balanced when the potential at the galvanometer's two terminals is equal, so no current flows through it ( $I_g = 0$ ).

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Wheatstone bridge is used to measure :$
(a) e.m.f.
(b) potential
(c) resistance
(d) current
›Reveal solutionSolution

A Wheatstone bridge is a four-arm resistance network used to accurately measure an unknown resistance by balancing it against three known resistances.

The Wheatstone bridge consists of four resistors arranged in a diamond/quadrilateral, with a galvanometer connected across one diagonal and a battery across the other. By adjusting the known resistances until the galvanometer shows zero deflection (balanced condition, $P/Q = R/S$ ), the unknown resistance can be calculated precisely from the …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. In the circuit, it is given that AB = 6 Ω, BC = 3 Ω, CD = 6 Ω, DA = 12 Ω and G = 10 Ω. Current through the galvanometer will be$
(a) 8.7 mA
(b) 7.8 mA
(c) 8.7 A
(d) 0 A
›Reveal solutionSolution

The Wheatstone-bridge balance condition AB/BC = AD/DC is satisfied exactly, so no current flows through the galvanometer.

For the bridge A-B-C-D with the galvanometer across the B-D diagonal, the bridge is balanced when

$\dfrac{AB}{BC}=\dfrac{AD}{DC}$

Here $AB=6\,\Omega,\ BC=3\,\Omega,\ AD=12\,\Omega,\ DC=6\,\Omega$ :

$\dfrac{6}{3}=2,\qquad \dfrac{12}{6}=2$ …

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Part (a)

Definition

Deriving j⃗=αE⃗\vec j=\alpha\vec Ej​=αE (Drude model)

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Deriving $\vec j=\alpha\vec E$ (Drude model)

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