A current flows through a cylindrical conductor of radius $R$. The current density at a point in the conductor is $j = \alpha r$ (along its axis), where $\alpha$ is a constant and $r$ is the distance from the axis of the conductor. The current flowing through the portion of the conductor from $r = 0$ to $r = \frac{R}{2}$ is proportional to: (A) $R$ (B) $R^2$ (C) $R^3$ (D) $R^4$

A current flows through a cylindrical conductor of radius R. The current density at a point in the conductor i

Question

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CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★

✓ Free question

Concept understanding — Microscopic Model of Current and Current Density

Building I from first principles. Consider a conductor of cross-sectional area A carrying n free electrons per unit volume, all moving with the same drift velocity $v_d$ . In a small time interval dt, every electron within a thin slice of thickness $dx=v_d\,dt$ crosses the cross-section; the number of electrons in that slice is $n\cdot A\,dx = n\,A\,v_d\,dt$ , and since each carries charge e, the charge crossing in time dt is $dQ=neAv_d\,dt$ . Dividing by dt gives the central microscopic-model result,

I = neAv_d

directly connecting the everyday, measurable current I to the microscopic quantities n (electron density), e (electron charge), A (cross-sectional area) and $v_d$ (drift velocity).

Current density. Dividing current by cross-sectional area defines the current density, $J=I/A$ , SI unit $\text{A/m}^2$ ; substituting the result above gives $J=nev_d$ . In general, current density is treated as a vector $\vec J=ne\vec v_d$ , pointing along the direction of (conventional) current flow at a given point -- distinct from the total current I through a surface, which is the scalar quantity $I=\vec J\cdot\vec A$ built by dotting the current-density vector with the chosen surface's area vector.

Microscopic form of Ohm's law. Substituting the drift-velocity expression $\vec v_d=-(e\tau/m)\vec E$ into $\vec J=ne\vec v_d$ gives $\vec J=-(ne^2\tau/m)\vec E$ ; taking the direction of conventional current density along $\vec E$ (as is standard, since the negative sign is an artefact of tracking the electrons rather than the conventional positive-charge picture) gives the compact relation

\vec J = \sigma\vec E

where $\sigma=ne^2\tau/m$ is the material's conductivity. This relation is called the microscopic form of Ohm's law, and it is the true, fundamental starting point from which the everyday, macroscopic law $V=IR$ is later derived by integrating over a specific wire's geometry (length and area). It directly shows WHY a material conducts well or poorly: high electron density n, long mean free time $\tau$ (few collisions) and low electron mass m all push conductivity up, which is exactly the physical basis for why metals conduct far better than semiconductors.

The current is found by integrating the current density over the cross-sectional area. Since $j = \alpha r$ , the current through a radius $R/2$ scales as $R^3$ , making option (C) correct.

The key here is to understand that current density $j$ is the current per unit area. When $j$ varies with $r$ , you cannot simply multiply by area — you must integrate. The problem gives $j = \alpha r$ , meaning the current density increases linearly from zero at the centre to a maximum at the surface. We want the total current through only the inner half of the conductor's cross-section (from $r=0$ to $r=R/2$ ).

Set up the integral for current.

The current through a tiny ring of radius $r$ and thickness $dr$ is $dI = j \cdot dA$ , where $dA$ is the area of that ring. For a ring, $dA = 2\pi r \, dr$ .

So $dI = (\alpha r) \cdot (2\pi r \, dr) = 2\pi \alpha \, r^2 \, dr$ .
Integrate from the centre to $R/2$ .

I = \int_{0}^{R/2} 2\pi \alpha \, r^2 \, dr = 2\pi \alpha \left[ \frac{r^3}{3} \right]_{0}^{R/2} = 2\pi \alpha \cdot \frac{(R/2)^3}{3} = 2\pi \alpha \cdot \frac{R^3}{24} = \frac{\pi \alpha R^3}{12}.

Identify the proportionality. The result is $I = \frac{\pi \alpha}{12} R^3$ . Since $\pi$ and $\alpha$ are constants, $I \propto R^3$ .

Watch out

A common mistake is to treat $j$ as uniform and simply multiply by the area $\pi (R/2)^2$ . That would give $I = \alpha r \cdot \pi (R/2)^2$ , which is wrong because $j$ itself depends on $r$ — you cannot pick a single value of $r$ for the whole area. Always integrate when $j$ is not constant.

Tip

Notice that $j \propto r$ means the current density is zero at the axis and grows outward. The inner half carries less current than you might guess because the density there is lower. The cubic dependence on $R$ comes from the $r^2$ inside the integral — one power from the ring's circumference and one from $j$ itself.

✓Final answer

R^3

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 15 on this concept.

CBSE 2026Set A1 markMCQ
$Q. In a metallic conductor, the current is carried by (A) free electrons (B) protons (C) free electrons and protons (D) bound electrons$
›Reveal solutionSolution

Metals conduct via their sea of free electrons.

In a metallic conductor the outermost (valence) electrons of the atoms are loosely bound and become delocalised, forming a 'sea' of free electrons. When a potential difference is applied, these free electrons drift opposite to the field, constituting the electric current.

Protons and bound (core) electrons are fixed in the lattice and do not move, so they carry no current.

✓Final answer
$(A) free electrons.$
CBSE 2026Set A1 markMCQ
$Q. The relationship between current density J, specific conductance σ and electric field intensity E is (A) E = σJ (B) E = σ/J (C) J = E/σ (D) J = σE$
›Reveal solutionSolution

The microscopic form of Ohm's law is J = σE.

Ohm's law in microscopic (point) form relates the current density $\vec J$ to the electric field $\vec E$ through the conductivity $\sigma$ :

$\vec J=\sigma\vec E$

Since conductivity is the reciprocal of resistivity ( $\sigma=1/\rho$ ), this is equivalent to $\vec E=\rho\vec J$ . The correct relation among the options is therefore $J=\sigma E$ .

✓Final answer
$(D) J = σE.$
CBSE 2026Set ANNUAL1 mark
$Q. Define current density.$
›Reveal solutionSolution

Current density is a vector describing how much current flows through a unit area, in the direction of current flow.

Current density (J) at a point in a conductor is defined as the amount of electric current flowing per unit area of cross-section held perpendicular to the direction of current flow at that point: J = I/A (for uniform current over a flat area A). It is a vector quantity, its direction being the direction in which positive charge is flowing (conventional current direction). Its SI unit is ampere per square metre (A/m^2). For a general (possibly non-uniform) current, the current through any small area element is dI = J . dA (dot product with the area vector).

✓Final answer
$Current density J = I/A, the current per unit cross-sectional area, a vector along the direction of current flow, SI unit A/m^2.$
CBSE 2025Set 55/4/11 markMCQ
$R$
›Reveal solutionSolution
The current is found by integrating the current density over the cross-sectional area. Since $j = \alpha r$ , the current through a radius $R/2$ scales as $R^3$ , making option (C) correct.

The key here is to understand that current density $j$ is the current per unit area. When $j$ varies with $r$ , you cannot simply multiply by area — you must integrate. The problem gives $j = \alpha r$ , meaning the current density increases linearly from zero at the centre to a maximum at the surface. We want the total current through only the inner half of the conductor's cross-section (from $r=0$ to $r=R/2$ ).

Set up the integral for current.

The current through a tiny ring of radius $r$ and thickness $dr$ is $dI = j \cdot dA$ , where $dA$ is the area of that ring. For a ring, $dA = 2\pi r \, dr$ .

So $dI = (\alpha r) \cdot (2\pi r \, dr) = 2\pi \alpha \, r^2 \, dr$ .

Integrate from the centre to $R/2$ .

$I = \int_{0}^{R/2} 2\pi \alpha \, r^2 \, dr = 2\pi \alpha \left[ \frac{r^3}{3} \right]_{0}^{R/2} = 2\pi \alpha \cdot \frac{(R/2)^3}{3} = 2\pi \alpha \cdot \frac{R^3}{24} = \frac{\pi \alpha R^3}{12}.$

Identify the proportionality. The result is $I = \frac{\pi \alpha}{12} R^3$ . Since $\pi$ and $\alpha$ are constants, $I \propto R^3$ .

Watch out

A common mistake is to treat $j$ as uniform and simply multiply by the area $\pi (R/2)^2$ . That would give $I = \alpha r \cdot \pi (R/2)^2$ , which is wrong because $j$ itself depends on $r$ — you cannot pick a single value of $r$ for the whole area. Always integrate when $j$ is not constant.

Tip

Notice that $j \propto r$ means the current density is zero at the axis and grows outward. The inner half carries less current than you might guess because the density there is lower. The cubic dependence on $R$ comes from the $r^2$ inside the integral — one power from the ring's circumference and one from $j$ itself.

✓Final answer
$R^3$
CBSE 2025Set JS1 mark
$6.0 \times 10^{15}$
›Reveal solutionSolution

An electron circulating $f=6.0\times10^{15}$ times per second carries a charge $e$ past any point each revolution, giving an equivalent current $I=ef=9.6\times10^{-4}$ A.

Concept. Current is charge crossing a point per unit time, $I=q/t$ . A single electron completing $f$ revolutions per second passes any point on its orbit $f$ times each second, so the charge crossing that point per second is $e\times f$ . Thus the equivalent current is

$I = e f.$

Solution.

$I=(1.6\times10^{-19}\ \text{C})(6.0\times10^{15}\ \text{s}^{-1})=9.6\times10^{-4}\ \text{A}.$

✓Final answer
$I = 9.6\times10^{-4}$
CBSE 2025Set D1 markMCQ
$Q. The unit of current density is (A) ampere (A) (B) coulomb (C) (C) ampere per square metre (A/m^2) (D) volt per metre (V/m)$
›Reveal solutionSolution

Current density J = I/A has SI unit A/m².

Current density is the current flowing per unit cross-sectional area of a conductor:

$J = \frac{I}{A}$

Its SI unit is therefore ampere per square metre (A m⁻²). The ampere is the unit of current, the coulomb of charge, and volt per metre of electric field — none of these is current density.

✓Final answer
$(C) ampere per square metre (A/m²).$
CBSE 2024Set A11 markMCQ
$Q. The current density is a$
(a) scalar and its SI unit is $A\,m^2$
(b) vector and its SI unit is $A/m^3$
(c) vector and its SI unit is $A/m^2$
(d) scalar and its SI unit is $A/m$
›Reveal solutionSolution

(c) vector and its SI unit is $A/m^2$

✓Final answer
$A/m^2$

Current density $\vec{j}$ is defined by $I=\int\vec{j}\cdot d\vec{A}$ ; it has both magnitude and direction (the direction of current flow), so it is a vector. Its SI unit is ampere per square metre, $A\,m^{-2}$ .
CBSE 2024Set ANNUAL1 markMCQ
$Q. Current density is a -$
(a) scalar quantity
(b) vector quantity
(c) dimensionless quantity
(d) none of these
›Reveal solutionSolution

Current density $\vec{J} = \vec{v}_d\, n e$ has both magnitude and direction — it is properly a vector quantity.

Current density is defined as the current flowing per unit cross-sectional area, in the direction of current flow: $\vec{J} = ne\vec{v}_d$ , where $\vec{v}_d$ is the drift velocity vector, $n$ the free-electron density, and $e$ the electron charge. Because it carries directional information (unlike current itself, which is a scalar obtained by dotting $\vec{J}$ with an area vector and integrating), current density is correctly classified as a vector quantity — this is the standard, physically-correct classification used throughout Ohm's law in vector form, $\vec{J} = \sigma \vec{E}$ .

✓Final answer
$\vec{J}=ne\vec{v}_d$
CBSE 2023Set B1 markMCQ
$Q. S.I. unit of current density is$
(i) Coulomb / meter
(ii) Ampere / meter²
(iii) Coulomb / meter²
(iv) Ampere / meter
›Reveal solutionSolution

Current density is current flowing per unit area of cross-section; its SI unit is A/m².

Current density (J) at a point in a conductor is defined as the amount of current flowing per unit area of cross-section held perpendicular to the direction of current flow:

$J = \frac{I}{A}$

Since current I is measured in amperes (A) and area A in square metres (m²), the SI unit of current density is ampere per square metre (A/m²). Options (i), (iii) and (iv) mix up charge (coulomb) or the wrong power of length with current, so they are incorrect.

✓Final answer
$(ii) Ampere/meter² — J = I/A, SI unit A/m².$
CBSE 2022Set I1 markMCQ
$Q. Which of the following is correct for current density? (A) J = I\cdotA (B) J = I/A (C) J = A/I (D) J = I^2 A$
›Reveal solutionSolution

Current density J = I/A (current per unit area).

Current density is the current flowing per unit cross-sectional area of the conductor, measured perpendicular to the flow:

$J = \dfrac{I}{A}$

Its SI unit is A/m². It is a vector in the direction of current flow.

✓Final answer
$(B) J = I/A.$
CBSE 2022Set I1 markMCQ
$Q. Ampere-hour is the unit of (A) power (B) charge (C) energy (D) potential difference$
›Reveal solutionSolution

Charge = current × time, so ampere-hour is a unit of electric charge.

Electric charge is $q = I\,t$ . Multiplying a current (ampere) by a time (hour) therefore gives a quantity of charge:

$1\ \text{A·h} = 1\ \text{A} \times 3600\ \text{s} = 3600\ \text{C}.$

Ampere-hour is used to rate battery capacity, which is a charge.

✓Final answer
$(B) charge.$
CBSE 2021Set A1 markMCQ
$Q. One ampere is equal to (A) 1 coulomb / 1 second (B) 1 coulomb \times 1 second (C) 1 volt \times 1 ohm (D) 1 ohm / 1 volt$
›Reveal solutionSolution

1 ampere = 1 coulomb per second.

Electric current is the rate of flow of charge, I = q/t. The SI unit ampere is therefore defined so that a current of one ampere flows when one coulomb of charge passes a cross-section each second:

$1\ \text{A} = \frac{1\ \text{C}}{1\ \text{s}}$

Options using volt/ohm describe Ohm's law (I = V/R) but the direct definition of the ampere in terms of base quantities is charge per unit time.

✓Final answer
$(A) 1 coulomb / 1 second.$

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