Q.State the microscopic form of Ohm's law.

Showing the 12 most recent of 15 on this concept.

1. 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. …

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2. 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 $\sigma$:

   $\vec J=\sigma\vec E$

   …

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3. 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-unif …

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4. CBSE 2025Set 55/4/11 markMCQ
   Q.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) $R^4$

   ›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 $r=R/2$).

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

   2. Integrate from the centre to $R/2$.

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

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

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5. CBSE 2025Set JS1 mark
   Q.Electron in the hydrogen atom is moving round the nucleus with $6.0 \times 10^{15}$ cycle per second. What will be the value of current at a point on circular path?

   ›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 cur …

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6. 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:

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

   …

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7. 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 $A/m^2$ …

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8. 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 …

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9. 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:

   $J = \frac{I}{A}$ …

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10. CBSE 2022Set I1 markMCQ
    Q.Which of the following is correct for current density? (A) J = I·A (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:

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

    …

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11. 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:

    …

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12. CBSE 2021Set A1 markMCQ
    Q.One ampere is equal to (A) 1 coulomb / 1 second (B) 1 coulomb × 1 second (C) 1 volt × 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:

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

    …

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