Q.Define S.I. Unit of self Inductance.

Showing the 12 most recent of 35 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.When the current changes from +2 A to −2 A in 0.05 second in a coil, an e.m.f. of 8 V is induced in it. The coefficient of self-induction of the coil is

   (a) 0.1 henry

   (b) 0.2 henry

   (c) 0.4 henry

   (d) 0.8 henry

   ›Reveal solutionSolution

   Using $e = L\,\Delta I/\Delta t$ with $\Delta I = 4$ A and $\Delta t = 0.05$ s gives $L = 0.1$ H.

   The current changes from $+2$ A to $-2$ A, so the magnitude of the change is

   $\Delta I = |(-2) - (2)| = 4\ \text{A}, \qquad \Delta t = 0.05\ \text{s}$

   The magnitude of the self-induced emf is …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Self inductance is called

   (a) electric force

   (b) electrical inertia

   (c) electric pressure

   (d) electric energy

   ›Reveal solutionSolution

   Self-inductance L makes a circuit resist a change in the current flowing through it (via a back-emf = -L dI/dt), analogous to inertia resisting a change in velocity.

   Whenever the current through a coil tends to change, the induced back-emf (from self-induction) opposes that change (Lenz's law). This behaviour - opposing change rather than opposing the current itself - is directly analogous to mechanical inertia, which oppos …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The self-inductance of a coil is measured by

   (a) Electrical inertia

   (b) Electrical friction

   (c) Induced emf

   (d) Induced current

   ›Reveal solutionSolution

   Self-inductance is defined via the induced emf a coil produces in itself when its own current changes - it behaves like the 'electrical inertia' of the circuit, but it is quantified through that induced emf.

   When the current I through a coil changes, the magnetic flux linked with the coil changes too, and by Faraday's law this changing self-flux induces an emf in the SAME coil that opposes the change in current (Lenz's law). This self-induced emf defines the self-inductance L of the coil:

   emf = -L * (dI/dt)

   …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The unit of inductance is

   (a) henry

   (b) weber

   (c) newton

   (d) ohm

   ›Reveal solutionSolution

   Inductance L relates induced emf to the rate of change of current, so its unit works out to volt-second per ampere - named the henry.

   From emf = -L*(dI/dt), we get L = emf / (dI/dt), so the unit of L is volt / (ampere/second) = volt.second/ampere. This combination is given the special SI name henry (H), after Joseph Henry. Weber is the u …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.In a solenoid number of turns per unit length are doubled, it's self-inductance:

   (a) Halved

   (b) Doubled

   (c) Remains constant

   (d) Becomes four times

   ›Reveal solutionSolution

   Self-inductance of a solenoid is proportional to the square of the number of turns per unit length, so doubling $n$ makes $L$ four times as large.

   …

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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average e.m.f. of 100 V induced, give an estimate of the self-inductance of the circuit.

   (a) L = 4 H

   (b) L = 20 H

   (c) L = 40 H

   (d) L = 2 H

   ›Reveal solutionSolution

   Using $|\varepsilon| = L\left|\dfrac{dI}{dt}\right|$, the self-inductance works out to 2 H.

   Given: Current changes from $I_i = 5.0\text{ A}$ to $I_f = 0.0\text{ A}$ in $\Delta t = 0.1\text{ s}$, with average induced emf $|\varepsilon| = 100\text{ V}$.

   Step 1 — rate of change of current:

   $\left|\frac{\Delta I}{\Delta t}\right| = \frac{5.0 - 0.0}{0.1} = 50\ \text{A/s}$

   …

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7. CBSE 2025Set X11 mark
   Q.When a ________ rod is inserted into a coil, its self-inductance increases.

   ›Reveal solutionSolution

   ferromagnetic (soft iron) Self-inductance $L = \mu_r \mu_0 n^2 A l$. Inserting a ferromagnetic (soft iron) core has a large relative permeability $\mu_r \gg 1$, which greatly increases the mag …

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8. CBSE 2025Set D1 markMCQ
   Q.The self-inductance of a solenoid depends on (A) The current flowing through its medium (B) The number of turns per unit length (C) The length of the solenoid (D) Both (B) and (C)

   ›Reveal solutionSolution

   Self-inductance depends on the solenoid's geometry (turns per unit length and length/area), not on the current.

   For a solenoid the self-inductance is

   $L = \mu_0 n^{2} A l$

   where n is the number of turns per unit length, A the cross-sectional area and l the length. So it depends on the number of turns per unit length and the length (and area) — both geometric factors.

   …

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9. CBSE 2025Set ANNUAL1 markMCQ
   Q.The self-inductance of a coil is 5 henry. A current of 1 ampere changes to 2 amperes within 5 seconds through the coil. The value of the induced e.m.f. is

   (i) 10 volts

   (ii) 0.1 volt

   (iii) 1 volt

   (iv) 100 volts

   ›Reveal solutionSolution

   emf = L dI/dt = 5 x (1 A / 5 s) = 1 V.

   …

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10. CBSE 2024Set A11 markMCQ
    Q.Current in a coil changes from 1.6 A to 0.2 A in 2 second inducing an emf of 2.8 V. The value of self-inductance of the coil is

    (a) 40 H

    (b) 28 H

    (c) 4 H

    (d) 56 H

    ›Reveal solutionSolution

    (c) 4 H …

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11. CBSE 2024Set A1 markMCQ
    Q.S.I. unit of self-induction is (A) coulomb (C) (B) volt (V) (C) ohm (Ω) (D) henry (H)

    ›Reveal solutionSolution

    The SI unit of self-inductance L is the henry (H).

    Self-inductance L links the flux (or back-emf) of a coil to the current through it:

    $\phi = LI \qquad\text{and}\qquad \varepsilon = -L\frac{dI}{dt}.$

    …

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12. CBSE 2024Set ANNUAL1 markMCQ
    Q.The SI unit of self-inductance of a coil is

    (a) farad

    (b) henry

    (c) weber

    (d) oersted

    ›Reveal solutionSolution

    Self-inductance L is defined by the induced EMF equation EMF = −L(dI/dt); its SI unit is the henry (H).

    When the current through a coil changes, the changing magnetic flux linked with the coil itself induces an EMF in it (self-induction). This is written as:

    EMF = −L (dI/dt)

    where L is the coefficient of self-inductance of the coil. Rearranging, L = −EMF/(dI/dt), so the unit of L is volt/(ampere/second) = volt·second/ampere. This combination is given the special SI name henry (H), where 1 H = 1 V·s/A.

    …

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