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EMI/AC flash cards

Master EMI/AC through 89 JEE Advanced-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

EMI/AC, question and answer

23 of this chapter's 89 cards, laid out open so you can read straight through. The remaining 66 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.Define magnetic flux Φ\Phi through a surface.

    Φ=BdA=BAcosθ\Phi = \int \vec{B}\cdot d\vec{A} = BA\cos\theta, where θ\theta is the angle between B\vec{B} and the area vector (normal to the surface). SI unit: weber (Wb) == T\cdotm2^2.

    Hint: Dot product of field and area vector.

  2. 2.State Faraday's law of electromagnetic induction.

    The induced emf equals the negative rate of change of magnetic flux: ε=dΦdt\varepsilon = -\dfrac{d\Phi}{dt}. For NN turns, ε=NdΦdt\varepsilon = -N\dfrac{d\Phi}{dt}.

    Hint: emf tracks how fast flux changes.

  3. 3.State Lenz's law and its physical basis.

    The induced current flows so as to oppose the change in flux that produced it (the source of the minus sign in Faraday's law). It is a consequence of conservation of energy.

    Hint: Nature opposes the change.

  4. 4.What are the three ways to change flux and hence induce an emf?

    Change BB (field strength), change AA (area of the loop), or change θ\theta (orientation of loop relative to field). Any combination changes Φ=BAcosθ\Phi = BA\cos\theta.

    Hint: BB, AA, or θ\theta.

  5. 5.Write the expression for motional emf of a rod of length ll moving with velocity vv perpendicular to field BB.

    ε=Blv\varepsilon = Blv. More generally ε=(v×B)dl\varepsilon = \int (\vec{v}\times\vec{B})\cdot d\vec{l}.

    Hint: BlvBlv for mutually perpendicular vectors.

  6. 6.A rod moves on rails in field BB; give the induced current, force needed, and power.

    Current I=BlvRI = \dfrac{Blv}{R}; retarding force on rod F=BIl=B2l2vRF = BIl = \dfrac{B^2l^2v}{R}; external power to keep constant vv: P=Fv=B2l2v2RP = Fv = \dfrac{B^2l^2v^2}{R}, all dissipated as heat.

    Hint: Power in = I2RI^2R.

  7. 7.What is the origin of motional emf at the microscopic level?

    The magnetic force qv×Bq\vec{v}\times\vec{B} on free charges in the moving conductor separates them, setting up an electric field until equilibrium; the resulting potential difference is the motional emf.

    Hint: Lorentz force on carriers.

  8. 8.What are eddy currents?

    Circulating currents induced in the body of a bulk conductor when the flux through it changes. They dissipate energy as heat and oppose the motion (Lenz's law).

    Hint: Loops of current inside solid metal.

  9. 9.Give two applications and one drawback of eddy currents.

    Applications: electromagnetic (induction) braking, induction furnaces, metal detectors, damping in galvanometers. Drawback: energy loss (heating) in transformer/motor cores, reduced by using laminated cores.

    Hint: Braking vs core loss.

  10. 10.Why are transformer and motor cores laminated?

    Lamination (thin insulated sheets) breaks up the paths of eddy currents, greatly increasing their resistance and reducing eddy-current heat losses.

    Hint: Thin sheets raise eddy resistance.

  11. 11.Define self-inductance LL.

    The property of a coil by which it opposes any change in its own current: Φ=LI\Phi = LI and ε=LdIdt\varepsilon = -L\dfrac{dI}{dt}. SI unit: henry (H).

    Hint: Flux-linkage per unit current.

  12. 12.Give the self-inductance of a long solenoid.

    L=μ0n2Al=μ0N2AlL = \mu_0 n^2 A l = \dfrac{\mu_0 N^2 A}{l}, where n=N/ln = N/l is turns per unit length, AA the cross-sectional area, ll the length.

    Hint: Depends on N2N^2 and geometry, not current.

  13. 13.Define mutual inductance MM between two coils.

    Φ2=MI1\Phi_2 = M I_1 and ε2=MdI1dt\varepsilon_2 = -M\dfrac{dI_1}{dt}. MM depends on geometry, number of turns, and coupling. Unit: henry. Note M12=M21M_{12} = M_{21}.

    Hint: emf in coil 2 from current change in coil 1.

  14. 14.Two coils of self-inductance L1,L2L_1,L_2 have mutual inductance MM. What is the coupling relation?

    M=kL1L2M = k\sqrt{L_1 L_2}, where the coupling coefficient 0k10 \le k \le 1; k=1k=1 means perfect (ideal) coupling.

    Hint: Geometric mean of the two self-inductances.

  15. 15.Give the energy stored in an inductor carrying current II.

    U=12LI2U = \tfrac{1}{2}LI^2. This energy is stored in the magnetic field.

    Hint: Analogous to 12CV2\tfrac12 CV^2 for a capacitor.

  16. 16.Write the magnetic energy density in a field BB.

    u=B22μ0u = \dfrac{B^2}{2\mu_0} (energy per unit volume). Compare electric: uE=12ε0E2u_E = \tfrac12\varepsilon_0 E^2.

    Hint: Field energy per volume.

  17. 17.Write the growth of current in an LR circuit switched to a battery of emf ε\varepsilon.

    I(t)=εR(1et/τ)I(t) = \dfrac{\varepsilon}{R}\left(1 - e^{-t/\tau}\right), with time constant τ=L/R\tau = L/R.

    Hint: Rising exponential to ε/R\varepsilon/R.

  18. 18.Write the decay of current in an LR circuit when the battery is shorted out.

    I(t)=I0et/τI(t) = I_0\, e^{-t/\tau}, with τ=L/R\tau = L/R and I0I_0 the initial current.

    Hint: Decaying exponential.

  19. 19.What is the time constant of an LR circuit, and its physical meaning?

    τ=L/R\tau = L/R (seconds). It is the time for the current to reach 63%\approx 63\% of its final value during growth (or fall to 37%37\% during decay).

    Hint: LL over RR.

  20. 20.During current growth in an LR circuit, at t=τt=\tau what fraction of the final current is reached?

    1e10.6321 - e^{-1} \approx 0.632, i.e. about 63.2%63.2\% of the steady value ε/R\varepsilon/R.

    Hint: 11/e1 - 1/e.

  21. 21.Why can't current in an inductor change instantaneously?

    An instantaneous change would require infinite dI/dtdI/dt, hence infinite back-emf LdI/dtL\,dI/dt. So inductor current is continuous; it acts as a short at long times (DC) and opposes sudden changes.

    Hint: Back-emf forbids jumps in II.

  22. 22.Write the equation of a sinusoidal alternating emf and current.

    ε=ε0sinωt\varepsilon = \varepsilon_0\sin\omega t, I=I0sin(ωt+ϕ)I = I_0\sin(\omega t + \phi), where ω=2πf\omega = 2\pi f is angular frequency and ϕ\phi the phase difference.

    Hint: ω=2πf\omega = 2\pi f.

  23. 23.Define the rms value of an alternating current.

    The rms (root-mean-square) value is the steady DC current that produces the same average heating: Irms=I2I_{rms} = \sqrt{\langle I^2\rangle}. For sinusoid Irms=I0/2I_{rms} = I_0/\sqrt{2}.

    Hint: Equivalent DC for heating.

Open the interactive deck for the other 66 cards, with self-grading so the ones you keep missing come back.

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