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EMI/AC flash cards

Master EMI/AC through 101 NEET-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

29 of this chapter's 101 cards, laid out open so you can read straight through. The remaining 72 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.What is magnetic flux through a surface?

    Magnetic flux is the number of field lines crossing an area: ϕB=BA=BAcosθ\phi_B = \vec{B}\cdot\vec{A} = BA\cos\theta, where θ\theta is the angle between B\vec{B} and the area's normal.

    Hint: Dot product of field and area vectors.

  2. 2.What is the SI unit of magnetic flux?

    The weber (Wb). 1 Wb=1 Tm21\ \text{Wb} = 1\ \text{T}\cdot\text{m}^2. Flux is a scalar quantity.

    Hint: Named after a German physicist; tesla times metre-squared.

  3. 3.When is magnetic flux through a coil maximum and when is it zero?

    Flux is maximum when the plane of the coil is perpendicular to B\vec{B} (normal parallel to B\vec{B}, θ=0\theta = 0), and zero when the plane is parallel to B\vec{B} (θ=90\theta = 90^\circ).

    Hint: Think about cosθ\cos\theta at 00^\circ and 9090^\circ.

  4. 4.State Faraday's first law of electromagnetic induction.

    Whenever the magnetic flux linked with a circuit changes, an emf is induced in the circuit; the induced emf lasts as long as the change in flux continues.

    Hint: Change in flux drives the emf.

  5. 5.State Faraday's second law (the quantitative law of induction).

    The magnitude of the induced emf equals the rate of change of magnetic flux linkage: ε=dϕBdt\varepsilon = -\dfrac{d\phi_B}{dt}. For NN turns, ε=NdϕBdt\varepsilon = -N\dfrac{d\phi_B}{dt}.

    Hint: emf equals negative rate of flux change.

  6. 6.What does the negative sign in ε=dϕBdt\varepsilon = -\dfrac{d\phi_B}{dt} represent?

    It represents Lenz's law: the induced emf (and current) opposes the change in flux that produces it. It is a statement of energy conservation.

    Hint: Nature opposes the change.

  7. 7.State Lenz's law.

    The direction of the induced current is always such that it opposes the change in magnetic flux that causes it.

    Hint: Induced effects oppose their cause.

  8. 8.How is Lenz's law a consequence of conservation of energy?

    If the induced current aided the change, flux would grow without limit, giving free energy. Because it opposes the change, work must be done against the induced effect, and that work becomes the electrical energy generated.

    Hint: Opposition ensures no free energy.

  9. 9.A bar magnet's north pole approaches a coil. What is the induced polarity and current direction (as seen by the magnet)?

    The near face of the coil becomes a north pole to repel the approaching magnet, so the induced current flows anticlockwise as seen from the magnet's side.

    Hint: Coil repels the approaching pole.

  10. 10.A bar magnet is pulled away from a coil (north pole facing it). What happens?

    The near face becomes a south pole to attract and oppose the receding magnet; induced current flows clockwise as seen from the magnet. Flux is decreasing, so induced current tries to maintain it.

    Hint: Coil attracts the receding pole.

  11. 11.What is motional emf?

    The emf induced across a conductor of length ll moving with velocity vv perpendicular to a field BB: ε=Blv\varepsilon = Blv. It arises from the magnetic force on the free charges in the moving rod.

    Hint: BlvBlv for a rod cutting field lines.

  12. 12.Derive the force on a moving charge that leads to motional emf.

    A charge qq in the rod moving with velocity vv in field BB feels F=qvBF = qvB. This pushes charges to the ends, setting up an emf ε=Blv\varepsilon = Blv until the electric force balances the magnetic force.

    Hint: Magnetic force qvBqvB separates charges.

  13. 13.A rod of length ll moves at speed vv on rails in field BB; circuit resistance is RR. What is the induced current?

    Induced emf ε=Blv\varepsilon = Blv, so current I=BlvRI = \dfrac{Blv}{R}.

    Hint: I=ε/RI = \varepsilon/R with ε=Blv\varepsilon = Blv.

  14. 14.What retarding force acts on a rod carrying induced current II in field BB, and what power is dissipated?

    Retarding force F=BIl=B2l2vRF = BIl = \dfrac{B^2l^2v}{R}. Power dissipated P=Fv=B2l2v2R=I2RP = Fv = \dfrac{B^2l^2v^2}{R} = I^2R, equal to the mechanical power supplied.

    Hint: Force opposes motion; P=Fv=I2RP = Fv = I^2R.

  15. 15.What is the emf induced in a rod of length ll rotating with angular velocity ω\omega about one end in field BB?

    ε=12Bωl2\varepsilon = \dfrac{1}{2}B\omega l^2. The rod sweeps area at rate 12ωl2\tfrac{1}{2}\omega l^2 per second.

    Hint: Half B omega l-squared.

  16. 16.What are eddy currents?

    Eddy currents are circulating induced currents produced in the body of a bulk conductor when the magnetic flux through it changes. They flow in closed loops within the metal.

    Hint: Swirling currents in solid metal.

  17. 17.Give two useful applications of eddy currents.

    Electromagnetic (eddy-current) braking in trains, induction furnaces for melting metals, electromagnetic damping in galvanometers, and speedometers.

    Hint: Braking and induction heating.

  18. 18.Why are the cores of transformers and motors laminated?

    Lamination (thin insulated sheets) increases resistance to eddy-current paths, reducing eddy currents and the associated heat/energy loss.

    Hint: Thin sheets cut down circulating currents.

  19. 19.What is self-induction?

    Self-induction is the phenomenon in which a changing current in a coil induces an emf in the same coil, opposing the change in its own current.

    Hint: A coil opposing its own current change.

  20. 20.Define self-inductance LL and give its unit.

    Self-inductance relates flux linkage to current: Nϕ=LIN\phi = LI, so L=NϕIL = \dfrac{N\phi}{I}. Its SI unit is the henry (H). Also ε=LdIdt\varepsilon = -L\dfrac{dI}{dt}.

    Hint: Flux linkage per unit current; unit henry.

  21. 21.What is 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 is cross-sectional area, and ll is length.

    Hint: Proportional to n2n^2 and volume AlAl.

  22. 22.What is mutual induction?

    Mutual induction is the production of an induced emf in one coil due to a changing current in a neighbouring coil, via the shared changing flux.

    Hint: One coil induces emf in another.

  23. 23.Define mutual inductance MM.

    MM relates the flux linkage of coil 2 to the current in coil 1: N2ϕ2=MI1N_2\phi_2 = M I_1, so the induced emf ε2=MdI1dt\varepsilon_2 = -M\dfrac{dI_1}{dt}. Unit: henry (H).

    Hint: Coupling constant between two coils.

  24. 24.What is the mutual inductance of two long coaxial solenoids?

    M=μ0n1n2Al=μ0N1N2AlM = \mu_0 n_1 n_2 A l = \dfrac{\mu_0 N_1 N_2 A}{l}, where AA is the (inner) cross-sectional area and ll the common length.

    Hint: Similar to self-L but with two turn densities.

  25. 25.How is mutual inductance related to the two self-inductances (ideal coupling)?

    For perfect (unity) coupling, M=L1L2M = \sqrt{L_1 L_2}. In general M=kL1L2M = k\sqrt{L_1 L_2} where 0k10 \le k \le 1 is the coupling coefficient.

    Hint: Geometric mean, scaled by coupling factor.

  26. 26.What is the energy stored in an inductor carrying current II?

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

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

  27. 27.What is the magnetic energy density in a field BB?

    u=B22μ0u = \dfrac{B^2}{2\mu_0} (energy per unit volume). This is the magnetic analogue of the electric energy density 12ε0E2\tfrac12\varepsilon_0 E^2.

    Hint: BB-squared over 2μ02\mu_0.

  28. 28.How does an AC generator produce alternating emf?

    A coil rotates in a magnetic field, so the flux ϕ=NBAcosωt\phi = NBA\cos\omega t changes sinusoidally, inducing emf ε=NBAωsinωt=ε0sinωt\varepsilon = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t.

    Hint: Rotating coil gives sinusoidal flux.

  29. 29.What is the peak emf of an AC generator with NN turns, area AA, field BB, angular speed ω\omega?

    ε0=NBAω\varepsilon_0 = NBA\omega. The instantaneous emf is ε=ε0sinωt\varepsilon = \varepsilon_0\sin\omega t.

    Hint: All four factors multiplied together.

Open the interactive deck for the other 72 cards, with self-grading so the ones you keep missing come back.

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