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Faraday's Laws — Induced EMF, flux change

EMI

7

JEE Qs

8%

Hard

75

min

Master the calculation of magnetic flux for varying geometries and rigorously apply Lenz's Law to determine the direction of induced current, as both magnitude and direction are frequently tested.

🧮 Key Formulas

Φ_B = ∫ B ⋅ dA
Φ_B = B A cosθ (for uniform B and planar area A)
ε = -dΦ_B/dt
I_induced = ε/R
q_induced = ∫ I_induced dt = -ΔΦ_B / R
ε_motional = (B l v) (for conductor of length l moving with velocity v perpendicular to uniform magnetic field B)

✅ Key Points for JEE

  • 1Faraday's Law states that the magnitude of the induced electromotive force (EMF) is equal to the time rate of change of magnetic flux through the loop.
  • 2The negative sign in Faraday's Law signifies Lenz's Law, which states that the direction of the induced EMF (and hence current) is such that it opposes the change in magnetic flux that produced it.
  • 3Magnetic flux (Φ_B) can change due to: (a) change in magnetic field (B), (b) change in area (A) of the loop cutting the flux lines, or (c) change in the angle (θ) between B and the area vector.
  • 4Induced EMF can exist even in an open circuit, but induced current requires a closed conducting loop and resistance.

⚠️ Common Mistakes

  • Incorrectly determining the direction of induced current/EMF by misapplying Lenz's Law (e.g., opposing the flux itself instead of the *change* in flux).
  • Errors in calculating the magnetic flux, especially when B is non-uniform, the area is not planar, or the angle between B and area vector is not accounted for properly.
  • Confusing induced EMF with induced current; current flows only if there's a closed circuit, while EMF is induced regardless.
  • Neglecting the chain rule when differentiating flux with respect to time if B, A, or θ are functions of time (e.g., d(BAcosθ)/dt).

📝 Practice Questions

See all

Q42.Regarding self-inductance: A. The self-inductance of the coil depends on its geometry. B. Self-inductance does not depend on the permeability of the medium. C. Self-induced e.m.f. opposes any change in the current in a circuit. D. Self-inductance is electromagnetic analogue of mass in mechanics. E. Work needs to be done against self-induced e.m.f. in establishing the current. Choose the correct answer from the options given below: (1) A, B, C, E only (2) B, C, D, E only (3) A, C, D, E only (4) A, B, C, D only

2025·ConceptualEasy

Q49. A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field B exists into the page. The bar starts to move from the vertex at time t = 0 with a constant velocity. If the induced EMF is E ∝tn , then value of n is _.

2025·NumericalHard

Q35.A rectangular metallic loop is moving out of a uniform magnetic field region to a field free region with a constant speed. When the loop is partially inside the magnate field, the plot of magnitude of induced emf (ε) with time (t) is given by 2025 (22 Jan Shift 2) JEE Main Previous Year Paper (1) (2) (3) (4)

2025·Graph basedMedium

Q48.A parallel plate capacitor of area A = 16 cm2 and separation between the plates 10 cm , is charged by a DC current. Consider a hypothetical plane surface of area A0 = 3.2 cm2 inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6A. At the same instant the displacement current through A0 is ________ mA .

2025·NumericalMedium

Q31.A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10πrads−1 about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim ? (π = 3.14) (1) 0.5024 V (2) V (3) 0.2512V V (4) 0.1256V V

2025·MCQMedium

Q27.A coil of area A and N turns is rotating with angular velocity ω in a uniform magnetic field →B about an axis perpendicular to →B. Magnetic flux φ and induced emf ε across it, at an instant when →B is parallel to the plane of coil, are : (1) φ = AB, ε = 0 (2) φ = 0, ε = 0 (3) φ = 0, ε = NABω (4) φ = AB, ε = NABω

2025·MCQMedium

NCERT Chapters

  • Class 12 Physics Part 1 Ch 6: Electromagnetic Induction