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PhysicsHardClass 12

Motional EMF + Lenz's Law

EMI

48

JEE Qs

20%

Hard

75

min

Master Lenz's Law by practicing numerous direction-finding problems, as it's the most common pitfall and crucial for solving problems involving induced current and forces.

🧮 Key Formulas

e = (v x B) . L (for a straight rod of length vector L moving with velocity v in magnetic field B)
e = B L v (when B, L, v are mutually perpendicular)
e = 1/2 B omega L^2 (for a rod of length L rotating with angular velocity omega in magnetic field B perpendicular to plane of rotation)
Induced Current: I = e / R_eq
Magnetic force on induced current: F_m = I (L x B)
Power dissipated: P = e I = I^2 R_eq = e^2 / R_eq

✅ Key Points for JEE

  • 1Motional EMF arises from the Lorentz force (q(v x B)) on charge carriers within a conductor moving in a magnetic field, separating charges to create a potential difference.
  • 2Lenz's Law states that the direction of the induced current (or EMF) is such that it opposes the change in magnetic flux that produced it. This is a direct consequence of the conservation of energy.
  • 3The power supplied by an external agent to maintain constant velocity of a conductor inducing EMF is entirely dissipated as heat in the resistance of the circuit (assuming no other energy forms).
  • 4Motional EMF can be derived either from Lorentz force considerations (e = integral (v x B).dl) or by using Faraday's Law (e = -dΦ/dt) by considering the rate of change of area/flux.
  • 5For complex geometries or non-uniform fields, apply e = integral (v x B).dl along the length of the conductor. The 'effective length' concept is crucial for calculating EMF.

⚠️ Common Mistakes

  • Incorrectly applying Lenz's Law to determine the direction of induced current, often confusing 'opposing the cause' with 'opposing the field itself'.
  • Failing to correctly identify the components of velocity, magnetic field, and length that are mutually perpendicular when using simplified Blv formula, or misapplying vector cross products.
  • Neglecting to consider the external force required to maintain constant velocity against the magnetic retarding force, leading to incorrect energy balance calculations.
  • Confusing the roles of Fleming's Right-Hand Rule (for induced current) and Left-Hand Rule (for force on current) or simply guessing directions.

📝 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 Ch 6: Electromagnetic Induction