Motional EMF — E = Bvl, rod on rails
EMI
7
JEE Qs
8%
Hard
60
min
Always visualize the vectors (velocity, magnetic field, length) in 3D space to correctly apply the right-hand rules for force and induced EMF direction, especially in complex scenarios.
🧮 Key Formulas
✅ Key Points for JEE
- 1Motional EMF arises due to the Lorentz force (F = q(vec(v) x vec(B))) acting on the free charge carriers within a conductor moving through a magnetic field, pushing them to one end.
- 2The direction of the induced EMF (and hence the positive terminal) is given by the direction of (vec(v) x vec(B)), or by using the Right-Hand Rule (RHR) for the force on a positive charge.
- 3Lenz's Law dictates that the induced current will flow in a direction that opposes the change in magnetic flux, which often translates to a magnetic force opposing the conductor's motion.
- 4Conservation of energy is paramount: the work done by the external force pulling the conductor equals the electrical energy dissipated as heat in the circuit, plus any change in kinetic energy of the conductor.
- 5For the simplified E=Bvl formula, ensure that the magnetic field (B), velocity (v), and length of the conductor (l) are all mutually perpendicular. If not, use the vector cross product formulation.
⚠️ Common Mistakes
- ✕Incorrectly determining the direction of induced EMF or current, leading to sign errors or wrong force directions.
- ✕Applying the simplified E = Bvl formula when B, v, or l are not mutually perpendicular, failing to use the vector cross product properly.
- ✕Confusing motional EMF with flux linkage EMF (Faraday's Law) or failing to see them as two ways of analyzing the same phenomenon (Lorentz force vs. changing flux).
- ✕Neglecting the total resistance of the circuit (rod + rails + external resistor) when calculating induced current and power.
📝 Practice Questions
See allQ42.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
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 _.
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)
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 .
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
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
NCERT Chapters
- Class 12 Physics Ch 6: Electromagnetic Induction