Mutual Inductance
EMI
7
JEE Qs
8%
Hard
75
min
Master the definition of mutual inductance and its dependency on geometric factors; practice calculating M for standard configurations using the flux linkage method.
🧮 Key Formulas
✅ Key Points for JEE
- 1Mutual inductance (M) is a scalar constant for a given pair of coils and their relative orientation; it depends only on their geometry and the permeability of the core material, not on the currents flowing through them.
- 2It quantifies the magnetic flux linkage in one coil due to the current flowing in a *neighboring* coil.
- 3The reciprocity theorem states that the mutual inductance of coil 2 with respect to coil 1 (M₂₁) is numerically equal to the mutual inductance of coil 1 with respect to coil 2 (M₁₂), i.e., M₁₂ = M₂₁ = M.
- 4The induced EMF in a coil due to mutual induction is proportional to the *rate of change* of current in the other coil, not the current itself, and follows Lenz's Law (negative sign).
- 5The coefficient of coupling 'k' (where 0 ≤ k ≤ 1) indicates how effectively the magnetic flux produced by one coil links with the other coil. For ideal coupling, k=1.
⚠️ Common Mistakes
- ✕Mistaking mutual inductance as dependent on the current flowing through either coil, instead of recognizing it as a purely geometric property.
- ✕Incorrectly applying Faraday's Law, specifically omitting the negative sign in the induced EMF formula or failing to correctly differentiate the current with respect to time.
- ✕Confusing the definitions and applications of mutual inductance (M) with self-inductance (L) or resistance (R).
- ✕Struggling with accurately calculating the magnetic flux linkage for specific geometric configurations (e.g., solenoids, coils) to determine M.
📝 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