Physics · Electromagnetic Induction · NEET
Read what is GIVEN. If they give flux per turn and current, use L = NΦ/I. If a rod or wheel rotates, use EMF = ½BωR². If a coil rotates and they ask maximum EMF, use ε₀ = NBAω. If they give L and I and ask energy, use U = ½LI². If flux changes over a time, use EMF = N ΔΦ/Δt. NEET rarely mixes two formulas in one question, so match the given quantities to one formula.
Different points on the rod move at different speeds (the tip is fast, the centre is at rest). BvL uses one speed v, so it only works when the whole rod moves at the same v. For rotation, integrate along the length; the average effect gives EMF = ½BωR², where R is the rod length (radius). Trick: for the cycle-wheel PYQ, EMF between centre and rim = ½ × 0.1 × 10 × 0.5² = 0.125 V.
Φ here is the flux through ONE turn (flux per turn). The N multiplies it to give total flux linkage NΦ. So L = (total flux linkage)/current = NΦ/I. In the 2016 solenoid PYQ: N = 1000, Φ = 4×10⁻³ Wb per turn, I = 4 A, so L = 1000 × 4×10⁻³ / 4 = 1 H.
If the question asks the MAXIMUM (peak) EMF of a steadily rotating coil (AC generator), use ε₀ = NBAω directly. Convert revolutions per second to ω using ω = 2πf. If instead the coil is turned once by a fixed angle (like 90°) in a given time, use average EMF = N ΔΦ/Δt with ΔΦ = BA(cosθ_final - cosθ_initial).
No net EMF for a loop fully outside the field region, because no flux passes through it (Φ = 0, so dΦ/dt = 0). But a loop that ENCLOSES the changing-field region does get EMF = (dB/dt)πr², using the area of the field region πr², not the loop area. This is the 2016 two-loop trap: loop 1 gets EMF, loop 2 gets zero.
A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4 × 10⁻³ Wb. The self-inductance of the solenoid is:
A cycle wheel of radius 0.5 m is rotated with constant angular velocity of 10 rad/s in a region of magnetic field of 0.1 T which is perpendicular to the plane of the wheel. The EMF generated between its centre and the rim is:
The magnetic energy stored in an inductor of inductance 4 µH carrying a current of 2 A is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Five cover almost everything: EMF = N ΔΦ/Δt (Faraday), L = NΦ/I (self-inductance), EMF = ½BωR² (rotating rod/wheel), ε₀ = NBAω (peak AC EMF), and U = ½LI² (energy in inductor). Add EMF = BvL for a sliding rod and M ∝ R₂²/R₁ for coplanar loops.
Electromagnetic Induction usually gives 1-2 questions each year, and they are almost always direct formula-based numericals. That makes EMI a high-return, low-effort scoring chapter if you memorise the five core formulas and their units.
Match the GIVEN data to the formula. Flux per turn + current means L = NΦ/I. A rotating rod means ½BωR². A rotating coil asking maximum EMF means NBAω. Given L and I asking energy means ½LI². This 'given-to-formula' matching solves most NEET EMI PYQs in seconds.
They wrongly apply EMF = BvL, but a rotating rod does not have a single speed. The correct formula is EMF = ½BωR². This one substitution error is the most common EMI mistake in NEET.
