Physics · Electromagnetic Induction · NEET
Because it is much easier. The field of a current loop is simple ONLY at its centre: B = μ0 I / (2R). If you put current in the big loop, its centre field is almost uniform over the whole tiny small loop (since R2 << R1), so flux = B x area of small loop = easy. If you tried the reverse (current in the small loop), its field spreads out and is very NOT uniform over the big loop, so the integral is hard. We use the easy direction and rely on M12 = M21 (reciprocity) to know it works both ways.
The big loop's field changes from point to point, but over a region that is tiny compared to R1 the change is negligible. Since R2 << R1, every point of the small loop sits very close to the centre of the big loop, where B = μ0 I1 / (2R1). So we treat B as constant across the small loop. This is an approximation and it is only valid when R1 >> R2 — that condition is the whole reason the formula is clean.
Flux = B x A, and A is the area over which we assumed B is uniform — that is the small loop's area, π R2². We could not do this with the big loop's area because B is not uniform over the big loop. So the small area πR2² appears, multiplied by the big loop's centre field μ0 I1/(2R1).
For two circular loops sitting in the SAME plane with the same centre, coplanar and concentric mean the same thing here, and the formula M = μ0 π R2²/(2R1) applies. 'Coaxial' usually means two loops (or solenoids) on the same axis but in parallel planes, separated by a distance — that gives a different formula (a dipole-style 1/x³ field). Read the geometry in the question carefully.
The BIG radius R1 is in the denominator (it makes the weak central field, μ0 I/2R1). The SMALL radius R2 is squared in the numerator (it is the catching area, πR2²). So M = μ0 π R2² / (2 R1). Small radius squared on top, big radius on the bottom.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
M = μ0 π R2² / (2 R1), where R1 is the large loop radius and R2 is the small loop radius. It comes from flux through the small loop = [μ0 I1/(2R1)] x πR2², divided by I1.
The henry (H). 1 H = 1 weber per ampere = 1 volt-second per ampere. M tells you how many volts of emf appear in one loop for a 1 ampere-per-second change of current in the other.
No. M depends only on geometry (the radii R1, R2 and how the loops are placed) and the medium (μ0 for air). The current cancels out when you divide flux by I1. This is why M is a fixed property of the pair of loops.
Mutual inductance is reciprocal: the flux linkage per unit current is the same in both directions. We compute it the easy way (current in big loop, uniform field over small loop) and reciprocity guarantees the same M if the current were in the small loop instead.
emf = M x (dI1/dt) = [μ0 π R2²/(2R1)] x (rate of change of current in the big loop). If the big loop's current is steady, dI1/dt = 0 and no emf is induced.
