Physics · Magnetism And Matter · NEET
Not physically, but magnetically YES. A solenoid carries real current in wire loops. A bar magnet has no external current, but its atoms have tiny circulating electron currents (Ampere's hypothesis). When you add up all these atomic loops, they behave like one long solenoid. The proof: move a compass needle around a bar magnet and around a current-carrying solenoid — the deflections are the same at matching points. So for NEET, treat them as producing identical fields.
N is the TOTAL number of turns in the solenoid, not turns per metre. If the solenoid has n turns per unit length and length 2l (matching a magnet of half-length l), then total turns N = n × 2l. So m = NIA = (n·2l)·I·(πa²). Do not confuse N (total) with n (per metre) — mixing them is the most common mistake in this derivation.
μ₀nI is the field INSIDE the solenoid (uniform, along the axis). B = (μ₀/4π)(2m/r³) is the FAR field measured along the axis, far outside the solenoid, at distance r ≫ l. NCERT integrates the loop field over the whole solenoid length and, for r ≫ l, it simplifies to exactly the bar-magnet axial dipole formula. Same result confirms the two are magnetically equivalent.
Cutting a solenoid transversely gives two shorter solenoids, each still with a north face and south face. Cutting a bar magnet transversely gives two smaller bar magnets, each with its own N and S pole. This parallel is exactly why the analogy holds — you can never isolate a single pole (monopole) in either case.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It means a bar magnet and a solenoid produce the same magnetic field pattern, so a bar magnet can be modelled as a solenoid with magnetic moment m = NIA. Its far axial field is B = (μ₀/4π)(2m/r³), identical to the bar-magnet dipole formula.
m = NIA, where N = total number of turns, I = current, and A = cross-sectional area (A = πa² for radius a). If only n turns per metre and length 2l are given, use N = n × 2l first.
For r ≫ l (far on the axis), B = (μ₀/4π)(2m/r³). This is exactly the axial field of a bar magnet, which confirms the equivalence.
NEET rarely asks the pure derivation, but it tests the RESULT: the m = NIA moment and the dipole field formulas B_axial = (μ₀/4π)(2m/r³) and B_equatorial = (μ₀/4π)(m/r³). Master those two formulas.
